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202203151002 Exercise 3.3

Starting from

\langle x\rangle =\displaystyle{\int \psi^*(x)x\psi (x)\,\mathrm{d}x}

and using equation (3)

\displaystyle{\psi (x)=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\phi (p)e^{ipx/\hbar}\,\mathrm{d}p}

to express \psi (x) in terms of \phi (p), deduce equation (33):

\displaystyle{x=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}}.

Extracted from D. S. Saxon. (2012). Elementary Quantum Mechanics.

Some useful formulae. (Relationship between wave functions in configuration space \psi (x) and in momentum space \phi (p))

In differential forms,

\displaystyle{x\exp (ipx/\hbar )=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\exp (ipx/\hbar)}

\displaystyle{p\exp (-ipx/\hbar)=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}x}\exp (-ipx/\hbar )}

or, in integral forms,

\psi (x)=\displaystyle{\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\phi (p)\exp (ipx/\hbar )\,\mathrm{d}p}

\phi (p)=\displaystyle{\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\psi (x)\exp (ipx/\hbar )\,\mathrm{d}x}

Derivation.

\begin{aligned} \langle x\rangle & = \int \psi^*x\psi (x)\,\mathrm{d}x \\ & = \frac{1}{2\pi\hbar}\iiint\mathrm{d}x\,\mathrm{d}p\,\mathrm{d}p'\,\big( \phi^*(p')\exp (ip'x/\hbar )\big) x\big( \phi (p)\exp (ipx/\hbar )\big)\\ \dots &\enspace \textrm{by the fact }x\exp (ipx/\hbar )=-\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\exp (ipx/\hbar )\enspace \dots \\ \langle x\rangle & = \frac{1}{2\pi\hbar}\iiint\mathrm{d}x\,\mathrm{d}p\,\mathrm{d}p'\phi^*(p')\exp (ip'x/\hbar )\phi (p)\bigg( -\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\exp (ipx/\hbar ) \bigg) \\ \dots &\enspace \textrm{integrating wrt }x\textrm{ by parts }\enspace \dots \\ \langle x\rangle & = \frac{1}{2\pi\hbar}\iiint\mathrm{d}x\,\mathrm{d}p\,\mathrm{d}p'\,\phi^*(p')\exp [ix(p'-p)/\hbar ]\,\frac{\hbar}{i}\frac{\mathrm{d}\phi (p)}{\mathrm{d}p} \\ \dots &\enspace \textrm{doing }x\textrm{ integration }\enspace\dots \\ & = \iint \mathrm{d}p\,\mathrm{d}p'\,\phi^*(p')\frac{\hbar}{i}\frac{\mathrm{d}\phi (p)}{\mathrm{d}p}\delta (p'-p) \\ & = \int\mathrm{d}p\,\phi^*(x)\frac{\hbar}{i}\frac{\mathrm{d}}{\mathrm{d}p}\phi (p) \\ \end{aligned}

Afterword. (Using bra-ket/Dirac notation)

Reference: Question 86824 answered by joshphysics on Nov 17, 2013 (AT)physics.stackexchange(DOT)com

For any physically admissible state functions |\psi\rangle, which necessarily vanish at infinity, we see that

\begin{aligned} & \quad \langle p|[\hat{x},\hat{p}]|\psi\rangle \\ & = \langle p|\hat{x}\hat{p}-\hat{p}\hat{x}|\psi\rangle \\ & = \langle p|\hat{x}\hat{p}|\psi\rangle - \langle p|\hat{p}\hat{x}|\psi\rangle \\ & = \langle p|\hat{x}\hat{p}|\psi\rangle - p\langle p|\hat{x}|\psi\rangle \\ \end{aligned}

By the canonical commutation relation between \hat{x} and \hat{p}:

[\hat{x},\hat{p}]=i\hbar\hat{I}

where \hat{I} is the identity operator, we have

\begin{aligned} \langle p|i\hbar I|\psi\rangle & = \langle p|\hat{x}\hat{p}|\psi\rangle - p\langle p|\hat{x}|\psi\rangle \\ p\langle p|\hat{x}|\psi\rangle & = \langle p|\hat{x}\hat{p}|\psi\rangle - i\hbar\langle p|\psi\rangle \\ \end{aligned}

Manipulating the first term on RHS:

\begin{aligned} &\quad \langle p|\hat{x}\hat{p}|\psi \rangle \\ & = \int\mathrm{d}x\,\langle p|\hat{x}|x\rangle \langle x|\hat{p}|\psi\rangle \\ & = \int \mathrm{d}x\, x\exp (ipx/\hbar )\bigg( -i\hbar\frac{\mathrm{d}}{\mathrm{d}x}\psi (x)\bigg) \\ & = i\hbar\int\mathrm{d}x\,\frac{\mathrm{d}}{\mathrm{d}x}\big( (x\exp (ipx/\hbar ))\psi (x) \big) \\ & = i\hbar\int\mathrm{d}x\exp (ipx/\hbar )\psi (x)+i\hbar\int\mathrm{d}x\, x\bigg(\frac{ip}{\hbar}\bigg) \exp (ipx/\hbar )\psi (x) \\ & = i\hbar\psi (p)-p\int\mathrm{d}x\, x\exp (ipx/\hbar )\psi (x) \\ & = i\hbar\psi (p)+i\hbar p\frac{\mathrm{d}}{\mathrm{d}p}\int\mathrm{d}x\, \exp (ipx/\hbar )\psi (x)\\ & = i\hbar\psi (p)+i\hbar p\frac{\mathrm{d}}{\mathrm{d}p}\psi (p) \end{aligned}

thus the result

\displaystyle{p\langle p|\hat{x}|\psi\rangle = i\hbar p\frac{\mathrm{d}}{\mathrm{d}p}\psi (p)}.