Show that

$\mathit{\Psi }\left(x,t\right)\mathbf{=}{\left(\frac{m\omega }{\pi \sout{h}}\right)}^{\mathbf{1}\mathbf{/}\mathbf{4}}\mathbf{}\mathit{e}\mathit{x}\mathit{p}\left[-\frac{m\omega }{2\sout{h}}\left(x^2+\frac{a^2}{2}\left(1+e^{-2i\omega t}\right)+\frac{i\sout{h}t}{m}-2ax{e}^{{}^{-i\omega t}}\right)\right]$

satisfies the time-dependent Schrödinger equation for the harmonic oscillator potential (Equation 2.43). Here a is any real constant with the dimensions of length. 46

(b) Find$\mathbf{|}\mathbf{\Psi }\left(x,t\right){\mathbf{|}}^{\mathbf{2}}$ and describe the motion of the wave packet.

(c) Compute <x> and <p> and check that Ehrenfest's theorem (Equation 1.38) is satisfied.

The Ehrenfest theorem is a specific example of a more general relationship between the expectation of any quantum mechanical operator and the expectation of the commutator of that operator with the Hamiltonian of the system, where A is the expectation value of some quantum mechanical operator.

(a)

Here is an illustration of a precise solution to the harmonic oscillator potential part of the time-dependent Schrodinger equation:

$\Psi (\mathrm{x},\mathrm{t})={\left(\frac{\mathrm{m\omega }}{\mathrm{\pi }\sout{\mathrm{h}}}\right)}^{1/4}exp[-\frac{\mathrm{m\omega }}{2\sout{\mathrm{h}}}({\mathrm{x}}^2+\frac{{\mathrm{a}}^2}{2}\left(1+{\mathrm{e}}^{-2\mathrm{i\omega t}}\right)+\frac{\mathrm{i}\sout{\mathrm{h}}\mathrm{t}}{\mathrm{m}}-2{\mathrm{axe}}^{{}^{-\mathrm{i\omega t}}})]$

Where a is a constant.

Prove that this wave function satisfies the Schrodinger equation. The first derivative w.r.t t is:

$\frac{\partial \Psi }{\partial t}=\left[-\frac{1}{2}m{a}^2{\omega }^2{e}^{-2i\omega t}+\frac{1}{2}\sout{h}\omega +max{\omega }^2{e}^{-i\omega t}\right]\Psi$

The first derivative w.r.t x is:

$$\begin{gathered} \frac{\partial \Psi }{\partial x}=\left[\left(-\frac{m\omega }{2\sout{h}}\right)\left(2x-2a{e}^{-i\omega t}\right)\right]\psi \\ =-\frac{m\omega }{\sout{h}}\left(x-a{e}^{-i\omega t}\right)\Psi \end{gathered}$$

Also, the second derivative w.r.t x is:

$$\begin{gathered} \frac{{\partial }^2\Psi }{\partial x}=-\frac{m\omega }{\sout{h}}\Psi \left(x-a{e}^{-i\omega t}\right)\frac{\partial \Psi }{\partial x} \\ =\left[-\frac{m\omega }{\sout{h}}+{\left(\frac{m\omega }{\sout{h}}\right)}^2{\left(x-a{e}^{-i\omega t}\right)}^2\right]\Psi \end{gathered}$$

$$\begin{gathered} \mathrm{Assume}, \\ \left[\mathrm{s}\right]=-\frac{\sout{\mathit{h}}}{\mathit{2}\mathit{m}}\frac{{\partial }^2}{\partial {\mathrm{x}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2\mathrm{\Psi } \end{gathered}$$

Thus:

$$\begin{gathered} \left[\mathrm{s}\right]=-\frac{\sout{h}}{\mathit{2}m}\left[-\frac{m\omega }{\sout{h}}+{\left(\frac{m\omega }{\sout{h}}\right)}^2{\left(x-a{e}^{-i\omega t}\right)}^2\right]\Psi +\frac{1}{2}m{\omega }^2{x}^2\Psi \\ =\left[\frac{1}{2}\sout{\mathit{h}}\omega -\frac{1}{2}m{\omega }^2{\left(x-2ax{e}^{i\omega t}\right)}^2+\frac{1}{2}m{\omega }^2{x}^2\right]\Psi \\ =\left[\frac{1}{2}\sout{h}\omega +max{\omega }^2{e}^{-i\omega t}-\frac{1}{2}m{\omega }^2{a}^2{e}^{-i\omega t}\right]\Psi \\ =i\sout{h}\frac{\partial \Psi }{\partial t} \end{gathered}$$

Thus, the LHS and the So, LHS and RHS, of the Schrodinger equation are the same, that is:

$-\frac{{\sout{h}}^2}{\mathit{2}m}\frac{{\partial }^2\Psi }{\partial {\mathrm{x}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2\mathrm{\Psi }=i\sout{\mathit{h}}\frac{\mathit{\partial }\mathit{\Psi }}{\mathit{\partial }\mathit{t}}$

(b)

Calculate the modulus square of the wave function,

$$\begin{gathered} {\left|\psi \right|}^2=\sqrt{\frac{m\omega }{\pi \sout{h}}}exp-\frac{m\omega }{2\sout{h}}\left(x^2+\frac{a^2}{2}\left(1+e^{2i\omega t}\right)-\frac{i\sout{h}t}{m}-2ax{e}^{i\omega t}\right) \\ \times exp\left(x^2+\frac{a^2}{2}\left(1+e^{{}^{2i\omega t}}\right)+-\frac{i\sout{h}t}{m}-2ax{e}^{i\omega t}\right) \\ {\left|\psi \right|}^2=\sqrt{\frac{m\omega }{\pi \sout{h}}}exp-\frac{m\omega }{2\sout{h}}\left[2x^2+a^2+a^2\cos \left(2\omega t\right)-4ax\cos \left(\omega t\right)\right] \end{gathered}$$

but:

$a^2\left[1+\cos \left(2\omega t\right)\right]=2a^2{\cos }^2\left(\omega t\right)$

So,

$$\begin{gathered} {\left|\psi \right|}^2=\sqrt{\frac{m\omega }{\pi \sout{h}}}exp-\frac{m\omega }{\sout{h}}\left[x^2-2ax\cos \left(\omega t\right)+a^2{\cos }^2\left(\omega t\right)\right] \\ {\left|\psi \right|}^2=\sqrt{\frac{m\omega }{\pi \sout{h}}}exp-\frac{m\omega }{\sout{h}}{\left(x-a\cos \left(\omega t\right)\right)}^2 \end{gathered}$$

Therefore, the wave packet center has a sinusoidal shape $\left(\cos \left(\omega t\right)\right)$ , and also it has a fixed Gaussian shape. The oscillations have an amplitude of a and an angular frequency of $\omega$.

(c)

The expectation value of x is:

let y = x -a cos $\left(\omega t\right)$and thus dx = dy,

Now substitute

Now substitute from part (b) with and make the substitution with y,

So,

So we have two integrals, the first of which is zero, and the function is an odd function, as well as an odd function integration. $-\infty$to $\infty$is zero. To find the second integral, we use:

${\int }_{-\infty }^{\infty }{e}^{-by}dy=\sqrt{\frac{\pi }{b}}$

Therefore,

Hence, the value ofis a cos$\left(\omega t\right)$.

The expectation value of the momentum is:

Thus, the value of is $-ma\omega \sin \left(\omega t\right)$.

Now check if the theorem is satisfied.

Here,

Hence we conclude that Ehrenfest's theorem is satisfied.