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ψ =1(1,ψ)+2(2,ψ) with the two coefficients (1,ψ),(2,ψ) C.
∈
Let us discuss two particular types of evolutions.
First, letus discuss the Schr¨odingerequationwith diagonalHamiltonmatrix,i.e., with vanish-
ing off-diagonal elements,
E 0
H = 1 . (51)
ij 0 E
2
(cid:18) (cid:19)
In this case, the Schr¨odinger equation decouples and reduces to
βˆ‚ βˆ‚
i~ (1,ψ(t))=E (1,ψ(t)) , i~ (2,ψ(t))=E (2,ψ(t)) , (52)
1 2
βˆ‚t βˆ‚t
resulting in
βˆ’iE1t/~ βˆ’iE2t/~
(1,ψ(t))=ae , (2,ψ(t))=be , (53)
with a,b C, a2+ b2 =1. These solutions correspondto stationary states which do not change
∈ | | | |
in time; i.e., the probability to find the system in the two states is constant
(1,ψ)2 = a2 , (2,ψ)2 = b2 . (54)
| | | | | | | |
Second, let us discuss the Schr¨odingerequationwithwith non-vanishingbut equaloff-diagonal
elements A and with equal diagonal elements E of the Hamiltonian matrix; i.e.,
βˆ’
E A
H ij = A βˆ’ E . (55)
(cid:18) βˆ’ (cid:19)
In this case, the Schr¨odinger equation reads
βˆ‚
i~ (1,ψ(t)) = E(1,ψ(t)) A(2,ψ(t)) , (56)
βˆ‚t βˆ’
βˆ‚
i~ (2,ψ(t)) = E(2,ψ(t)) A(1,ψ(t)) . (57)
βˆ‚t βˆ’
Theseequationscanbesolvedinanumberofways. Forexample,takingthesumandthedifference
of the two, one obtains
βˆ‚
i~ ((1,ψ(t))+(2,ψ(t))) = (E A)((1,ψ(t))+(2,ψ(t))) , (58)
βˆ‚t βˆ’
βˆ‚
i~ ((1,ψ(t)) (2,ψ(t))) = (E+A)((1,ψ(t)) (2,ψ(t))) . (59)
βˆ‚t βˆ’ βˆ’
The solution are again two stationary states
(1,ψ(t))+(2,ψ(t)) = ae (i/~)(E A)t , (60)
βˆ’ βˆ’
(i/~)(E+A)t
(1,ψ(t)) (2,ψ(t)) = be . (61)
βˆ’
βˆ’
16
βœ“βœ βœ“βœ
|1 ✘✘✘✘✘✘✘✘ βœ”βœ’βœ‚H βœ”βœ”βœ’N |2
✘i ▲❝ i
βœ‘ βœ‘β
βœ“ H P✏ βœ”βœ” βœ‚βœ‚ βœ”βœ” β–² ❝ ❝❝
P P β–² βœ“βœ
βœ’β† β†βœ‘ P P βœ”P βœ”βœ“βœ” Hβœ‚ ✏ βœ˜βœ” βœβœ˜βœ˜βœ˜βœ˜β–²βœ˜ β–²βœ˜βœ˜βœ˜ βœ‚βœ‚H
βœ“βœ” β–² βœ’ βœ‘
❆
❆ βœ” (cid:0)(cid:0)βœ’βœ‘ H P P β–² βœ‚
❆ ❆ βœ“βœ”βœ” (cid:0) βœ’βœ‘P P P Pβœ“β–²β–² βœ‚
✏ ✏
N H
βœ’βœ‘ βœ’βœ‘
Figure 2: The two equivalent geometric arrangements of the ammonia (NH ) molecule.
3
Thus,
a b
(1,ψ(t)) = e (i/~)(E A)t+ e (i/~)(E+A)t , (62)
βˆ’ βˆ’ βˆ’
2 2
a b
(2,ψ(t)) = e (i/~)(E A)t e (i/~)(E+A)t . (63)
βˆ’ βˆ’ βˆ’
2 βˆ’ 2
Assume now that initially, i.e., at t = 0, the system was in state ,1) =,ψ(t = 0)). This
assumption corresponds to (1,ψ(t = 0)) = 1 and (2,ψ(t = 0)) = 0. What is the probability that
the system will be found in the state 2 at the time t>0, or that it will still be found in the state
1 at the time t>0? Setting t=0 in equations (62) and (63) yields
a+b a b
(1,ψ(t=0))= =1 , (2,ψ(t=0))= βˆ’ =0 , (64)
2 2
and thus a= b=1. Equations (62) and (63) can now be evaluated at t >0 by substituting 1 for
a and b,
e(i/~)At+e (i/~)At
At
(i/~)Et βˆ’ (i/~)Etcos
(1,ψ(t)) = e =e , (65)
βˆ’ 2 βˆ’ ~
(cid:20) (cid:21)
e(i/~)At (i/~)At
e At
βˆ’(i/~)Et βˆ’ βˆ’(i/~)Etsin
(2,ψ(t)) = e βˆ’ =ie . (66)
2 ~
(cid:20) (cid:21)
Finally, the probability that the system is in state ,1) and ,2) is
At At