Show that the scalar (or inner) product of position \(x\) and momentum \(p\) can be written as an exponential in terms of \(x\) and \(p\)

Method

 * 1) Show by induction that \([\hat{x},\hat{p}^n]= ni\hbar\hat{p}^{n-1}\)
 * 2) Show that \([\hat{x},\hat{p}^2]= 2i\hbar\hat{p}\)
 * 3) Use the commutation relation \([A,BC] = [A,B]C + B[A,C]\) to expand out the commutator: \([\hat{x},\hat{p}^2]=[\hat{x},\hat{p}]\hat{p} + \hat{p}[\hat{x},\hat{p}]\)
 * 4) Substitute out the standard commutator \([\hat{x},\hat{p}]=i\hbar\): \([\hat{x},\hat{p}^2]=2i\hbar\hat{p}\)
 * 5) Expand \([\hat{x},\hat{p}^n]\): \([\hat{x},\hat{p}^n]=[\hat{x},\hat{p}]\hat{p}^{n-1} + \hat{p}[\hat{x},\hat{p}^{n-1}]\)
 * 6) Assume that \([\hat{x},\hat{p}^{n-1}]= (n-1)i\hbar\hat{p}^{n-2}\) and substitute this into the equation: \([\hat{x},\hat{p}^n]=[\hat{x},\hat{p}]\hat{p}^{n-1} + (n-1)i\hbar\hat{p}^{n-1}\)
 * 7) Substitute out the standard commutator: \([\hat{x},\hat{p}^n]=i\hbar\hat{p}^{n-1} + (n-1)i\hbar\hat{p}^{n-1}\)
 * 8) Simplify: \([\hat{x},\hat{p}^n]=n i\hbar\hat{p}^{n-1}\)
 * 9) Show that \(e^{\frac{-i a\hat{p}}{\hbar}} |x\rangle = |x+a\rangle\)
 * 10) Show that \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=ae^{\frac{-i a \hat{p}}{\hbar}}\)
 * 11) Find the Taylor expansion of \(e^{\frac{-i a\hat{p}}{\hbar}}\): \(\sum_{m=0}^{\infty}\frac{1}{m!} \left (\frac{-i a \hat{p}}{\hbar} \right )^m\)
 * 12) Substitute the Taylor expansion into the commutator: \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=[\hat{x},\sum_{m=0}^{\infty}\frac{1}{m!} \left (\frac{-i a \hat{p}}{\hbar} \right )^m]\)
 * 13) Simplify, removing the zeroth term in the expansion: \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=\sum_{m=1}^{\infty}\frac{1}{m!} \left (\frac{-i a}{\hbar} \right )^m [\hat{x},\hat{p}^m]\)
 * 14) Substitute in the previously proven identity \([\hat{x},\hat{p}^n]= ni\hbar\hat{p}^{n-1}\): \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=\sum_{m=1}^{\infty}\frac{1}{m!} \left (\frac{-i a}{\hbar} \right )^m mi\hbar\hat{p}^{m-1}\)
 * 15) Simplify: \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=a\sum_{m=1}^{\infty}\frac{1}{(m-1)!} \left (\frac{-i a}{\hbar} \right )^{m-1} \hat{p}^{m-1}\)
 * 16) Adjust the limits of the sum so it starts at 0: \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=a\sum_{m=0}^{\infty}\frac{1}{(m)!} \left (\frac{-i a}{\hbar} \right )^{m} \hat{p}^{m}\)
 * 17) Substitute the Taylor expansion out again: \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=ae^{\frac{-i a \hat{p}}{\hbar}}\)
 * 18) Show that \(\hat{x}e^{\frac{-i a\hat{p}}{\hbar}}|x\rangle = (x + a)e^{\frac{-i a\hat{p}}{\hbar}}|x\rangle\)
 * 19) Multiply out the commutator in \([\hat{x},e^{\frac{-i a\hat{p}}{\hbar}}]=ae^{\frac{-i a \hat{p}}{\hbar}}\): \(\hat{x}e^{\frac{-i a\hat{p}}{\hbar}}-e^{\frac{-i a\hat{p}}{\hbar}}\hat{x}=ae^{\frac{-i a \hat{p}}{\hbar}}\)
 * 20) Multiply through by \(|x\rangle\): \(\hat{x}e^{\frac{-i a\hat{p}}{\hbar}}|x\rangle-e^{\frac{-i a\hat{p}}{\hbar}}\hat{x}|x\rangle=ae^{\frac{-i a \hat{p}}{\hbar}}|x\rangle\)
 * 21) Substitute the momentum operator out of the second term: \(\hat{x}e^{\frac{-i a\hat{p}}{\hbar}}|x\rangle-e^{\frac{-i a\hat{p}}{\hbar}}x|x\rangle=ae^{\frac{-i a \hat{p}}{\hbar}}|x\rangle\)
 * 22) Add \(e^{\frac{-i a\hat{p}}{\hbar}}x|x\rangle\) to each side and simplify: \(\hat{x}e^{\frac{-i a\hat{p}}{\hbar}}|x\rangle=(a+x)e^{\frac{-i a \hat{p}}{\hbar}}|x\rangle\)
 * 23) Use the identity \(\hat{x}|x + a\rangle = (x + a)|x + a\rangle\) to show that \(e^{\frac{-i a\hat{p}}{\hbar}} |x\rangle = |x+a\rangle\)
 * 24) Solve \(\langle x+a|p \rangle\).
 * 25) Substitute \(e^{\frac{-i a\hat{p}}{\hbar}} |x\rangle = |x+a\rangle\) into the expression: \(\langle x+a|p \rangle=\langle x|e^{\frac{i a\hat{p}}{\hbar}}|p\rangle\)
 * 26) Use the identity for the Fourier transforms of operators, \(f(\hat{O})|O\rangle = f(O)|O\rangle\) to remove the momentum operator from the expression: \(\langle x+a|p \rangle=e^{\frac{i ap}{\hbar}}\langle x|p\rangle\)
 * 27) Set \(x\) to 0: \(\langle a|p \rangle=e^{\frac{i ap}{\hbar}}\langle 0|p\rangle\)
 * 28) Normalise, setting our normalisation \(N\) to \(\langle 0|p\rangle\).
 * 29) Write the delta function for \(p\) in Dirac notation: \(\delta(p-p')=\langle p'|p \rangle\)
 * 30) Put this in integral form as described here: \(\delta(p-p')=\int da \langle p'|a \rangle \langle a|p \rangle\)
 * 31) Substitute in our unnormalised expression: \(\delta(p-p')=\int da\,Ne^{\frac{-i ap'}{\hbar}} N e^{\frac{i ap}{\hbar}}\)
 * 32) Simplify: \(\delta(p-p')= N^2\int da\,e^{\frac{ia(p - p')}{\hbar}}\)
 * 33) Solve the integral, using the identity \(\int da \, e^{ia(p - p')}=2 \pi \delta(p - p')\): \(\delta(p-p')=2\pi\hbar N^2 \delta(p - p')\)
 * 34) Divide through by \(\delta(p-p')\): \(1=2\pi\hbar N^2\)
 * 35) Solve for \(N\): \(N=\sqrt{\frac{1}{2 \pi \hbar}}\)
 * 36) Substitute \(N\) into our expression for \(\langle a|p \rangle\), and substitute \(x\) for \(a\): \(\langle x|p \rangle=\frac{e^{i xp/\hbar}}{\sqrt{2 \pi \hbar}}\)