Question

$$\begin{gathered} \mathbf{Find}\mathbf{}\mathbf{the}\mathbf{}\mathbf{momentum}\mathbf{-}\mathbf{space}\mathbf{}\mathbf{wave}\mathbf{}\mathbf{function}\mathbf{}\mathbf{}{\mathbf{\varphi }}_{\mathbf{n}}\mathbf{(}\mathbf{p}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{}\mathbf{for}\mathbf{}\mathbf{the}\mathbf{}{\mathbf{n}}^{\mathbf{th}}\mathbf{}\mathbf{stationary}\mathbf{}\mathbf{state}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the} \\ \mathbf{}\mathbf{infinite}\mathbf{}\mathbf{square}\mathbf{}\mathbf{well}\mathbf{.}\mathbf{}\mathbf{Graph}\mathbf{}\mathbf{}{\left|{\varphi }_1(p,t)\right|}^{\mathbf{2}\mathbf{}}\mathbf{and}\mathbf{}{\left|{\varphi }_2(p,t)\right|}^{\mathbf{2}\mathbf{}}\mathbf{}\mathbf{,}\mathbf{}\mathbf{as}\mathbf{}\mathbf{functions}\mathbf{}\mathbf{of}\mathbf{}\mathbf{p}\mathbf{}\mathbf{(}\mathbf{pay}\mathbf{}\mathbf{particular}\mathbf{}\mathbf{attention}\mathbf{}\mathbf{to}\mathbf{}\mathbf{the}\mathbf{} \\ \mathbf{points}\mathbf{}\mathbf{p}\mathbf{=}\mathbf{\pm }\mathbf{n\pi h}\mathbf{/}\mathbf{a}\mathbf{)}\mathbf{}\mathbf{.}\mathbf{}\mathbf{Use}\mathbf{}{\mathbf{\varphi }}_{\mathbf{n}}\mathbf{(}\mathbf{p}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{}\mathbf{}\mathbf{to}\mathbf{}\mathbf{calculate}\mathbf{}\mathbf{the}\mathbf{}\mathbf{expectation}\mathbf{}\mathbf{value}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{p}}^{\mathbf{2}}\mathbf{}\mathbf{.}\mathbf{} \\ \mathbf{Compare}\mathbf{}\mathbf{your}\mathbf{}\mathbf{answer}\mathbf{}\mathbf{to}\mathbf{}\mathbf{Problem}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{.} \end{gathered}$$

Short answer

The final result for the momentum-space time dependent wave function:${\varphi }_n(p,t)=\sqrt{\frac{\mathrm{\pi a}}{\sout{h}}}\frac{n{e}^{-i\omega nt}}{n^2{\mathrm{\pi }}^2-{\mathrm{a}}^2{\mathrm{p}}^2/\sout{{\mathrm{h}}^2}}(1-{(-1)}^n{e}^{ipa/\sout{h}}).$

The expectation value of.

Step-by-step solution

Fourier transform  

A Fourier transform (FT) is a mathematical transformation that breaks down geographically or temporally dependent functions into spatially or temporally independent functions. First observe the${\mathit{n}}^{\mathbf{t}\mathbf{h}}$in the coordinate representation for the infinite square well of width a: localid="1656312964496" ${\mathit{\psi }}_{\mathbf{n}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{a}}}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\frac{\mathbf{n}\mathbf{\pi x}}{\mathbf{a}}\mathbf{\right)}\mathbf{.}$

Now, the time-dependent solution to the given problem is just the stationary state multiplied by the time propagator:

localid="1656316815015" ${\mathbf{\psi }}_{\mathbf{n}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\sqrt{\frac{\mathbf{2}}{\mathbf{a}}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}{\mathbf{i\omega }}_{\mathbf{n}}\mathbf{t}}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{n\pi x}}{\mathbf{a}}\mathbf{\right)}\mathbf{.}$

Where the ${\omega }_n$is the abbreviation in the following context:

$$\begin{gathered} E_n=\sout{h}{\omega }_n \\ =\frac{n^2{\mathrm{\pi }}^2{\sout{\mathrm{h}}}^2}{2m{a}^2} \end{gathered}$$

Now, the time dependent solution in the momentum representation is given by the Fourier transform of the coordinate basis, which connects the two basis. Therefore:

$$\begin{gathered} {\varphi }_n(p,t)=\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}{\int }_{-\infty }^{\infty }{\psi }_n(x,t)e^{-ipx/\sout{h}}dx \\ =\frac{1}{\sqrt{2\mathrm{\pi }\sout{\mathrm{h}}}}\sqrt{\frac{2}{a}}{e}^{-i{\omega }_{n}t}{\int }_0^a\left(\frac{n\mathrm{\pi x}}{a}\right)e^{-ipx/\sout{h}}dx \end{gathered}$$

The limits of the integration are from to since the wave function outside the box vanishes. Now isolate the integral and try to evaluate it.

$l={\int }_0^a\sin \left(Nx\right)e^{-kx}dx$

Where the abbreviations $N=n\mathrm{\pi }/\mathrm{a}$and $k=p/\sout{h}$.

Find the total solution to the integral l

Now rewrite the function in the terms of exponential function as follows:

$\sin \left(Nx\right)=\frac{1}{2i}(e^{iNx}-e^{-iNx})$

Therefore, the integrall has the following form:

$$\begin{gathered} l={\int }_0^a\sin \left(Nx\right)e^{-ipx/\sout{h}}dx \\ ={\int }_0^a\frac{1}{2i}(e^{iNx}-e^{-iNx})e^{-ikx}dx \\ =\frac{1}{2i}{\int }_0^a(e^{-i(k-N)x}-e^{-i(k+N)x})dx \\ =l_{-}-l_{+} \end{gathered}$$

The integral lhas the following form:

$$\begin{gathered} l_{\pm }=\frac{1}{2i}{\int }_0^a{e}^{-i(K\pm N)x}dx \\ =\frac{1}{2i}{\overline{)\frac{e^{-i(K\pm N)x}}{-i(K\pm N)}}}_0^a \\ =\frac{1}{2(K\pm N)}(e^{-i(K\pm N)}-1) \end{gathered}$$

Therefore, the total solution to the integral lis:

$$\begin{gathered} l_{\pm }=l_{-}-l_{+} \\ =\frac{1}{2(K-N)}(e^{-i(K-N)a}-1)-\frac{1}{2(K+N)}(e^{-i(K+N)a}-1) \\ =\frac{e^{-i(Ka-n\mathrm{\pi })}}{2(K-N)}-\frac{e^{-i(Ka+n\mathrm{\pi })}}{2(K+N)} \\ =\frac{{(-1)}^n{e}^{-iKa}}{2(K-n\mathrm{\pi }/\mathrm{a})}-\frac{{(-1)}^n{e}^{-iKa}}{2(K+n\mathrm{\pi }/\mathrm{a})} \end{gathered}$$

Find momentum-space representation with the special case when n = 1 .

Now, combine this result for the integral with the time propagator and pre-factors, just obtain (by the use of common denominator) the final result for the momentum-space time dependent wave function: ${\Phi }_n(p,t)=\sqrt{\frac{\pi a}{\hslash }}\frac{n{e}^{-i{\omega }_{n}t}}{n^2{\pi }^2-a^2{p}^2/{\hslash }^2}(1-(-1{)}^n{e}^{-ipa/\sout{h})}$

Next, now that a general momentum-space representation is there find the special cases when n = 1 and n = 2 .

For n = 1 :

$$\begin{gathered} {\varphi }_1(p,t)=\sqrt{\frac{\mathrm{\pi a}}{\sout{h}}}\frac{e^{-i{E}_{1}t/\sout{h}}}{{\mathrm{\pi }}^2-{\mathrm{a}}^2{\mathrm{p}}^2/\sout{{\mathrm{h}}^2}}(1+e^{-ipa/\sout{h}}) \\ =\sqrt{\frac{\mathrm{\pi a}}{\sout{h}}}\frac{e^{-i{E}_{1}t/\sout{h}}}{{\mathrm{\pi }}^2-{\mathrm{a}}^2{\mathrm{p}}^2/\sout{{\mathrm{h}}^2}}{e}^{-ipa/2\sout{h}}(e^{ipa/2\sout{h}}-e^{-ipa/2\sout{h}}) \\ =2\sqrt{\frac{\mathrm{\pi a}}{\sout{h}}}\frac{e^{-i{E}_{1}t/\sout{h}}}{{\mathrm{\pi }}^2-{\mathrm{a}}^2{\mathrm{p}}^2/\sout{{\mathrm{h}}^2}}{e}^{{}^{-ipa/2\sout{h}}}\cos \left(\frac{pa}{2\sout{h}}\right) \end{gathered}$$

Therefore, the square modulus can be simply obtained, a pure imaginary exponential function in the numerator is obtained:

$\overline{){\varphi }_2\left\{{\overline{)p,t}}^2=\frac{4\mathrm{\pi a}}{\sout{h}}\right.\frac{{\cos }^2(pa/2\sout{h}}{{({\mathrm{\pi }}^2-{\mathrm{a}}^2{\mathrm{p}}^2/\sout{{\mathrm{h}}^2})}^2}}$

Obtain the result for n = 2 and plot graph.

To obtain the result for n = 2, proceed in the similar manner as before, the only difference would now be that the factor ${(-1)}^n$will now yield a positive result, which in total will give a sin function. Since the steps are similar, they shall be skipped, so the final result is:

${\left|{\varphi }_2(p,t)\right|}^2=\frac{16\mathrm{\pi a}}{\sout{h}}\frac{{\sin }^2(pa/2\sout{h})}{{(4{\mathrm{\pi }}^2-{\mathrm{p}}^2{\mathrm{a}}^2/\sout{{\mathrm{h}}^2})}^2}$

And the graph for the latter function has been shown below:

Find the expectation value of the  p2operator

To find the expectation value of the $p^2$operator, first investigate how the momentum-space wave function depends on the parity of the number n. In the example for n = 1 and n = 2 the result obtained is sine and cosine functions. These numbers are by no means special; the same argument would hold for any other odd or even number n . Therefore, depending on the parity of the number n results can be obtained as sine or cosine function, as seen above. Next, it is important to notice that in momentum space, the expectation value of the $p^2$operator is simply:

$\left(p^2\right)={\int }_{-\infty }^{\infty }{p}^2{\left|\varphi \left(p\right)\right|}^{2}dp$

Which in our case is simply by the use of the above facts about the parity of nfor the equation (1):

Where the trig function is equal to cosine if n is odd and to sine if is even. The latter integral is not easy to evaluate, but it can be evaluated by a few different methods. One of the possibilities is to use complex analysis and evaluate the contour integral, second being partial fractions. Evaluate it for simplicity by the use of mathematics, since this is purely a mathematical step which brings no new important physics to the table. The integral, as some might've expected yields the same answer for either sine or cosine. The value of the integral is:

This is in agreement with the result of the problem 2.4 .