Question

(a) Write down the time-dependent "Schrödinger equation" in momentum space, for a free particle, and solve it. Answer: $\mathbf{exp}\left(-i{p}^{2}t/2m\sout{h}\right)\mathbf{\Phi }\left(p,0\right)\mathbf{.}$

(b) Find role="math" localid="1656051039815" $\mathbf{\Phi }\mathbf{(}\mathbf{p}\mathbf{,}\mathbf{0}\mathbf{)}$for the traveling gaussian wave packet (Problem 2.43), and construct $\mathbf{\Phi }\mathbf{(}\mathbf{p}\mathbf{,}\mathbf{t}\mathbf{)}$for this case. Also construct ${\left|\Phi (p,t)\right|}^{\mathbf{2}}$, and note that it is independent of time.

(c) Calculaterole="math" localid="1656051188971" androle="math" localid="1656051181044" by evaluating the appropriate integrals involving$\mathbf{\Phi }$, and compare your answers to Problem 2.43.

(d) Show thatrole="math" localid="1656051421703" $<H>\mathbf{=}{<p>}^{\mathbf{2}}\mathbf{/}\mathbf{2}\mathbf{m}\mathbf{+}{<H>}_{\mathbf{0}}$(where the subscript denotes the stationary gaussian), and comment on this result.

Short answer

Required answer are

$\left(a\right)\mathrm{\Phi }\left(p,t\right)=e^{-i{p}^{2}t/2m\sout{h}}\mathrm{\Phi }\left(p,0\right)$

$\left(b\right)\left\{\begin{array}{l}\mathrm{\Phi }\left(p,0\right)=\frac{1}{{\left(2\mathrm{\pi a}{\sout{\mathrm{h}}}^2\right)}^{1/4}}{e}^{-{\left(I-p/\sout{h}\right)}^2}/4a\\ \mathrm{\Phi }\left(p,t\right)=\frac{1}{{\left(2\mathrm{\pi a}{\sout{\mathrm{h}}}^2\right)}^{1/4}}{e}^{-{\left(I-p/\sout{h}\right)}^2/4a}{e}^{-p{p}^{2}t/2m\sout{h}}\\ {\left|\mathrm{\Phi }\left(p,t\right)\right|}^2=\frac{1}{\sqrt{2\mathrm{\pi a}}\sout{h}}{e}^{-{\left(I-p/\sout{h}\right)}^2/2a}\end{array}\right.$

Step-by-step solution

Hamiltonian and Schrodinger equation for the free particle

The Hamiltonian for a free particle is:

$\mathit{H}\mathbf{=}\frac{{\mathbf{p}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$

and Schrodinger equation for the free particle is:

$\mathit{i}\sout{\mathbf{h}}\frac{\mathbf{\partial }\mathbf{\psi }}{\mathbf{\partial }\mathbf{t}}\mathbf{=}\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{\psi }}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}$

Step 2(a): first and second derivative of wave function

The wave function $\mathrm{\psi }\left(x,t\right)$can be written in the momentum space, as:

$\mathrm{\psi }\left(x,t\right)=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\infty }^{\infty }{e}^{ipx/\sout{h}}\mathrm{\Phi }\left(p,t\right)dp$

take the first-time derivative of this equation, to get:

$\frac{\partial \mathrm{\psi }}{\partial t}=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\infty }^{\infty }{e}^{ipx/\sout{h}}\frac{\partial \mathrm{\Phi }}{\partial t}dp$

and the second time derivative, to get:

$\frac{{\partial }^2\mathrm{\psi }}{\partial x^2}=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\infty }^{\infty }\left(-\frac{p^2}{{\sout{h}}^2}\right)e^{ipx/\sout{h}}\mathrm{\Phi }dp$

substitute into (1) to get:$\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\infty }^{\infty }{e}^{ipx/\sout{h}}\left[i\sout{h}\frac{\partial \mathrm{\Phi }}{\partial t}\right]dp=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\infty }^{\infty }{e}^{ipx/\sout{h}}\left[\frac{p^2}{2m}\mathrm{\Phi }\right]dp$

by comparing both sides, we can see that the terms in the square brackets in both sides must equal, so:

$i\sout{h}\frac{\partial \mathrm{\Phi }}{\partial t}=\frac{p^2}{2m}\mathrm{\Phi }$

this called the Schrodinger equation in the momentum space. We can write it as:

$\frac{1}{\mathrm{\Phi }}d\mathrm{\Phi }=-\frac{i{p}^2}{2m\sout{h}}dt$

integrate both sides from to , and from to , to get:

$$\begin{gathered} \mathrm{In}\left(\frac{\mathrm{\Phi }\left(p,t\right)}{\mathrm{\Phi }\left(p,0\right)}\right)=-\frac{i{p}^{2}t}{2m\sout{h}} \\ \mathrm{\Phi }\left(p,t\right)=e^{-i{p}^{2}t/2m\sout{h}}\mathrm{\Phi }\left(p,0\right) \end{gathered}$$

Step 3(b): solve for Φ(p,t)

From problemthe traveling gaussian wave packet is given by,

$\mathrm{\psi }\left(x,0\right)=A{e}^{-a{x}^2}{e}^{ilx}A={\left(\frac{2a}{\mathrm{\pi }}\right)}^{1/4}$

we need to construct , using:

$\mathrm{\Phi }\left(x,0\right)=\frac{1}{\sqrt{\mathit{2}\pi \sout{\mathit{h}}}}{\int }_{-\infty }^{\infty }{e}^{-ipx/\sout{h}}\mathrm{\psi }\left(x,0\right)dx$

as:

$$\begin{gathered} \mathrm{\Phi }\left(p,0\right)=\frac{1}{\sqrt{2\pi \sout{h}}}{\left(\frac{2a}{\pi }\right)}^{1/4}{\int }_{-\infty }^{\infty }{e}^{-ipx/\sout{h}}{e}^{-a{x}^2}{e}^{ilx}dx \\ =\frac{1}{{\left(2\mathrm{\pi a}{\sout{\mathrm{h}}}^2\right)}^{1/4}}{e}^{-{\left(-ipx/\sout{h}\right)}^2/4a} \\ \mathrm{\Phi }\left(p,0\right)=\frac{1}{{\left(2\pi a{\sout{h}}^2\right)}^{1/4}}{e}^{-{\left(-ipx/\sout{h}\right)}^2/4a} \end{gathered}$$

using the results of part (a), we can construct $\mathrm{\Phi }\left(p,t\right)$as:

$\Phi \left(p,t\right)=\frac{1}{\left(-...2\right)1/4}{e}^{-{\left(-ipx/\sout{h}\right)}^2}{e}^{-i{p}^{2t/2m\sout{h}}}$

thus:

${\left|\Phi \left(p,t\right)\right|}^2=\frac{1}{\sqrt{2\mathrm{\pi a}}\sout{h}}{e}^{-{\left(I-p/\sout{h}\right)}^2/2a}$

Step 4(c): find the expectation value of the momentum

In this part we need to find the expectation value of the momentum as:

let $y=\left(p/\sout{h}\right)-I,\mathrm{so}p=\sout{h}\left(y+I\right)\mathrm{and}dp=\sout{h}dy$

so, we have two integrals, the first one is odd and the integration of the odd function from $-\infty$to $\infty$is zero, so:

but,

${\int }_0^{\infty }{e}^{-y^2/2a}dy=\sqrt{\frac{\pi a}{2}}$

SO,

localid="1656056598257"

now we need to find the expectation value of the momentum squared, as:

so, we have three integrals, the first and the last one is even, where the second one is odd, the value of the odd function is zero,

the results are consistent with problem $2.43$.

Step 5(d) : solve for <H>

In this part we need to show that:

where the subscript 0 indicates the stationary Gaussian. The Hamiltonian is:

$H=\frac{p^2}{2m}$

thus:

but:

SO: