HW #5. 1. Harmonic Oscillator. (a) (b) HW5.nb 1. . We first calculate a a. We rewrite the Hamiltonian H. i p, x. Therefore, 1

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1 HW5.nb HW #5. Harmonic Oscillator (a) We rewrite the Hamiltonian H m x using a x i, a x i. We first calculate a a x i x i x i, x m Ω x. m Ω Therefore, Ω a a x Ω m, and hence H Ωa a. (b) The ground state condition a 0 0 can be written in the osition reresentation as x a 0 x x i 0 x i i d x 0 0, d x and hence x d d x Ψ 0 x 0. This equation can be solved easily and we find Ψ 0 x N e x. To normalize the wave function, we comute e x d x Π. Therefore, the correctly normalized ground state wave function is Ψ 0 x Π 4 e x. The shae of the wave function is Ψ 0 x_ : 4 Π E x

2 HW5.nb PlotΨ 0 x. m, Ω,, x, 3, 3; (c) The first excited state is given by a 0, and its osition reresentation by x x a 0 x i i d x 0 d x x x Π 4 e x Π 4 x e x Its shae is Ψ x_ : 4 Π x E x PlotΨ x. m, Ω,, x, 3, 3; Check that it is roerly normalized: IntegrateΨ x, x,,, Assumtions Re 0

3 HW5.nb 3 The second excited state is given by a, and its osition reresentation by x x a 0 x i i d x d x Its shae is Simlify m x Ω x Ψ x m x Ω 4 Π 4 DΨ x, x m x Ω Ψ x_ : m x Ω 4 Π 4 PlotΨ x. m, Ω,, x, 3, 3; Check that it is roerly normalized: IntegrateΨ x, x,,, Assumtions Re 0 (d) From the definitions of the annihilation and creation oerators, we can solve for x, x a a. Starting with the exectation values, x n x n n a a n n n n n n. Because of the orthonormality of the Hamiltonian eigenstates n m n, m,

4 HW5.nb 4 x 0. Moving on to the variance, x n a a n n a a a a n, n N a, a n n. x n. From the definitions of the annihilation and creation oerators, we can solve for, i a a. Together with the exression for x from the revious roblem, you can easily verify x, i. Starting with the exectation values, n n i n a a n, 0. Moving on to the variance, n a a n n a a a a n, n N a, a n n. n. Therefore, x n. The ground state n 0 is a minimum uncertainty state, while the excited states have larger uncertainties.

5 HW5.nb 5 (e) There are many ways to show this. First of all, using the relation a n n n, we can show that a n 0 n n by recursion. It obviously holds for n. If it holds for n, the n th one is a n 0 a a n 0 a n n n n n n n, and it holds again. Therefore, it holds for all n. Using the definition f e f a we can write it as Then f 0 e f ne f. n n a n 0 e f, a f a ne f n n e f, n n n where the summation is now taken only from i because i 0 term vanishes by a 0 0. Continuing on, a f n e f me f, n n m 0 m where the dummy variable was chgned to m n. Pulling one factor of f out of the sum, f m a f f m 0 f m me f f f. m Another way to show the same result is by first showing the relation a, a n na n. It obviously holds for n. If it holds for n, the n th one is a, a n a, a a n a, a a n a a, a n a n a na n n a n and hence it holds as well. Therefore, it holds for any n. Starting with the definition f e f a 0 e f f n n a n 0e f, f a f a n n a n 0e f f n n a, a n 0e f, Here, we used the fact a 0 0. Using the relation shown above, f a f n n n n a n 0e f f n n n a n 0e f, where the summation is now taken only from i because i 0 term vanishes by a, 0. Changing the dummy variable to m n, f a f m a m 0e f, m 0 m a f f m 0 f m m a m 0e f f f. Finally, we verify the normalization, f f e f f m m 0 m n 0 am a n 0. Consider a acting on the left and a acting on the right, again by orthonormality we must have m n, so f f e f f n n n n nn e f f n n e n f e f

6 HW5.nb 6 Finally, we verify the normalization, f f e f f m m 0 m n 0 am a n 0. Consider a acting on the left and a acting on the right, again by orthonormality we must have m n, so f f e f f n n n n nn e f f n n e n f e f f f. (f) Using the result from (d), f x f f a a f f f Re f. Here, we used the fact f a f f, obtained by taking the hermitian conjugate of a f f f. Similarly, using the result from (e), f f f i a a f i f f Im f. Now on the variance, f x f f a a f f a a a a a a f f a a, a a a a f f f f f f f and hence x. Similarly, f f f a a f f a a a a a a f f a a, a a a a f f f f f f f and hence. Therefore, x and hence the coherent state is a minimum uncertainty state for any f. (g) [otional] We take the solution from Eq.

7 HW5.nb 7 Ψfx_ 4 Π Ex x f and aly the oerator a as in art (b) f Absf ; resa x Ψfx i i x. The desired result is resd f Ψfx;. We see that they are indeed the same Simlifyresa resd Ψfx ; 0 (h) The Heisenberg equation of motion is i d d t x x, H x, m i m, i d, H, d t x i x. There are many ways to solve these couled equations. One way is to use the exonential of a matrix. Write the equations in the matrix form, d d t x Ω Ω 0 x 0 x. Therefore, x t ex 0 i x i Ω t 0 exσ i 0 i Ω t x 0. Here, Σ is one of the Pauli matrices. The exonential factor can be worked out using its Taylor exansion, exσ i Ω t n Σ n i Ω t n. It is easy to check that Σ, and hence Σ even, Σ odd Σ. Therefore, exσ i Ω t, even n i Ω tn, odd n Σ i Ω t n cos Ω t i Σ sin Ω t cos Ω t sin Ω t sin Ω t cos Ω t. We find the solution x t cos Ω t sin Ω t sin Ω t cos Ω t x 0 x0cos Ω t 0sin Ω t x0sin Ω t 0cos Ω t. Using this solution, we calculate the exectation values, xt f xt f f x0cos Ω t 0 sin Ω t f Re f cos Ω t Im f sin Ω t

8 HW5.nb 8 xt Re f cos Ω t Im f sin Ω t. t f t f f x0sin Ω t 0cos Ω t f Re f sinω t Im f cos Ω t t Re f sinω t Im f cos Ω t. Indeed, these solutions corresond to the classical ones. (i) [otional] The Schrödinger equation gives n, t e i H t n e i Ωnt n e i Ωnt n. Using what we showed above, f n, n its time evolution is f, t e i H t n n n f e i Ω t n n nei Ωnt ne i Ω t f e i Ω t i Ω t e f, t f e i Ω t e i Ω t. where the coherent state in the last exression has the eigenvalue a f e i Ω t f e i Ω t f e i Ω t. (j) The robability density on x is x f e i Ω t, i.e. Ρfx_, t_ ConjugateΨfxΨfx. f x0 ExI Ω t;. For the convenience in lotting, we set all the constants to Ρfcx_, t_ SimlifyComlexExandΡfx, t. m,, Ω, x0 xcost Π. Now we animate over time interval t 0, Π with gas t Π 4: Grahics Animation ; MoviePlotΡfcx, t, x, 4, 4, t, 0, Π, Π 4, AsectRatio.5, PlotRange 4, 4, 0, 0.6;

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11 HW5.nb Perfectly oscillatory Gaussian! Here m ; however, as one would exect by the Corresondence Princile, as m, the Gaussian aroaches a delta distribution.. Time Evolution of Gaussian Wave Packet [otional] (a) The result for q Ψ Φq from Hwk 4 is Φq_ d Π 4 I q x0 Ex Ex q d ;. It is given that Ψ, t e i q tm qq Ψq so Ψx, t x Ψ, t e i q tm x qq Ψq This gives e i q x e i q tm Φqq. Π myassumtions m 0, 0, d 0, x0 0,, q, t Reals; Ψx_, t_ IntegrateEx I q t m Ex I q x Φq, q,,, Assumtions myassumtions Π d t m x m xx0 t x0 d m t Π 4 t d d m (Note that I defined myassumtions for use later so Mathematica understands the nature of the constants here.) This result looks a little odd with the i s floating about and t in the denominators, but Ψx, 0 is indeed just the Ψx given in Hwk 4: TrigToExSimlifyComlexExandΨx, 0, Assumtions myassumtions x xx0 4 d d Π 4

12 HW5.nb (b) As we know xt Ψ, t x Ψ, t Ψx, t x Ψx, t x which gives meanx IntegrateComlexExandConjugateΨx, tx Ψx, t, x,,, Assumtions myassumtions t x0 m This is the exected result! (c) The second moment x t Ψ, t x Ψ, t Ψx, t x Ψx, t x which gives meanx IntegrateComlexExandConjugateΨx, tx Ψx, t, x,,, Assumtions myassumtions 4 d 4 m t 4 d t m x0 4 d m So the disersion square x x x is disx Simlifymeanx meanx, Assumtions myassumtions d t 4 d m Again, as exected, the disersion increases with time; secifically, the diserson square increases quadratically, as shown below with all constants set to. Plotdisx. d,, m, t, 0, 0;

13 HW5.nb 3 (d) There is only one Heisenberg equation of motion since, H 0, so like in art (h) d i d t x x, H x, i m m, which clearly gives xt x0 m t or xt x on the x basis. t m i x For the second moment, d d t x x d x so d x d t d t x x0 m t m m x0 m t or x t m t x0 x t t m m i t x m i m x0t x 0 x x x. (e) Again like in art (h), xt Ψ xt Ψ Ψx, 0 xtψx, 0 x which gives meanxh IntegrateComlexExandConjugateΨx, 0 x Ψx, 0 t m I x x,,, Assumtions myassumtions Ψx, 0, t x0 m Similarly, for x t meanxh IntegrateComlexExand ConjugateΨx, 0 t x x Ψx, 0 t m I m I x x x,,, Assumtions myassumtions 4 d 4 m t 4 d t m x0 4 d m. Thus, for x t we find Ψx, 0 Ψx, 0 x Ψx, 0,

14 HW5.nb 4 disxh Simlifymeanxh meanxh, Assumtions myassumtions d t 4 d m These results in the Heisenberg icture are just as we calculated in arts (b) and (c) for the Schroedinger icture! To check: meanx meanxh disx disxh 0 0