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Proposition 9.13 Suppose \( A \) is an unbounded operator on \( \mathbf{H} \) and that \( B \) is a bounded operator defined on all of \( \mathbf{H} \) . Let \( A + B \) denote the operator with \( \operatorname{Dom}\left( {A + B}\right) = \operatorname{Dom}\left( A\right) \) and given by \( \left( {A + B}\right) \psi ... | In particular, the sum of an unbounded self-adjoint operator and a bounded self-adjoint operator (defined on all of \( \mathbf{H} \) ) is self-adjoint on the domain of the unbounded operator. Proof. See Exercise 3. ∎ | No |
Proposition 9.14 Let \( A \) be a closed operator and \( \lambda \) an element of \( \mathbb{C} \) . Suppose that there exists \( \varepsilon > 0 \) such that\n\n\[ \n\parallel \left( {A - {\lambda I}}\right) \psi \parallel \geq \varepsilon \parallel \psi \parallel \n\]\n\n(9.2)\n\nfor all \( A \) in \( \operatorname{D... | Proof. Assume that \( {\phi }_{n} \) is a sequence in the range of \( A - {\lambda I} \) converging to some \( \phi \) . Then \( {\phi }_{n} = \left( {A - {\lambda I}}\right) {\psi }_{n} \), for some sequence \( {\psi }_{n} \) in \( \operatorname{Dom}\left( A\right) \) . Applying (9.2) with \( \psi = {\psi }_{n} - {\ps... | Yes |
Example 9.15 Let \( \left\langle {e}_{j}\right\rangle \) be an orthonormal basis for \( \mathbf{H} \) and let \( \left\langle {\lambda }_{j}\right\rangle \) be an arbitrary sequence of real numbers. Define an operator \( A \) on \( \mathbf{H} \) with \( \operatorname{Dom}\left( A\right) \) equal to the space of finite ... | Proof. Note that for any sequence \( \left\langle {a}_{j}\right\rangle \) of coefficients satisfying the condition on the right-hand side of (9.3), we have \( \mathop{\sum }\limits_{j}{\left| {a}_{j}\right| }^{2} < \infty \) and, thus, the\n\nsum \( \mathop{\sum }\limits_{j}{a}_{j}{e}_{j} \) converges in \( \mathbf{H} ... | Yes |
Theorem 9.17 If \( A \) is an unbounded self-adjoint operator on \( \mathbf{H} \), the spectrum of \( A \) is contained in the real line. | Proof. Consider a complex number \( \lambda = a + {ib} \) with \( b \neq 0 \) . Since \( A \) is symmetric, the proof of Lemma 7.8 applies, giving\n\n\[ \langle \left( {A - {\lambda I}}\right) \psi ,\left( {A - {\lambda I}}\right) \psi \rangle \geq {b}^{2}\langle \psi ,\psi \rangle \]\n\n(9.5)\n\nfor all \( \psi \in \o... | Yes |
Proposition 9.18 If \( A \) is an unbounded self-adjoint operator on \( \mathbf{H} \), then the following hold.\n\n1. A number \( \lambda \in \mathbb{R} \) belongs to the spectrum of \( A \) if and only if there exists a sequence \( {\psi }_{n} \) of nonzero vectors in \( \operatorname{Dom}\left( A\right) \) such that\... | Proof. For Point 1, if a sequence as in (9.6) existed, then as in the proof of Proposition 7.7, \( A - {\lambda I} \) could not have a bounded inverse, so \( \lambda \) must be in the spectrum of \( A \) . Conversely, suppose no such sequence exists. Then there is some \( \varepsilon > 0 \) such that\n\n\[ \parallel \l... | No |
Proposition 9.20 Let \( A \) be an unbounded self-adjoint operator on \( \mathbf{H} \). If \( A \) is non-negative, then the spectrum of \( A \) is contained in \( \lbrack 0,\infty ) \). More generally, if \( A \) is bounded below by \( c \), then the spectrum of \( A \) is contained in \( \lbrack c,\infty ) \). | Proof. Suppose \( A \) is bounded below by \( c \) and \( \lambda \) is a point in the spectrum of \( A \). If \( {\psi }_{n} \) be a sequence as in Point 1 of Proposition 9.18, with the \( {\psi }_{n} \) ’s normalized to be unit vectors, then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| \left\langle {{\... | Yes |
Proposition 9.23 If \( A \) is a symmetric operator on \( \mathbf{H} \), then \( A \) is selfadjoint if and only if\n\n\[ \operatorname{Range}\left( {A - {iI}}\right) = \operatorname{Range}\left( {A + {iI}}\right) = \mathbf{H}. \] | Proof. Suppose first that \( A \) is self-adjoint. Then by Theorem 9.21, the ranges of \( A - {iI} \) and \( A + {iI} \) are dense in \( \mathbf{H} \). On the other hand,\n\n\[ \parallel \left( {A - {iI}}\right) \psi {\parallel }^{2} \geq \parallel \psi {\parallel }^{2} \]\n\n(9.9)\n\nby (the proof of) Lemma 7.8, with ... | Yes |
Theorem 9.24 Suppose that \( A \) is a symmetric operator on \( \mathbf{H} \) and that \( \langle \psi ,{A\psi }\rangle \geq 0 \) for all \( \psi \in \operatorname{Dom}\left( A\right) \) . Then \( A \) is essentially self-adjoint if and only if \( A + I \) has dense range. Equivalently, \( A \) is essentially self-adjo... | Proof. Assume first that \( A \) is essentially self-adjoint. Then \( {\left( A + I\right) }^{ * } = \) \( {A}^{ * } + I = {A}^{cl} + I \) . It is easily seen that \( {A}^{cl} \) is also positive definite, and so\n\n\[ \left\langle {\psi ,\left( {{A}^{cl} + I}\right) \psi }\right\rangle = \langle \psi ,\psi \rangle + \... | Yes |
Example 9.25 Suppose that \( A \) is a symmetric operator on \( \mathbf{H} \) that has an orthonormal basis of eigenvectors. That is to say, suppose there is an orthonormal basis \( \left\{ {e}_{j}\right\} \) for \( \mathbf{H} \) such that for each \( j \), we have \( {e}_{j} \in \operatorname{Dom}\left( A\right) \) an... | Proof. For any \( j,\left( {A - {iI}}\right) {e}_{j} = \left( {{\lambda }_{j} - i}\right) {e}_{j} \). Since \( {\lambda }_{j} \) is real, we have a nonzero multiple of \( {e}_{j} \) belonging to Range \( \left( {A - {iI}}\right) \), for each \( j \). This shows that \( \operatorname{Range}\left( {A - {iI}}\right) \) is... | Yes |
Suppose \( \mathbf{H} \) is a Hilbert space direct sum of a sequence of separable Hilbert spaces \( {\mathbf{H}}_{j} \) :\n\n\[ \mathbf{H} = {\bigoplus }_{j = 1}^{\infty }{\mathbf{H}}_{j} \]\n\nSuppose also that \( {A}_{j} \) is a bounded self-adjoint operator on \( {\mathbf{H}}_{j} \), for each \( j \) . Define a subs... | Proof. Since \( {A}_{j} \) is self-adjoint, the ranges of \( {A}_{j} - {iI} \) and \( {A}_{j} + {iI} \) are dense in \( {\mathbf{H}}_{j} \) . Thus, the closure of the range of \( A - {iI} \) contains each \( {\mathbf{H}}_{j} \) and is therefore dense in \( \mathbf{H} \), and similarly for \( A + {iI} \) . This shows th... | Yes |
Proposition 9.27 Let \( \operatorname{Dom}\left( A\right) \subset {L}^{2}\left( \left\lbrack {0,1}\right\rbrack \right) \) be the space of continuously differentiable functions \( f \) on \( \left\lbrack {0,1}\right\rbrack \) satisfying \[ \psi \left( 0\right) = \psi \left( 1\right) = 0. \] For \( \psi \in \operatornam... | Proof of symmetry. Using integration by parts we see that for all \( \phi \) and \( \psi \) in \( \operatorname{Dom}\left( A\right) \) we have \[ {\int }_{0}^{1}\overline{\phi \left( x\right) }\frac{d\psi }{dx}{dx} = \overline{\phi \left( 1\right) }\psi \left( 1\right) - \overline{\phi \left( 0\right) }\psi \left( 0\ri... | Yes |
Lemma 9.28 If \( \phi \) is a continuously differentiable function on \( \left\lbrack {0,1}\right\rbrack \), then \( \phi \in \operatorname{Dom}\left( {A}^{ * }\right) \) and \( {A}^{ * }\phi = - i\hslash {d\phi }/{dx} \) . | Proof. If \( \phi \) is continuously differentiable, then for any \( \psi \) in \( \operatorname{Dom}\left( A\right) \), we may integrate by parts as in (9.12). Since \( \psi \) is zero at both ends of the interval, the boundary terms vanish and we obtain\n\n\[ \langle \phi ,{A\psi }\rangle = i\hslash {\int }_{0}^{1}\f... | Yes |
Proposition 9.29 Let \( P \) be the densely defined operator with \( \operatorname{Dom}\left( P\right) = \) \( {C}_{c}^{\infty }\left( \mathbb{R}\right) \subset {L}^{2}\left( \mathbb{R}\right) \) and given by \( {P\psi } = - i\hslash {d\psi }/{dx} \) . Then \( P \) is essentially self-adjoint. | Proof. Our strategy is to apply Corollary 9.22. Since \( P \) is symmetric, we expect that \( {P}^{ * } \) will be given by the formula \( - i\hslash d/{dx} \), on some suitable domain inside \( {L}^{2}\left( \mathbb{R}\right) \) . Thus, if \( \psi \in \ker \left( {{P}^{ * } + {iI}}\right) \), this should mean that \( ... | Yes |
Proposition 9.30 Suppose \( V : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a measurable function. Let \( V\left( \mathbf{X}\right) \) be the unbounded operator with domain\n\n\[ \n\operatorname{Dom}\left( {V\left( \mathbf{X}\right) }\right) = \left\{ {\psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \mid V\left( \math... | Proof. Define a subset \( {E}_{m} \) of \( {\mathbb{R}}^{n} \) by\n\n\[ \n{E}_{m} = \left\{ {\mathbf{x} \in {\mathbb{R}}^{n}\left| \right| V\left( \mathbf{x}\right) \mid < m}\right\} , \n\]\n\nso that \( { \cup }_{m}{E}_{m} = {\mathbb{R}}^{n} \) . Then for any \( \psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \), the f... | Yes |
Proposition 9.32 For each \( j = 1,2,\ldots, n \), define a domain \( \operatorname{Dom}\left( {P}_{j}\right) \subset \) \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) as follows:\n\n\[ \operatorname{Dom}\left( {P}_{j}\right) = \left\{ {\psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \mid {k}_{j}\widehat{\psi }\left( \math... | The domain \( \operatorname{Dom}\left( {P}_{j}\right) \) of \( {P}_{j} \) can also be described as the set of all \( \psi \in \) \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) such that \( \partial \psi /\partial {x}_{j} \), computed in the distribution sense, belongs to \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) . For a... | Yes |
Lemma 9.33 Suppose \( \psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \) has the property that \( \partial \psi /\partial {x}_{j} \), computed in the distribution sense, is equal to an \( {L}^{2} \) function \( \phi \) . Then \( \widehat{\phi }\left( \mathbf{k}\right) = \) \( i{k}_{j}\widehat{\psi }\left( \mathbf{k}\rig... | Proof. Suppose \( \partial \psi /\partial {x}_{j} \), computed in the distribution sense, is equal to the \( {L}^{2} \) function \( \phi \) (see Definition A.28). Then by the unitarity of the Fourier transform (Theorem A.19) and its behavior with respect to differentiation (Proposition A.17), we have\n\n\[ \langle \chi... | No |
Proposition 9.34 Define a domain \( \operatorname{Dom}\left( \Delta \right) \) as follows:\n\n\[ \operatorname{Dom}\left( \Delta \right) = \left\{ {\psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \left| {\;{\left| \mathbf{k}\right| }^{2}\widehat{\psi }\left( \mathbf{k}\right) \in {L}^{2}\left( {\mathbb{R}}^{n}\right) }\... | The proof of Proposition 9.34 is extremely similar to that of Proposition 9.32 and is omitted. | No |
Proposition 9.35 Suppose \( \psi \left( \mathbf{x}\right) = g\left( \mathbf{x}\right) f\left( \left| \mathbf{x}\right| \right) \), where \( g \) is a smooth function on \( {\mathbb{R}}^{n} \) and \( f \) is a smooth function on \( \left( {0,\infty }\right) \) . Suppose also that \( f \) satisfies\n\n\[ \mathop{\lim }\l... | Proof. To apply Proposition 9.34, we need to compute \( \langle \psi ,{\Delta \chi }\rangle \), for each \( \chi \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . We choose a large cube \( C \), centered at the origin and such that the support of \( \chi \) is contained in the interior of \( C \) . Then we consi... | No |
Theorem 9.37 (Kato-Rellich Theorem) Suppose that \( A \) and \( B \) are unbounded self-adjoint operators on \( \mathbf{H} \). Suppose that \( \operatorname{Dom}\left( A\right) \subset \operatorname{Dom}\left( B\right) \) and that there exist positive constants \( a \) and \( b \) with \( a < 1 \) such that\n\n\[ \para... | Proof. We use the trivial variant of Theorem 9.21 given in Exercise 8. Choose a positive real number \( \mu \) large enough that \( a + b/\mu < 1 \), which is possible because we assume \( a < 1 \). Then for any \( \psi \in \operatorname{Dom}\left( A\right) \), we have\n\n\[ \left( {A + B + {i\mu I}}\right) \psi = \lef... | Yes |
Theorem 9.38 Suppose \( n \) is at most 3 and \( V : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a measurable function that can be decomposed as a sum of two real-valued, measurable functions \( {V}_{1} \) and \( {V}_{2} \), with \( {V}_{1} \) belonging to \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) and \( {V}_{2} \)... | Proof. We apply the Kato-Rellich theorem with \( A = - {\hslash }^{2}\Delta /{2m} \) and \( B = \) \( V\left( \mathbf{X}\right) \). Assume \( \psi \in \operatorname{Dom}\left( \Delta \right) \) and fix some \( \varepsilon > 0 \). By Exercise 14, there exists a constant \( {c}_{\varepsilon } \) such that\n\n\[ \left| {\... | No |
Theorem 9.39 Suppose \( n \) is at most 3 and \( V : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a measurable function that can be decomposed as a sum of three real-valued, measurable functions \( {V}_{1},{V}_{2} \), and \( {V}_{3} \), with \( {V}_{1} \) belonging to \( {L}^{2}\left( {\mathbb{R}}^{n}\right) ,{V}_{2} ... | The proof of this result would take us too far afield and is omitted. See Theorem X.29 in Volume II of [34]. Note that we assume only that \( {V}_{3} \) is non-negative and locally square-integrable; \( {V}_{3} \) can tend to \( + \infty \) arbitrarily fast at infinity. Again, the same result applies in \( {\mathbb{R}}... | No |
Proposition 9.40 Fix \( \mathbf{a} \) and \( \mathbf{b} \) in \( {\mathbb{R}}^{n} \) and let \( \mathbf{a} \cdot \mathbf{X} + \mathbf{b} \cdot \mathbf{P} \) denote the operator given by\n\n\[ \left( {\mathbf{a} \cdot \mathbf{X} + \mathbf{b} \cdot \mathbf{P}}\right) \psi \left( \mathbf{x}\right) = \left( {\mathbf{a} \cd... | Proof. We use the same strategy as in Sect. 9.7, namely we explicitly solve the equation \( {A}^{ * }\psi = \pm {i\psi } \) and find that there are no nonzero, square-integrable solutions.\n\nThe case \( \mathbf{b} = 0 \) is not hard to analyze and is left as an exercise (Exercise 20). Assume, then, that \( \mathbf{b} ... | No |
Theorem 9.41 Define an operator \( \widehat{H} \) with \( \operatorname{Dom}\left( \widehat{H}\right) = {C}_{c}^{\infty }\left( \mathbb{R}\right) \) by the formula \[ \widehat{H} = - \frac{{\hslash }^{2}}{2m}\frac{{d}^{2}}{d{x}^{2}} - {x}^{4} \] Then \( \widehat{H} \) is not essentially self-adjoint. | In preparation for the proof, let us define a function \( p\left( x\right) \) on \( \mathbb{R} \) such that \[ \frac{p{\left( x\right) }^{2}}{2m} - {x}^{4} = {i\alpha } \] that is, \[ p\left( x\right) = \sqrt{2m}\sqrt{{x}^{4} + {i\alpha }} \] (9.27) Here we take the square root that is in the first quadrant. The functi... | No |
Lemma 9.42 If \( {\psi }_{\alpha } \) is given by\n\n\[ \n{\psi }_{\alpha }\left( x\right) = \frac{1}{\sqrt{p\left( x\right) }}\exp \left\{ {\frac{i}{\hslash }{\int }_{0}^{x}p\left( y\right) {dy}}\right\} ,\n\]\n\nthen \( {\psi }_{\alpha } \) belongs to \( {L}^{2}\left( \mathbb{R}\right) \) and the function\n\n\[ \n- \... | Proof. Let us consider the integral of \( p \) ,\n\n\[ \n{\int }_{0}^{x}p\left( y\right) {dy} = \sqrt{2m}{\int }_{0}^{x}\sqrt{{y}^{4} + {i\alpha }}{dy}.\n\]\n\nUsing the power series for \( {\left( 1 + x\right) }^{a} \) we see that for large \( y \) ,\n\n\[ \n\sqrt{{y}^{4} + {i\alpha }} = {y}^{2}\sqrt{1 + {i\alpha }/{y... | Yes |
Proposition 10.1 Suppose \( \mu \) is a projection-valued measure on \( \left( {X,\Omega }\right) \) with values in \( \mathcal{B}\left( \mathbf{H}\right) \) and \( f : X \rightarrow \mathbb{C} \) is a measurable function (not necessarily bounded). Define a subspace \( {W}_{f} \) of \( \mathbf{H} \) by\n\n\[ \n{W}_{f} ... | Note that since \( {\mu }_{\psi } \) is a finite measure for all \( \psi \), if \( f \) is bounded then the domain of \( {\int }_{X}{fd\mu } \) is all of \( \mathbf{H} \) . Thus, in the bounded case, the definition of \( {\int }_{X}{fd\mu } \) in Proposition 10.1 agrees with our earlier definition (in Chap. 7) of the i... | Yes |
Proposition 10.2 Let \( f \) be a measurable function on \( X \) and let \( {W}_{f} \) be as in (10.2). Then the following results hold.\n\n1. The space \( {W}_{f} \) is a dense subspace of \( \mathbf{H} \) and the map \( {Q}_{f} : {W}_{f} \rightarrow \mathbb{C} \) given by\n\n\[ \n{Q}_{f}\left( \psi \right) = {\int }_... | Proof. It is easy to see that \( {W}_{f} \) is closed under scalar multiplication. To show that it is closed under addition, note that since \( \mu \left( E\right) \) is self-adjoint and satisfies \( \mu {\left( E\right) }^{2} = \mu \left( E\right) \), we have\n\n\[ \n{\mu }_{\phi + \psi }\left( E\right) = \parallel \m... | Yes |
Proposition 10.8 Let \( A \) be a self-adjoint operator on \( \mathbf{H} \). Then the spectral subspaces \( {V}_{E} \) associated to \( A \) have the following properties.\n\n1. If \( E \) is a bounded subset of \( \mathbb{R} \), then \( {V}_{E} \subset \operatorname{Dom}\left( A\right) ,{V}_{E} \) is invariant under \... | Proof. Point 1 holds because the function \( f\left( \lambda \right) = \lambda \) is bounded on \( E \). (See the proof of Proposition 10.3.) Point 2 then holds because, as in the proof of Proposition 10.3, the restriction of \( A \) to \( {V}_{E} \) coincides with the restriction to \( {V}_{E} \) of the operator \( f\... | Yes |
Theorem 10.10 (Spectral Theorem, Multiplication Operator Form) Suppose \( A \) is a self-adjoint operator on \( \mathbf{H} \). Then there is a \( \sigma \)-finite measure space \( \left( {X,\mu }\right) \), a measurable, real-valued function \( h \) on \( X \), and a unitary map \( U : \mathbf{H} \rightarrow {L}^{2}\le... | These theorems are also proved in Sect. 10.4. | No |
Example 10.12 Let \( \mathbf{H} = {L}^{2}\left( {\mathbb{R}}^{n}\right) \) and let \( {U}_{\mathbf{a}}\left( t\right) \) be the translation operator given by\n\n\[ \left( {{U}_{\mathbf{a}}\left( t\right) \psi }\right) \left( \mathbf{x}\right) = \psi \left( {\mathbf{x} + t\mathbf{a}}\right) \]\n\nThen \( U\left( \cdot \... | Proof. It is easy to see that \( {U}_{\mathbf{a}}\left( \cdot \right) \) is a one-parameter unitary group. To see that \( {U}_{\mathbf{a}}\left( \cdot \right) \) is strongly continuous, consider first the case in which \( \psi \) is continuous and compactly supported. Since a continuous function on a compact metric spa... | Yes |
If \( {U}_{\mathbf{a}}\left( \cdot \right) ,\mathbf{a} \in {\mathbb{R}}^{n} \), is the strongly continuous one-parameter unitary group in Example 10.12, then each \( \psi \in {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) is in the domain of the infinitesimal generator \( A \) of \( {U}_{\mathbf{a}}\left( \cdot \ri... | Proof. The formula for the infinitesimal generator is easy to establish for \( \psi \) in \( {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . The essential self-adjointness of \( A \) is a special case of Proposition 13.5 (the proof of which is similar to the proof of Proposition 9.29). | No |
Lemma 10.17 Let \( U\left( \cdot \right) \) be a strongly continuous one-parameter unitary group and let \( A \) be its infinitesimal generator. If \( \psi \in \operatorname{Dom}\left( A\right) \), then for all \( t \in \mathbb{R} \), the vector \( U\left( t\right) \psi \) belongs to \( \operatorname{Dom}\left( A\right... | Proof. We compute that\n\n\[ \frac{U\left( {t + h}\right) \psi - U\left( t\right) \psi }{h} = U\left( t\right) \frac{\left\lbrack U\left( h\right) \psi - \psi \right\rbrack }{h}. \]\n\n(10.19)\n\nSince \( \psi \in \operatorname{Dom}\left( A\right) \), the limit as \( h \) tends to zero of (10.19) exists and is equal to... | Yes |
Lemma 10.18 For any strongly continuous one-parameter unitary group \( U\left( \cdot \right) \), the infinitesimal generator \( A \) is densely defined. | Proof. Given any continuous function \( f \) of compact support, define an operator \( {B}_{f} \) by setting\n\n\[ {B}_{f} = {\int }_{-\infty }^{\infty }f\left( \tau \right) U\left( \tau \right) {d\tau } \]\n\nHere, the operator-valued integral is the unique bounded operator such that\n\n\[ \left\langle {\phi ,{B}_{f}\... | Yes |
Lemma 10.22 If \( A \) and \( B \) are commuting elements of \( \mathcal{B}\left( \mathbf{H}\right) \), then\n\n\[ R\left( {AB}\right) \leq R\left( A\right) R\left( B\right) \] | Proof. If \( A \) is any bounded operator, the proof of Lemma 8.1 shows that for any real number \( T \) with \( T > R\left( A\right) \), we have\n\n\[ \mathop{\lim }\limits_{{m \rightarrow \infty }}\frac{\begin{Vmatrix}{A}^{m}\end{Vmatrix}}{{T}^{m}} = 0 \]\n\nIf \( A \) and \( B \) are two commuting bounded operators ... | Yes |
Theorem 10.23 If \( A \in \mathcal{B}\left( \mathbf{H}\right) \) is normal, then for any polynomial \( p \) in two variables, we have\n\n\[ \sigma \left( {p\left( {A,{A}^{ * }}\right) }\right) = \{ p\left( {\lambda ,\bar{\lambda }}\right) \mid \lambda \in \sigma \left( A\right) \} . \] | Proof of Theorem 10.23 in the Matrix Case. For matrices, the spectrum is nothing but the set of eigenvalues. If \( A \) commutes with \( {A}^{ * } \), then for any \( \lambda \in \mathbb{C} \) ,\n\n\[ \left\langle {\left( {{A}^{ * } - \bar{\lambda }I}\right) \psi ,\left( {{A}^{ * } - \bar{\lambda }I}\right) \psi }\righ... | Yes |
Lemma 10.25 Suppose \( A \in \mathcal{B}\left( \mathbf{H}\right) \) is normal.\n\n1. If \( \psi \) is an \( \varepsilon \) -almost eigenvector for \( A \) with eigenvalue \( \lambda \), then \( \psi \) is an \( \varepsilon \) -almost eigenvector for \( {A}^{ * } \) with eigenvalue \( \bar{\lambda } \) .\n\n2. A number ... | Proof. Point 1 follows immediately from (10.24), which holds for bounded normal operators, not just matrices. For Point 2, suppose that an \( \varepsilon \) -almost eigenvector with eigenvalue \( \lambda \) exists for all \( \varepsilon > 0 \) . Then \( A - {\lambda I} \) cannot have a bounded inverse, and so \( \lambd... | Yes |
Lemma 10.26 Suppose \( A \in \mathcal{B}\left( \mathbf{H}\right) \) is normal. Then for each polynomial \( p \) in two variables and each number \( \lambda \in \mathbb{C} \), there is a constant \( C \) such that if \( \psi \) is an \( \varepsilon \) -almost eigenvector for \( A \) with eigenvalue \( \lambda \), then \... | Proof. We decompose \( p\left( {A,{A}^{ * }}\right) - p\left( {\lambda ,\bar{\lambda }}\right) I \) into a linear combination of terms of the form \( {A}^{k}{\left( {A}^{ * }\right) }^{l} - {\lambda }^{k}{\bar{\lambda }}^{l} \) and we estimate such terms by induction on \( k + l \) . If \( k = 1 \) and \( l = 0 \), the... | Yes |
Lemma 10.27 Let \( A \in \mathcal{B}\left( \mathbf{H}\right) \) be normal, let \( p \) be a polynomial in two variables, and let \( \mu \) be an element of the spectrum of \( p\left( {A,{A}^{ * }}\right) \) . Then for all \( \varepsilon > 0 \), there exists a nonzero closed subspace \( {W}^{\varepsilon } \) of \( \math... | Proof. Fix some \( \mu \) in the spectrum of \( p\left( {A,{A}^{ * }}\right) \) and let \( B = p\left( {A,{A}^{ * }}\right) - {\mu I} \) . Then \( B \) is normal and 0 belongs to the spectrum of \( B \) . Using Point 2 of Lemma 10.25 and Lemma 10.26, we see that 0 belongs to the spectrum of the self-adjoint operator \(... | Yes |
Theorem 10.28 (Cayley Transform) If \( A \) is a self-adjoint operator on \( \mathbf{H} \), let \( U \) be the operator defined by\n\n\[ \n{U\psi } = \left( {A + {iI}}\right) {\left( A - iI\right) }^{-1}\psi .\n\]\n\nThen the following results hold.\n\n1. The operator \( U \) is a unitary operator on \( \mathbf{H} \).\... | Proof. The resolvent operator \( {\left( A - iI\right) }^{-1} \) must be injective, because\n\n\[ \n\left( {A - {iI}}\right) {\left( A - iI\right) }^{-1}\psi = \psi \n\]\n\nfor all \( \psi \in \mathbf{H} \). Furthermore, \( {\left( A - iI\right) }^{-1} \) maps \( \mathbf{H} \) onto \( \operatorname{Dom}\left( A\right) ... | Yes |
Proposition 10.29 Let \( A \) be a self-adjoint operator on \( \mathbf{H} \), let \( U \) be the unitary operator in Theorem 10.28, and let \( D : {S}^{1} \smallsetminus \{ 0\} \rightarrow \mathbb{R} \) be as in (10.29). Then\n\n\[ A = D\left( U\right) \]\n\n\( \left( {10.33}\right) \)\n\nwhere \( D\left( U\right) \) i... | Proof. Suppose \( E \) is a Borel subset of \( {S}^{1} \smallsetminus \{ 0\} \) such that the closure of \( E \) does not contain 1, and let \( {V}_{E} = \operatorname{Range}\left( {{\mu }^{U}\left( E\right) }\right) \) be the associated spectral subspace. Then the spectrum of \( {\left. U\right| }_{E} \) is contained ... | Yes |
Theorem 10.30 Define a projection-valued measure \( {\mu }^{A} \) on \( \mathbb{R} \) by\n\n\[ \n{\mu }^{A}\left( E\right) = {\mu }^{U}\left( {C\left( E\right) }\right) \n\]\n\n(10.34)\n\nThen\n\n\[ \nA = {\int }_{\mathbb{R}}{\lambda d}{\mu }^{A}\left( \lambda \right) \n\]\n\n(10.35)\n\nwhere \( {\mu }^{U} \) is the pr... | Proof. If for any \( \psi \in \mathbf{H} \), we define \( {\mu }_{\psi }^{U}\left( E\right) = \left\langle {\psi ,{\mu }^{U}\psi }\right\rangle \) and similarly define \( {\mu }_{\psi }^{A} \), then we have\n\n\[ \n{\mu }_{\psi }^{A}\left( E\right) = {\mu }_{\psi }^{U}\left( {C\left( E\right) }\right) \n\]\n\nBy the ab... | Yes |
Proposition 11.1 Suppose that \( \psi \) is an eigenvector for \( {a}^{ * }a \) with eigenvalue \( \lambda \) . Then\n\n\[{a}^{ * }a\left( {a\psi }\right) = \left( {\lambda - 1}\right) {a\psi }\]\n\n\[{a}^{ * }a\left( {{a}^{ * }\psi }\right) = \left( {\lambda + 1}\right) {a}^{ * }\psi\]\n\nThus, either \( {a\psi } \) i... | Proof. Using the commutation relation (11.7), we find that\n\n\[{a}^{ * }a\left( {a\psi }\right) = \left( {a\left( {{a}^{ * }a}\right) - a}\right) \psi = \left( {\lambda - 1}\right) {a\psi }.\]\n\nA similar calculation applies to \( {a}^{ * }\psi \), using (11.8). ∎ | Yes |
Theorem 11.2 If \( {\psi }_{0} \) is a unit vector with the property that \( a{\psi }_{0} = 0 \), then the vectors\n\n\[ \n{\psi }_{n} \mathrel{\text{:=}} {\left( {a}^{ * }\right) }^{n}{\psi }_{0},\;n \geq 0, \]\n\nsatisfy the following relations for all \( n, m \geq 0 \) :\n\n\[ \n{a}^{ * }{\psi }_{n} = {\psi }_{n + 1... | Proof. The first result is the definition of \( {\psi }_{n + 1} \) and the second follows from Proposition 11.1 and the fact that \( {a}^{ * }a{\psi }_{0} = 0 \) . For the third result, if \( n \neq m \), we use the general result that eigenvectors for a self-adjoint operator (in our case, \( {a}^{ * }a \) ) with disti... | Yes |
Theorem 11.3 The ground state \( {\psi }_{0} \) of the harmonic oscillator is given by (11.12). The excited states \( {\psi }_{n} \) are given by\n\n\[ \n{\psi }_{n} = {H}_{n}{\psi }_{0} \n\]\n\n(11.13)\n\nwhere \( {H}_{n} \) is a polynomial of degree \( n \) given inductively by the formulas\n\n\[ \n{H}_{0}\left( \wid... | Proof. When \( n = 0 \) ,(11.13) reduces to \( {\psi }_{0} = {\psi }_{0} \) . Assuming that (11.13) holds for some \( n \), we compute \( {\psi }_{n + 1} \) as\n\n\[ \n{\psi }_{n + 1} = {a}^{ * }{\psi }_{n} = \frac{1}{\sqrt{2}}\left( {\widetilde{x}{H}_{n}\left( \widetilde{x}\right) C{e}^{-{\widetilde{x}}^{2}/2} - \frac... | Yes |
Theorem 11.4 The functions\n\n\[ \n{\psi }_{n}\left( x\right) = {H}_{n}\left( \widetilde{x}\right) {\psi }_{0}\left( \widetilde{x}\right) \]\n\n\[ \n= {H}_{n}\left( {\sqrt{\frac{m\omega }{\hslash }}x}\right) \sqrt{\frac{\pi m\omega }{\hslash }}\exp \left\{ {-\frac{m\omega }{2\hslash }{x}^{2}}\right\} \]\n\nform an orth... | The following result is the key to the proof.\n\nLemma 11.5 For all | No |
For all \( \alpha \in \mathbb{C} \), the partial sums of the series\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{\alpha }^{n}{\widetilde{x}}^{n}}{n!}{e}^{-{\widetilde{x}}^{2}/2} \]\n\nconverge in \( {L}^{2}\left( \mathbb{R}\right) \) to the function \( {e}^{\alpha \widetilde{x}}{e}^{-{\widetilde{x}}^{2}/2} \) ... | We need to show that\n\n\[ {\begin{Vmatrix}{e}^{\alpha \widetilde{x}}{e}^{-{\widetilde{x}}^{2}/2} - \mathop{\sum }\limits_{{n = 0}}^{N}\frac{{\alpha }^{n}{\widetilde{x}}^{n}}{n!}{e}^{-{\widetilde{x}}^{2}/2}\end{Vmatrix}}^{2} = \int {\left| \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\frac{{\alpha }^{n}{\widetilde{x}}^... | Yes |
Proposition 12.3 If \( A \) is a symmetric operator on \( \mathbf{H} \), then for all unit vectors \( \psi \in \operatorname{Dom}\left( A\right) \), we have \( {\Delta }_{\psi }A = 0 \) if and only if \( \psi \) is an eigenvector for \( A \) . | Proof. If \( {\Delta }_{\psi }A = 0 \), then from (12.3), we see that \( \left( {A - \langle A{\rangle }_{\psi }I}\right) \psi = 0 \) , meaning that \( \psi \) is an eigenvector for \( A \) with eigenvalue \( \langle A{\rangle }_{\psi } \) . Conversely, if \( {A\psi } = {\lambda \psi } \) for some \( \lambda \), then \... | Yes |
Theorem 12.4 Suppose \( A \) and \( B \) are symmetric operators and \( \psi \) is a unit vector belonging to \( \operatorname{Dom}\left( {AB}\right) \cap \operatorname{Dom}\left( {BA}\right) \) . Then\n\n\[{\left( {\Delta }_{\psi }A\right) }^{2}{\left( {\Delta }_{\psi }B\right) }^{2} \geq \frac{1}{4}{\left| \langle \l... | Proof. Define operators \( {A}^{\prime } \) and \( {B}^{\prime } \) by \( {A}^{\prime } \mathrel{\text{:=}} A - \langle \psi ,{A\psi }\rangle I \) and \( {B}^{\prime } \mathrel{\text{:=}} \) \( B - \langle \psi ,{B\psi }\rangle I \) . (We use the same domains for \( {A}^{\prime } \) and \( {B}^{\prime } \) as for \( A ... | Yes |
Corollary 12.5 Suppose \( A \) and \( B \) are symmetric operators satisfying \[ \left\lbrack {A, B}\right\rbrack = i\hslash I \] on \( \operatorname{Dom}\left( {AB}\right) \cap \operatorname{Dom}\left( {BA}\right) \). Then if \( \psi \in \operatorname{Dom}\left( {AB}\right) \cap \operatorname{Dom}\left( {BA}\right) \)... | Note that the factor of \( \hslash \) appearing on the right-hand side of (12.8) is really just \( \left| {\langle \psi ,\left\lbrack {A, B}\right\rbrack \psi \rangle }\right| \). Since, however, \( \psi \) is a unit vector and \( \left\lbrack {A, B}\right\rbrack = i\hslash I \), \( \psi \) drops out of the right-hand ... | Yes |
Proposition 12.6 If \( A \) and \( B \) are symmetric and \( \psi \) is a unit vector in \( \operatorname{Dom}\left( {AB}\right) \cap \operatorname{Dom}\left( {BA}\right) \), equality holds in (12.4) if and only if one of the following holds: (1) \( \psi \) is an eigenvector for \( A \) ,(2) \( \psi \) is an eigenvecto... | Proof. To get equality in (12.4), we must have equality in both (12.5) and (12.6). Equality in (12.5) occurs if and only if \( {A}^{\prime }\psi = 0 \) or \( {B}^{\prime }\psi = 0 \) or \( {A}^{\prime }\psi = c{B}^{\prime }\psi \) for some nonzero constant \( c \) . If \( {A}^{\prime }\psi \) is zero, \( \psi \) is an ... | Yes |
Theorem 12.8 Suppose \( A \) and \( B \) are self-adjoint operators on \( \mathbf{H} \). Suppose that for all \( a \in \mathbb{R} \) and \( \psi \in \operatorname{Dom}\left( A\right) \), we have that \( {e}^{iaB}\psi \) belongs to \( \operatorname{Dom}\left( A\right) \) and that\n\n\[ A{e}^{iaB}\psi = {e}^{iaB}{A\psi }... | Proof. See Exercise 5. - | No |
Corollary 12.9 For any \( j = 1,\ldots n \) and any unit vector \( \psi \in {L}^{2}\left( {\mathbb{R}}^{n}\right) \) with \( \psi \in \operatorname{Dom}\left( {X}_{j}\right) \cap \operatorname{Dom}\left( {P}_{j}\right) \), we have\n\n\[ \left( {{\Delta }_{\psi }{X}_{j}}\right) \left( {{\Delta }_{\psi }{P}_{j}}\right) \... | Proof. In the case that \( A = {X}_{j} \) and \( B = {P}_{j} \), we have \( \left( {{e}^{{iaB}/\hslash }\psi }\right) \left( \mathbf{x}\right) = \) \( \psi \left( {\mathbf{x} + a{\mathbf{e}}_{j}}\right) \), by Exercise 2 in Chap. 10. Thus, in this case,(12.22) says that\n\n\[ \left( {{x}_{j} + a}\right) \psi \left( {\m... | No |
Proposition 12.10 A unit vector \( \psi \in \operatorname{Dom}\left( X\right) \cap \operatorname{Dom}\left( P\right) \) satisfies\n\n\[ \left( {{\Delta }_{\psi }X}\right) \left( {{\Delta }_{\psi }P}\right) = \frac{\hslash }{2} \]\n\nif and only if \( \psi \) satisfies\n\n\[ \left( {X + {i\delta P}}\right) \psi = {\lamb... | Proof. All the relations in the proof of Theorem 12.7 are equalities, except for the inequality in the last line of (12.21). Equality will hold in that line if and only if one of \( \left( {X - {\alpha I}}\right) \psi \) and \( \left( {P - {\beta I}}\right) \psi \) is zero or \( \left( {P - {\beta I}}\right) \psi \) is... | Yes |
Proposition 12.11 If the parameter \( \delta \) in (12.24) is negative, there are no nonzero solutions to (12.24). If the parameter \( \delta \) is positive, there exists a unique (up to multiplication by a constant) solution \( {\psi }_{\delta ,\lambda } \) to (12.24) for every complex number \( \lambda \) . The funct... | Proof. The equation \( \left( {X + {i\delta P}}\right) \psi = {\lambda \psi } \) amounts to\n\n\[ {x\psi } + \delta \hslash \frac{d\psi }{dx} = {\lambda \psi }\left( x\right) \]\n\n(12.26)\n\nwhere \( \psi \) is assumed to be in the domain of \( P \), so that the distributional derivative of \( \psi \) is an \( {L}^{2}... | Yes |
If \( f\left( {x, p}\right) = {x}^{2} \), then the Weyl, Wick-ordered and anti-Wick-ordered quantizations of \( f \) are as follows:\n\n\[ \n{Q}_{\text{Weyl }}\left( {x}^{2}\right) = {X}^{2} \]\n\n\[ \n{Q}_{\text{Wick }}\left( {x}^{2}\right) = {X}^{2} - \frac{1}{2}\alpha \hslash I \]\n\n\[ \n{Q}_{\text{anti-Wick }}\lef... | Proof. The value for \( {Q}_{\text{Weyl }}\left( {x}^{2}\right) \) is apparent. To compute the Wick- and anti-Wick-ordered quantizations, we first write \( x \) as \( \left( {z + \bar{z}}\right) /2 \), so that\n\n\[ \n{x}^{2} = \frac{{\left( z + \bar{z}\right) }^{2}}{4} = \frac{1}{4}\left( {{z}^{2} + {2z}\bar{z} + {\ba... | Yes |
Proposition 13.3 The Weyl quantization-viewed as a linear map of the space of polynomials on \( {\mathbb{R}}^{2} \) into operators on \( {C}_{c}^{\infty }\left( \mathbb{R}\right) \) -is uniquely characterized by the following identity:\n\n\[ \n{Q}_{\text{Weyl }}\left( {\left( ax + bp\right) }^{j}\right) = {\left( aX + ... | Proof. The Weyl quantization is easily seen to satisfy the identity \n\n\[ \n{Q}_{\text{Weyl }}\left( {\left( {{a}_{1}x + {b}_{1}p}\right) \cdots \left( {{a}_{j}x + {b}_{j}p}\right) }\right) \n\] \n\n\[ \n= \frac{1}{j!}\mathop{\sum }\limits_{{\sigma \in {S}_{j}}}\sigma \left( {{a}_{1}X + {b}_{1}P,\ldots ,{a}_{j}X + {b}... | No |
Proposition 13.4 The Weyl quantization satisfies\n\n\[ \n{Q}_{\text{Weyl }}\left( {xg}\right) = {Q}_{\text{Weyl }}\left( x\right) {Q}_{\text{Weyl }}\left( g\right) - \frac{i\hslash }{2}{Q}_{\text{Weyl }}\left( \frac{\partial g}{\partial p}\right) \]\n\n(13.7)\n\n\[ \n= {Q}_{\text{Weyl }}\left( g\right) {Q}_{\text{Weyl ... | Proof. Suppose \( A = \left( {{a}_{1}X + {b}_{1}P}\right) \) and \( B = \left( {{a}_{2}X + {b}_{2}P}\right) \) . Then \( \left\lbrack {A, B}\right\rbrack \) is a multiple of \( I \), from which we can easily verify that\n\n\[ \nA{B}^{j} = {B}^{k}A{B}^{j - k} + k\left\lbrack {A, B}\right\rbrack {B}^{j - 1}, \]\n\nfor \(... | No |
For all \( \mathbf{a} \) and \( \mathbf{b} \) in \( {\mathbb{R}}^{n} \), the operators \( {U}_{\mathbf{a},\mathbf{b}}\left( t\right) \) on \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) given by\n\n\[ \left( {{U}_{\mathbf{a},\mathbf{b}}\left( t\right) \psi }\right) \left( \mathbf{x}\right) = {e}^{i{t}^{2}\hslash \left( {\... | Proof. It is apparent that \( {U}_{\mathbf{a},\mathbf{b}} \) is unitary for each \( \mathbf{a} \) and \( \mathbf{b} \), and it is a simple direct computation to show that it is indeed a unitary group. Strong continuity is proved in the usual way using a dense subspace, as in the proof of Example 10.12. When \( \psi \) ... | Yes |
Theorem 13.8 The map \( {Q}_{\text{Weyl }} \) is a constant multiple of a unitary map of \( {L}^{2}\left( {\mathbb{R}}^{2n}\right) \) onto \( \operatorname{HS}\left( {{L}^{2}\left( {\mathbb{R}}^{n}\right) }\right) \) . The inverse map \( {Q}_{\text{Weyl }}^{-1} : \operatorname{HS}\left( {{L}^{2}\left( {\mathbb{R}}^{n}\... | Proof. Proposition 13.6 gives a unitary identification of \( \operatorname{HS}\left( {{L}^{2}\left( {\mathbb{R}}^{n}\right) }\right) \) with \( {L}^{2}\left( {{\mathbb{R}}^{n} \times {\mathbb{R}}^{n}}\right) \) . Thus, it suffices to show that the map \( f \mapsto {\kappa }_{f} \) is a multiple of a unitary map. This r... | No |
Proposition 13.9 The Moyal product \( f \star g \) may be characterized in terms of the Fourier transform as\n\n\[ \n\widehat{\left( f \star g\right) }\left( {\mathbf{a},\mathbf{b}}\right) = {\left( 2\pi \right) }^{-n}\iint {e}^{-i\hslash \left( {\mathbf{a} \cdot {\mathbf{b}}^{\prime } - \mathbf{b} \cdot {\mathbf{a}}^{... | Proof. It is, of course, possible to obtain this formula using kernel functions. It is, however, easier to work with the (13.17), which can be shown (Exercise 7) to give the same result as Definition 13.7 when \( f \) is a Schwartz function. We assume standard properties of the Bochner integral for functions with value... | No |
Proposition 13.10 The Moyal product \( f \star g \) extends to a continuous map of \( {L}^{2}\left( {\mathbb{R}}^{2n}\right) \times {L}^{2}\left( {\mathbb{R}}^{2n}\right) \) into \( {L}^{2}\left( {\mathbb{R}}^{2n}\right) \) and the composition formula (13.29) holds for all \( f \) and \( g \) in \( {L}^{2}\left( {\math... | Proof. A standard inequality asserts that for any two Hilbert-Schmidt operators \( A \) and \( B \), we have\n\n\[ \parallel {AB}{\parallel }_{\mathrm{{HS}}} \leq \parallel A{\parallel }_{\mathrm{{HS}}}\parallel B{\parallel }_{\mathrm{{HS}}} \]\n\nIt follows that the product map \( \left( {A, B}\right) \mapsto {AB} \) ... | Yes |
Lemma 13.14 Consider an element \( A \) of \( \mathcal{D}\left( {\mathbb{R}}^{n}\right) \) expressed as \[ A = \mathop{\sum }\limits_{\mathbf{k}}{f}_{\mathbf{k}}\left( \mathbf{x}\right) {\left( \frac{\partial }{\partial \mathbf{x}}\right) }^{\mathbf{k}} \] where \( \mathbf{k} \) ranges over multi-indices, where the \( ... | Proof. For each multi-index \( \mathbf{k} \), let \( \left| \mathbf{k}\right| = {k}_{1} + \cdots + {k}_{n} \) . Suppose not all the \( {f}_{\mathbf{k}} \) ’s are zero, let \( N \) be the smallest non-negative integer for which \( {f}_{\mathbf{k}} \) is nonzero for some \( \mathbf{k} \) with \( \left| \mathbf{k}\right| ... | Yes |
Lemma 13.15 If \( A \) belongs to \( \mathcal{D}\left( {\mathbb{R}}^{n}\right) \) and \( A \) commutes with \( {X}_{j} \) and \( {P}_{j} \) for all \( j = 1,\ldots, n \), then \( A = {cI} \) for some \( c \in \mathbb{C} \) . | Proof. We may easily prove by induction that\n\n\[ \n{\left( \frac{\partial }{\partial {x}_{j}}\right) }^{k}\left( {{x}_{j}g\left( \mathbf{x}\right) }\right) = k{\left( \frac{\partial }{\partial {x}_{j}}\right) }^{k - 1}g\left( \mathbf{x}\right) + {x}_{j}{\left( \frac{\partial }{\partial {x}_{j}}\right) }^{k}g\left( \m... | Yes |
Lemma 13.16 For any \( f \in {\mathcal{P}}_{2} \), there exist \( {g}_{1},\ldots ,{g}_{j} \) and \( {h}_{1},\ldots ,{h}_{j} \) in \( {\mathcal{P}}_{2} \) such that\n\n\[ f = \left\{ {{g}_{1},{h}_{1}}\right\} + \cdots + \left\{ {{g}_{j},{h}_{j}}\right\} \] | Proof. See Exercise 12. ∎ | No |
Example 14.4 Let \( {A}_{j} \) be the usual position operator \( {X}_{j} \) acting on \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) and let \( {B}_{j} \) be the usual momentum operator \( {P}_{j} \) . Then the \( A \) ’s and \( B \) ’s satisfy the exponentiated commutation relations. | Proof. Since \( {X}_{j} \) is just multiplication by \( {x}_{j} \), it is easily verified that \( {e}^{{is}{X}_{j}} \) is just multiplication by \( {e}^{{is}{x}_{j}} \) . Meanwhile, the exponentiated momentum operators satisfy (Example 10.16) \[ \left( {{e}^{{it}{P}_{j}}\psi }\right) \left( \mathbf{x}\right) = \psi \le... | Yes |
Example 14.5 Let \( A \) be the operator in Sect. 12.2 and let \( B \) be the (unique self-adjoint extension of) the operator in that section. Then \( A \) and \( B \) do not satisfy the exponentiated commutation relations. | Proof. The operator \( A \) is multiplication by \( x \), and so the operator \( {e}^{isA} \) is just multiplication by \( {e}^{isx} \) . Meanwhile, the operator \( B \) is \( - i\hslash d/{dx} \) , with periodic boundary conditions. We will now demonstrate that \( {e}^{itB} \) consists of \ | No |
Proposition 14.10 For any operators satisfying the exponentiated commutation relations, the associated map \( Q \) in Definition 14.9 has the following properties.\n\n1. If \( f \in \mathcal{S}\left( {\mathbb{R}}^{2n}\right) \) is real valued, \( Q\left( f\right) \) is self-adjoint.\n\n2. For all \( \mathbf{a} \) and \... | Proof. For Point 1, we can re-express \( Q\left( f\right) \) as\n\n\[ \n{\left( 2\pi \right) }^{-n}{\int }_{{\mathbb{R}}^{2n}}\frac{1}{2}\left\lbrack {\widehat{f}\left( {\mathbf{a},\mathbf{b}}\right) {e}^{i\left( {\mathbf{a} \cdot \mathbf{A} + \mathbf{b} \cdot \mathbf{B}}\right) } + \widehat{f}\left( {-\mathbf{a}, - \m... | Yes |
Lemma 14.12 Suppose that \( \mu \) is a smooth, strictly positive density on \( {\mathbb{C}}^{n} \) and that \( F \) and \( G \) are sufficiently nice (but not necessarily holomorphic) functions on \( {\mathbb{C}}^{n} \) . Then\n\n\[ \n{\int }_{{\mathbb{C}}^{n}}\overline{F\left( \mathbf{z}\right) }\frac{\partial G}{\pa... | Proof. Let us approximate the integral over \( {\mathbb{C}}^{n} \) on the left-hand side of (14.24) by an integral over a large cube. By performing either the \( {x}_{j} \) - integral or the \( {y}_{j} \) -integral first, we can integrate by parts to push the derivatives with respect to \( {x}_{j} \) or \( {y}_{j} \) o... | Yes |
Lemma 14.13 Specialize Lemma 14.12 to the case in which \( F \) and \( G \) are holomorphic polynomials and \( \mu \) is the density \( {\mu }_{\hslash } \) given by\n\n\[ \n{\mu }_{\hslash }\left( \mathbf{z}\right) = \frac{1}{{\left( \pi \hslash \right) }^{n}}{e}^{-{\left| \mathbf{z}\right| }^{2}/\hslash }.\n\]\n\nThe... | Proof. In the case that \( F \) and \( G \) are holomorphic polynomials, \( \partial F/\partial {\bar{z}}_{j} = 0 \) , so the first term on the right-hand side of (14.24) is zero. Furthermore, \( \bar{F}{G\mu } \) decreases rapidly at infinity and so the boundary terms vanish in this case. Finally, we may compute \( \p... | Yes |
For each \( \mathbf{a} \in {\mathbb{C}}^{n} \), the operator \( {T}_{\mathbf{a}} \) defined by (14.30) is a unitary operator on the Segal-Bargmann space, and the map \( \mathbf{a} \mapsto {T}_{\mathbf{a}} \) is strongly continuous. These operators satisfy \[ {T}_{\mathbf{a}}{T}_{\mathbf{b}} = {e}^{i\hslash \operatornam... | It is evident that \( {T}_{\mathbf{a}}F\left( \mathbf{z}\right) \) is holomorphic as a function of \( \mathbf{z} \) for each fixed a. Meanwhile, for any \( F \in \mathcal{H}{L}^{2}\left( {{\mathbb{C}}^{n},{\mu }_{\hslash }}\right) \), we have \[ {\begin{Vmatrix}{T}_{\mathbf{a}}F\end{Vmatrix}}_{{L}^{2}\left( {{\mathbb{C... | No |
Proposition 14.17 For all \( F \in \mathcal{H}{L}^{2}\left( {{\mathbb{C}}^{n},{\mu }_{\hslash }}\right) \), we have\n\n\[ F\left( \mathbf{z}\right) = {\int }_{{\mathbb{C}}^{n}}{e}^{\mathbf{z} \cdot \overline{\mathbf{w}}/\hslash }F\left( \mathbf{w}\right) {\mu }_{\hslash }\left( \mathbf{w}\right) d\mathbf{w}. \] | Proof. We begin by establishing the result in the case \( \mathbf{z} = 0 \) . We have already established, in the proof of Proposition 14.15, that the Taylor series of \( F \) converges to \( F \) in \( \mathcal{H}{L}^{2}\left( {{\mathbb{C}}^{n},{\mu }_{\hslash }}\right) \), and the distinct monomials in this series ar... | Yes |
Theorem 14.18 Let \( V \) be the inverse of the map \( U : \mathcal{H}{L}^{2}\left( {{\mathbb{C}}^{n},{\mu }_{\hslash }}\right) \rightarrow \) \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) given by the Stone-von Neumann theorem, normalized so that \( V \) takes the function \( {\phi }_{0} \in {L}^{2}\left( {\mathbb{R}}^{... | Proof. By the unitarity of \( V \) and the \( \mathbf{z} = 0 \) case of Proposition 14.17, we have\n\n\[ {\left\langle {\phi }_{0},\psi \right\rangle }_{{L}^{2}\left( {\mathbb{R}}^{n}\right) } = {\left\langle V{\phi }_{0}, V\psi \right\rangle }_{\mathcal{H}{L}^{2}\left( {{\mathbb{C}}^{n},{\mu }_{\hslash }}\right) } = \... | Yes |
For any two numbers \( {E}_{1} \) and \( {E}_{2} \) with \( {E}_{1} > \mathop{\inf }\limits_{{x \in \mathbb{R}}}V\left( x\right) \) , there exists a constant \( C \) and a nonzero function \( A \in {C}_{c}^{\infty }\left( \mathbb{R}\right) \) with the following property. For every \( E \in \left\lbrack {{E}_{1},{E}_{2}... | Proof. For any \( E \in \left\lbrack {{E}_{1},{E}_{2}}\right\rbrack \), the classically allowed region for energy \( E \) contains the classically allowed region for energy \( {E}_{1} \) . We choose, then, \( A \) to be any nonzero element of \( {C}_{c}^{\infty }\left( \mathbb{R}\right) \) with support in the classical... | Yes |
Theorem 15.8 For any potential \( V \) and range \( \left\lbrack {{E}_{1},{E}_{2}}\right\rbrack \) of energies satisfying Assumption 15.3, there is a constant \( C \) such that the following holds. For any energy \( E \in \left\lbrack {{E}_{1},{E}_{2}}\right\rbrack \) satisfying Condition 15.1, there exists a nonzero f... | As noted already in Sect. 15.3, an estimate of the form \( \parallel \widehat{H}\psi - {E\psi }\parallel < \) \( \varepsilon \parallel \psi \parallel \) implies that there is a point \( \widetilde{E} \) in the spectrum of \( \widehat{H} \) with \( \mid E - \) \( \widetilde{E} \mid < \varepsilon \) . (See Exercise 4 in ... | No |
Lemma 15.9 Let \( {\psi }_{1} \) denote the scaled Airy function in (15.26), let \( {\widetilde{\psi }}_{1} \) denote the same function with the Airy function replaced by the right-hand side of (15.33), and let \( {\psi }_{2} \) denote the oscillatory WKB function in (15.27). If \( x - a \) is positive and of order \( ... | Proof of Lemma 15.9. We consider only the estimates for the derivatives of the functions involved. The analysis of the functions themselves is similar (but easier) and is left as an exercise to the reader (Exercise 11).\n\nWe begin by considering \( {\psi }_{1}^{\prime } - {\widetilde{\psi }}_{1}^{\prime } \) . With a ... | No |
An \( n \times n \) matrix \( U \in {M}_{n}\left( \mathbb{C}\right) \) is said to be unitary if \( {U}^{ * }U = U{U}^{ * } = I \) . A matrix \( U \) is unitary if and only if\n\n\[ \langle {Uv},{Uw}\rangle = \langle v, w\rangle \]\n\nfor all \( v, w \in {\mathbb{C}}^{n} \) . | The condition \( {\left( {U}^{ * }U\right) }_{jk} = {\delta }_{jk} \) is equivalent to the condition that the columns of \( U \) form an orthonormal set in \( {\mathbb{C}}^{n} \), as can be seen by direct computation. Geometrically, the condition \( {U}^{ * }U = I \) is equivalent to the condition that \( \left\langle ... | Yes |
An \( n \times n \) real matrix \( R \in {M}_{n}\left( \mathbb{R}\right) \) is said to be orthogonal if \( {R}^{tr}R = R{R}^{tr} = I \) . A matrix \( R \) is orthogonal if and only if\n\n\[ \langle {Rv},{Rw}\rangle = \langle v, w\rangle \]\n\nfor all \( v, w \in {\mathbb{R}}^{n} \) . | As in the unitary case, the condition \( {R}^{tr}R = I \) implies that \( R{R}^{tr} = I \) and that the columns of \( R \) form an orthonormal set in \( {\mathbb{R}}^{n} \) . Geometrically, a real matrix \( R \) is in \( \mathrm{O}\left( n\right) \) if and only if \( \left\langle {R{v}_{1}, R{v}_{2}}\right\rangle = \le... | Yes |
The groups \( \mathrm{O}\left( n\right) ,\mathrm{{SO}}\left( n\right) ,\mathrm{U}\left( n\right) \), and \( \mathrm{{SU}}\left( n\right) \) are compact. | Proof. The conditions defining these groups are obtained by setting certain continuous functions equal to a constant. The group \( \mathrm{{SU}}\left( n\right) \), for example, is defined by setting \( {\left( {U}^{ * }U\right) }_{jk} = {\delta }_{jk} \) for each \( j \) and \( k \) and by setting \( \det U = 1 \) . Th... | Yes |
The group \( \mathrm{U}\left( n\right) \) is connected. | Proof. If \( U \in {M}_{n}\left( \mathbb{C}\right) \) is unitary, then \( U \) has an orthonormal basis of eigenvectors with eigenvalues of absolute value 1 . Thus, there is another unitary matrix \( V \) (the change of basis matrix) such that\n\n\[ U = V\left( \begin{matrix} {e}^{i{\theta }_{1}} & & & \\ & {e}^{i{\the... | Yes |
The group \( \mathrm{{SU}}\left( 2\right) \) is simply connected. | Proof. We claim that\n\n\[ \mathrm{{SU}}\left( 2\right) = \left\{ {\left. \left( \begin{matrix} \alpha & - \bar{\beta } \\ \beta & \bar{\alpha } \end{matrix}\right) \right| \;\alpha ,\beta \in \mathbb{C},{\left| \alpha \right| }^{2} + {\left| \beta \right| }^{2} = 1}\right\} .\n\]\n\nIt is easy to see that each matrix ... | Yes |
Example 16.11 Let \( \mathcal{A} \) be an associative algebra and let \( \mathfrak{g} \) be a subspace of \( \mathcal{A} \) with the property that for all \( x, y \) in \( \mathfrak{g},{xy} - {yx} \) is again in \( \mathfrak{g} \) . Then the bracket\n\n\[ \left\lbrack {x, y}\right\rbrack \mathrel{\text{:=}} {xy} - {yx}... | In Example 16.11, we may take, for example, \( \mathfrak{g} = \mathcal{A} \) . It is evident that this bracket satisfies Properties 1, 2, and 3 of a Lie algebra, and the Jacobi identity is easily verified by direct calculation. As it turns out, every Lie algebra is isomorphic to a Lie algebra of this type. (This claim ... | Yes |
Theorem 16.15 The matrix exponential has the following properties for all \( X, Y \in {M}_{n}\left( \mathbb{C}\right) \) . | Properties 1, 2, and 3 are easily verified using term-by-term computation. Property 6 follows from Property 5 by taking \( Y = - X \) and applying Property 1 . The proofs of Properties 4, 5, and 7 are outlined in Exercises 5, 6, and 7. | No |
If\n\n\[ \nX = \left( \begin{matrix} 0 & a \\ - a & 0 \end{matrix}\right) \n\]\n\nthen\n\n\[ \n{e}^{X} = \left( \begin{array}{rr} \cos a & \sin a \\ - \sin a & \cos a \end{array}\right) .\n\] | Proof. The eigenvalues of \( X \) are \( \pm {ia} \) and the corresponding eigenvectors are \( \left( {1, \pm i}\right) \) . Thus, we may calculate that\n\n\[ \n{e}^{X} = \left( \begin{array}{rr} 1 & 1 \\ i & - i \end{array}\right) \left( \begin{matrix} {e}^{ia} & 0 \\ 0 & {e}^{-{ia}} \end{matrix}\right) \frac{1}{\left... | Yes |
Theorem 16.18 If \( A\left( \cdot \right) \) is a one-parameter subgroup of \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) \), there exists a unique \( X \in {M}_{n}\left( \mathbb{C}\right) \) such that\n\n\[ A\left( t\right) = {e}^{tX} \]\n\nfor all \( t \in \mathbb{R} \) . | This is Theorem 2.13 in [21]. | No |
Proposition 16.20 For any matrix Lie group \( G \), the Lie algebra \( \mathfrak{g} \) of \( G \) has the following properties.\n\n1. The zero matrix 0 belongs to \( \mathfrak{g} \).\n\n2. For all \( X \) in \( \mathfrak{g},{tX} \) belongs to \( \mathfrak{g} \) for all real numbers \( t \).\n\n3. For all \( X \) and \(... | Proof. Points 1 and 2 are elementary, and Point 3 follows from the Lie product formula, using the assumption that \( G \) is closed. Point 4 follows from Property 3 in Theorem 16.15. To verify Point 5, we observe that the commutator \( \left\lbrack {X, Y}\right\rbrack \) may be computed as\n\n\[ \left\lbrack {X, Y}\rig... | Yes |
Example 16.21 Let \( \mathrm{{gl}}\left( {n;\mathbb{C}}\right) ,\mathrm{{gl}}\left( {n;\mathbb{R}}\right) ,\mathrm{{sl}}\left( {n;\mathbb{C}}\right) \), and \( \mathrm{{sl}}\left( {n;\mathbb{R}}\right) \) denote the Lie algebras of \( \mathrm{{GL}}\left( {n;\mathbb{C}}\right) ,\mathrm{{GL}}\left( {n;\mathbb{R}}\right) ... | Proof. Let us consider, for example, the case of \( \mathfrak{{sl}}\left( {n;\mathbb{C}}\right) \) . By Property 4 of Theorem 16.15, if \( \operatorname{trace}\left( X\right) = 0 \), then\n\n\[ \det \left( {e}^{tX}\right) = {e}^{t\operatorname{trace}\left( X\right) } = {e}^{0} = 1 \]\n\nso that \( {e}^{tX} \in \mathrm{... | No |
The Lie algebras \( \mathrm{u}\left( n\right) \) and \( \mathrm{{su}}\left( n\right) \) of \( \mathrm{U}\left( n\right) \) and \( \mathrm{{SU}}\left( n\right) \) are given by\n\n\[ \mathfrak{u}\left( n\right) = \left\{ {X \in {M}_{n}\left( \mathbb{C}\right) \mid {X}^{ * } = - X}\right\} \]\n\n\[ \mathfrak{{su}}\left( n... | Proof. If \( {X}^{ * } = - X \), then by Property 2 of Theorem 16.15,\n\n\[ {\left( {e}^{tX}\right) }^{ * } = {e}^{t{X}^{ * }} = {e}^{-{tX}} = {\left( {e}^{tX}\right) }^{-1}, \]\n\nshowing that \( {e}^{tX} \) is unitary. In the other direction, if \( {e}^{tX} \) is unitary for all \( t \in \mathbb{R} \), then \( {\left... | Yes |
Theorem 16.23 Suppose \( {G}_{1} \) and \( {G}_{2} \) are matrix Lie groups with Lie algebras \( {\mathfrak{g}}_{1} \) and \( {\mathfrak{g}}_{2} \), respectively, and suppose \( \Phi : {G}_{1} \rightarrow {G}_{2} \) is a Lie group homomorphism. Then there exists a unique linear map \( \phi : {\mathfrak{g}}_{1} \rightar... | To construct \( \phi \), note that since \( \Phi \) is a continuous homomorphism, the map \( t \mapsto \Phi \left( {e}^{tX}\right) \) is a one-parameter subgroup. By Theorem 16.18, there exists a unique \( Y \) such that \( \Phi \left( {e}^{tX}\right) = {e}^{tY} \) for all \( t \in \mathbb{R} \) . We then set \( \phi \... | Yes |
Corollary 16.24 Suppose that \( {G}_{1} \) and \( {G}_{2} \) are matrix Lie groups with Lie algebras \( {\mathfrak{g}}_{1} \) and \( {\mathfrak{g}}_{2} \), respectively. If \( {G}_{1} \) is isomorphic to \( {G}_{2} \), then \( {\mathfrak{g}}_{1} \) is isomorphic to \( {\mathfrak{g}}_{2} \) . | Proof. See Exercise 11. ∎ | No |
Theorem 16.25 Let \( G \) be a matrix Lie group with Lie algebra \( \mathfrak{g} \). Then there exists a neighborhood \( U \) of 0 in \( {M}_{n}\left( \mathbb{C}\right) \) and a neighborhood \( V \) of \( I \) in \( {M}_{n}\left( \mathbb{C}\right) \) such that the matrix exponential maps \( U \) diffeomorphically onto ... | See Theorem 2.27 in [21]. | No |
Corollary 16.26 Every matrix Lie group \( G \subset \mathsf{{GL}}\left( {n;\mathbb{C}}\right) \) is a real embedded submanifold of \( {M}_{n}\left( \mathbb{C}\right) \) with the dimension of \( G \) equal to the dimension of \( \mathfrak{g} \) as a real vector space. | The claim means, more precisely, that for each \( A \in G \), there exists a neighborhood \( U \) of \( A \) and a diffeomorphism \( \Phi \) of \( U \) with a neighborhood \( V \) of 0 in \( {\mathbb{R}}^{2{n}^{2}} \) such that \( \Phi \left( {U \cap G}\right) = V \cap {\mathbb{R}}^{d} \), where \( d = \dim \mathfrak{g... | No |
The Lie algebra \( \mathfrak{g} \) of a matrix Lie group \( G \) is the tangent space to \( G \) at \( I \) . That is to say, \( \mathfrak{g} \) coincides with the set of those \( X \) in \( {M}_{n}\left( \mathbb{C}\right) \) for which there exists a smooth curve \( \gamma : \mathbb{R} \rightarrow {M}_{n}\left( \mathbb... | Proof. If \( X \in \mathfrak{g} \), then \( X \) is the derivative of \( {e}^{tX} \) at \( t = 0 \), so \( \mathfrak{g} \) is contained in the tangent space at \( I \) . In the other direction, if \( \gamma \) is any smooth curve in \( {M}_{n}\left( \mathbb{C}\right) \) that lies entirely in \( G \) and passes through ... | No |
Corollary 16.28 If a matrix Lie group \( G \) is connected, then for all \( A \in G \) there exists a finite sequence \( {X}_{1},{X}_{2},\ldots ,{X}_{N} \) of elements of \( \mathfrak{g} \) such that\n\n\[ A = {e}^{{X}_{1}}{e}^{{X}_{2}}\cdots {e}^{{X}_{N}}. \] | Proof. If \( G \) is connected in the sense of Definition 16.6 (which really means that \( G \) is path connected), then \( G \) is certainly connected in the usual topological sense of having no nontrivial sets that are both open and closed. Let \( U \) denote the set of points in \( G \) that can be expressed as a pr... | Yes |
Corollary 16.29 Suppose that \( {G}_{1} \) and \( {G}_{2} \) are matrix Lie groups with Lie algebras \( {\mathfrak{g}}_{1} \) and \( {\mathfrak{g}}_{2} \), respectively. Suppose that \( {\Phi }_{1} : {G}_{1} \rightarrow {G}_{2} \) and \( {\Phi }_{2} : {G}_{1} \rightarrow {G}_{2} \) are Lie group homomorphisms, with ass... | Proof. The result follows from Corollary 16.28 and the condition \( {\Phi }_{j}\left( {e}^{X}\right) = \) \( {e}^{{\phi }_{j}\left( X\right) }, j = 1,2 \) . ∎ | No |
Theorem 16.30 Suppose that \( {G}_{1} \) and \( {G}_{2} \) are matrix Lie groups with Lie algebras \( {\mathfrak{g}}_{1} \) and \( {\mathfrak{g}}_{2} \), respectively, and suppose that \( \phi : {\mathfrak{g}}_{1} \rightarrow {\mathfrak{g}}_{2} \) is a Lie algebra homomorphism. If \( {G}_{1} \) is connected and simply ... | One way to prove this deep result is to make use of the Baker-Campbell-Hausdorff formula. (See, e.g., Chap. 3 of [21].) This formula states that for all sufficiently small \( X \) and \( Y \) in \( {M}_{n}\left( \mathbb{C}\right) \) we have\n\n\[ \n{e}^{X}{e}^{Y} = {e}^{X + Y + \frac{1}{2}\left\lbrack {X, Y}\right\rbra... | No |
Corollary 16.31 Suppose that \( {G}_{1} \) and \( {G}_{2} \) are matrix Lie groups with Lie algebras \( {\mathfrak{g}}_{1} \) and \( {\mathfrak{g}}_{2} \), respectively. If \( {G}_{1} \) and \( {G}_{2} \) are connected and simply connected and \( {\mathfrak{g}}_{1} \) is isomorphic to \( {\mathfrak{g}}_{2} \), then \( ... | Proof. Suppose \( \phi : {\mathfrak{g}}_{1} \rightarrow {\mathfrak{g}}_{2} \) is a Lie algebra isomorphism. Since \( {G}_{1} \) is connected and simply connected, there exists a Lie group homomorphism \( \Phi : {G}_{1} \rightarrow {G}_{2} \) related in the usual way to \( \phi \) . Since \( {G}_{2} \) is connected and ... | Yes |
The Lie algebras \( \mathrm{{su}}\left( 2\right) \) and \( \mathrm{{so}}\left( 3\right) \) are isomorphic, but the groups \( \mathrm{{SU}}\left( 2\right) \) and \( \mathrm{{SO}}\left( 3\right) \) are not isomorphic. | Proof. The Lie algebra \( \mathrm{{su}}\left( 2\right) \) of \( \mathrm{{SU}}\left( 2\right) \) is the space of \( 2 \times 2 \) skew-self-adjoint matrices with trace zero. Explicitly,\n\n\[ \mathrm{{su}}\left( 2\right) = \left\{ {\left. \left( \begin{matrix} {ia} & b + {ic} \\ - b + {ic} & - {ia} \end{matrix}\right) \... | Yes |
Example 16.34 Let \( \Phi : \mathrm{{SU}}\left( 2\right) \rightarrow \mathrm{{SO}}\left( 3\right) \) be the unique Lie group homomorphism for which the associated Lie algebra homomorphism \( \phi \) satisfies \( \phi \left( {E}_{j}\right) = {F}_{j}, j = 1,2,3 \) . Then \( \ker \Phi = \{ I, - I\} \) and \( \left( {\math... | Proof. Since \( {E}_{1} \) is diagonal, it is easy to see that \( {e}^{{2\pi }{E}_{1}} = - I \) in \( {SU}\left( 2\right) \) . On the other hand, by a trivial extension of Example 16.16, we have\n\n\[ \n{e}^{a{F}_{1}} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & \cos a & - \sin a \\ 0 & \sin a & \cos a \end{array}\righ... | No |
Proposition 16.39 Suppose \( G \) is a connected matrix Lie group with Lie algebra \( \mathfrak{g} \) . Suppose that \( \Pi : G \rightarrow \mathrm{{GL}}\left( V\right) \) is a finite-dimensional representation of \( G \) and \( \pi : \mathfrak{g} \rightarrow \mathrm{{gl}}\left( V\right) \) is the associated Lie algebr... | Proof. Suppose \( W \subset V \) is invariant under \( \pi \left( X\right) \) for all \( X \in \mathfrak{g} \) . Then \( W \) is invariant under \( \pi {\left( X\right) }^{m} \) for all \( m \) . Since \( V \) is finite dimensional, any subspace of it is automatically a closed subset and thus \( W \) is invariant under... | Yes |
Theorem 16.40 (Schur’s Lemma) If \( {V}_{1} \) and \( {V}_{2} \) are two irreducible representations of a group or Lie algebra, then the following hold.\n\n1. If \( \Phi : {V}_{1} \rightarrow {V}_{2} \) is an intertwining map, then either \( \Phi = 0 \) or \( \Phi \) is an isomorphism.\n\n2. If \( \Phi : {V}_{1} \right... | Proof. It is easy to see that \( \ker \Phi \) is an invariant subspace of \( {V}_{1} \) . Since \( {V}_{1} \) is irreducible, this means that either \( \ker \Phi = {V}_{1} \), in which case \( \Phi = 0 \) , or \( \ker \Phi = \{ 0\} \), in which case \( \Phi \) is injective. Similarly, the range of \( \Phi \) is invaria... | Yes |
Proposition 16.42 Let \( \Pi : G \rightarrow \mathrm{{GL}}\left( V\right) \) be a finite-dimensional representation of a connected matrix Lie group \( G \), and let \( \pi \) be the associated representation of the Lie algebra \( \mathfrak{g} \) of \( G \) . Let \( \langle \cdot , \cdot \rangle \) be an inner product o... | Proof. Suppose first that \( \Pi \left( A\right) \) is unitary for all \( A \in G \) . Then for all \( X \in \mathfrak{g} \) and \( t \in \mathbb{R} \) we have\n\n\[ \Pi {\left( {e}^{tX}\right) }^{ * } = \Pi {\left( {e}^{tX}\right) }^{-1} = \Pi \left( {e}^{-{tX}}\right) = {e}^{-{t\pi }\left( X\right) }.\]\n\nOn the oth... | Yes |
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