Chapter 6, Problem 6.31P

To determine

Prove the operator is a function of given function.

Answer to Problem 6.31P

The operator is a function of given function.

Explanation of Solution

Write the expression for the momentum space wave function.

$\begin{array}{c}\varphi \left(p\right)=\frac{1}{\sqrt{2\pi \hslash }}{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-ipx/\hslash }\psi \left(x\right)dx}\\ =\frac{\lambda }{\sqrt{2\pi \hslash }}{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-\lambda \left|x\right|-ipx/\hslash }dx}\\ =\sqrt{\frac{2\lambda }{\pi \hslash }}\left(\frac{\lambda }{{\left(p/\hslash \right)}^2+{\lambda }^2}\right)\end{array}$        (I)

Here, $p$ is the momentum, $\varphi \left(p\right)$ is the momentum space wave function, $\lambda$ is the wavelength, $\hslash$ is Planck’s constant, $\psi \left(x\right)$ is wave function, $x$ is position.

Write the expression for the function.

$f_p\left(x\right)=\frac{e^{-ipx/\hslash }}{\sqrt{2\pi \hslash }}$        (II)

Here, $f\left(x\right)$ is the function.

Write the expression for the given operator.

$\widehat{T}\left(a\right)\psi \left(x\right)={\displaystyle {\int }_{-\infty }^{\infty }{e}^{-iap/\hslash }{f}_p\left(x\right)\varphi \left(p\right)dp}$        (III)

Here, $\widehat{T}\left(a\right)$ is the operator,

Conclusion:

Substitute, $\left(\frac{e^{-ipx/\hslash }}{\sqrt{2\pi \hslash }}\right)$ for $f_p\left(x\right)$, $\sqrt{\frac{2\lambda }{\pi \hslash }}\left(\frac{\lambda }{{\left(p/\hslash \right)}^2+{\lambda }^2}\right)$ for $\varphi \left(p\right)$ in equation (III) to find $\widehat{T}\left(a\right)$.

$\widehat{T}\left(a\right)\psi \left(x\right)={\displaystyle {\int }_{-\infty }^{\infty }{e}^{-iap/\hslash }\left(\frac{e^{-ipx/\hslash }}{\sqrt{2\pi \hslash }}\right)\left[\sqrt{\frac{2\lambda }{\pi \hslash }}\left(\frac{\lambda }{{\left(p/\hslash \right)}^2+{\lambda }^2}\right)\right]dp}$        (V)

Considering the expression of the momentum.

$p=q\lambda \hslash$        (VI)

Here, $q$ is the charge.

Rewrite the expression for the operator from equation (V) by using (VI).

$\begin{array}{c}\widehat{T}\left(a\right)\psi \left(x\right)=\frac{\sqrt{\lambda }}{\pi }{\displaystyle {\int }_{-\infty }^{\infty }{e}^{-iq\lambda \left(x-a\right)/q^2+1}dq}\\ =\frac{2\sqrt{\lambda }}{\pi }{\displaystyle {\int }_0^{\infty }\frac{\cos \left(q\lambda \left|x-a\right|\right)}{q^2+1}dq}\\ =\sqrt{\lambda }\left(e^{-\lambda \left|x-a\right|}\right)\end{array}$

Thus, the operator is a function of given function.