Quantum Mechanics Term 1 Problems Class 4

Consider the 1-D Simple Harmonic Oscillator with frequency $w$, If $\widehat{a}$ is the annihilation operator
a) Construct the eigenstates $|z⟩$
b) Normalise it
c) Find its time evolution
d) Describe the time evolution of the corresponding wavefunction in position space.

Hint: Baker-Campell-Haussdorff formula

$$e^{\widehat{X}}{e}^{\widehat{Y}}=e^{\widehat{Y}+[\widehat{X},\widehat{Y}]+\frac{1}{2}[\widehat{X},[\widehat{X},\widehat{Y}]]+\frac{1}{3!}[\widehat{X},[\widehat{X},[\widehat{X},\widehat{Y}]]]+\mathrm{\dots }}{e}^{\widehat{X}}$$ (1)

Other facts of the 1-D SHO

$\widehat{H}$ $=\mathrm{\hslash }w\left({\widehat{a}}^{+}\widehat{a}+{\displaystyle \frac{1}{2}}\right)$

$|n⟩$ $={\displaystyle \frac{{({\widehat{a}}^{+})}^n}{\sqrt{n!}}}|0⟩$

$\widehat{a}$ $={\displaystyle \frac{1}{\sqrt{2mw\mathrm{\hslash }}}}(mw\widehat{x}+i\widehat{p})$

a) We need to solve $\widehat{a}|z⟩=z|z⟩$. An important thing to note is that since $\widehat{a}$ is not hermitian, its eigenvalues does not have to be real. We also choose to write $|z⟩$ in terms of the eigenbasis of the SHO.

$|z⟩$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{c}_n(z)|n⟩$

$z|z⟩$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}z{c}_n(z)|n⟩$

$\widehat{a}|z⟩$ $={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{c}_n(z)\widehat{a}|n⟩$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{c_n(z)}{\sqrt{n!}}}\widehat{a}{({\widehat{a}}^{+})}^n|0⟩$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{c_n(z)}{\sqrt{n!}}}({({\widehat{a}}^{+})}^n\widehat{a}+[\widehat{a},{({\widehat{a}}^{+})}^n])|0⟩$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{c_n(z)}{\sqrt{n!}}}n{({\widehat{a}}^{+})}^{n-1}|0⟩$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\sqrt{n}{c}_n(z)|n-1⟩$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\sqrt{n+1}{c}_{n+1}(z)|n⟩$

$\Rightarrow z{c}_n(z)$ $=\sqrt{n+1}{c}_{n+1}(z)$

$\Rightarrow c_n(z)$ $=c{\displaystyle \frac{z^n}{\sqrt{n!}}}$

$\Rightarrow |z⟩$ $=c{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{z^n}{\sqrt{n!}}}{\displaystyle \frac{{({\widehat{a}}^{+})}^n}{\sqrt{n!}}}|0⟩$

$=c{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{z^n}{n!}}{({\widehat{a}}^{+})}^n|0⟩$

$=c{e}^{z{\widehat{a}}^{+}}|0⟩$

b) We need $⟨z|z⟩=1$.

$⟨z|$ $=c^{*}{e}^{z^{*}\widehat{a}}⟨0|$

$⟨z|z⟩$ $={|c|}^2⟨0|e^{z^{*}\widehat{a}}{e}^{z{\widehat{a}}^{+}}|0⟩$

We also note the following fact

$e^{z^{*}\widehat{a}}|0⟩$ $=(\widehat{I}+{\widehat{z}}^{*}\widehat{a}+\mathrm{\dots })|0⟩=|0⟩$

$\Rightarrow ⟨0|e^{z{\widehat{a}}^{+}}$ $=⟨0|$

So we use the hint with $\widehat{X}=z^{*}\widehat{a},\widehat{Y}=z{\widehat{a}}^{+}$.

$[\widehat{X},\widehat{Y}]$ $={|z|}^2[\widehat{a},{\widehat{a}}^{+}]={|z|}^2$

$\Rightarrow [\widehat{X},[\widehat{X},\widehat{Y}]]$ $=[z^{*}\widehat{a},{|z|}^2]=0$

$e^{z^{*}\widehat{a}}{e}^{z{\widehat{a}}^{+}}$ $=e^{z{\widehat{a}}^{+}+{|z|}^2}{e}^{z^{*}\widehat{a}}$

$=e^{{|z|}^2}{e}^{z{\widehat{a}}^{+}}{e}^{z^{*}\widehat{a}}$

$\Rightarrow ⟨z|z⟩$ $={|c|}^2{e}^{{|z|}^2}⟨0|e^{z{\widehat{a}}^{+}}{e}^{z^{*}\widehat{a}}|0⟩$

$={|c|}^2{e}^{{|z|}^2}$

$\Rightarrow c$ $=e^{-\frac{{|z|}^2}{2}}$

c) If $|\psi (t=0)⟩=|z⟩$,

$|\psi (t)⟩$ $=e^{-\frac{it}{\mathrm{\hslash }}\widehat{H}}|z⟩$

$=c(z)e^{-\frac{it}{\mathrm{\hslash }}\widehat{H}}{e}^{z{\widehat{a}}^{+}}|0⟩$

We know again from a 1-D SHO,

$\widehat{H}|0⟩$ $={\displaystyle \frac{\mathrm{\hslash }w}{2}}|0⟩$

$\Rightarrow e^{-\frac{it}{\mathrm{\hslash }}\widehat{H}}|0⟩$ $=e^{-\frac{iwt}{2}}|0⟩$

So we again use the hint to commute the exponentials. This time $\widehat{X}=-\frac{it}{\mathrm{\hslash }}\widehat{H},\widehat{Y}=z{\widehat{a}}^{+}$.

$[\widehat{X},\widehat{Y}]$ $=-{\displaystyle \frac{it}{\mathrm{\hslash }}}z\mathrm{\hslash }w[{\widehat{a}}^{+}\widehat{a},{\widehat{a}}^{+}]$

$=-itwz({\widehat{a}}^{+}[\widehat{a},{\widehat{a}}^{+}]+[{\widehat{a}}^{+},{\widehat{a}}^{+}]\widehat{a})$

$=-itwz{\widehat{a}}^{+}$

$]$ $=-itw.-itwz[{\widehat{a}}^{+}\widehat{a},{\widehat{a}}^{+}]$

$={(-itw)}^{2}z{\widehat{a}}^{+}$

We make a hypothesis that $[\widehat{X},[\widehat{X},\mathrm{\dots },[\widehat{X},\widehat{Y}]]\mathrm{\dots }]={(-itw)}^{n}z{\widehat{a}}^{+}$. We prove this using induction.

$[\widehat{X},[\widehat{X},\mathrm{\dots },[\widehat{X},\widehat{Y}]]\mathrm{\dots }]$ $=-{\displaystyle \frac{it}{\mathrm{\hslash }}}\mathrm{\hslash }w{(-itw)}^{n}z[{\widehat{a}}^{+}\widehat{a},{\widehat{a}}^{+}]$

$={(-itw)}^{n+1}z{\widehat{a}}^{+}$

$\Rightarrow \widehat{Y}+[\widehat{X},\widehat{Y}]+{\displaystyle \frac{1}{2!}}[\widehat{X},[\widehat{X},\widehat{Y}]]+\mathrm{\dots }$ $=z{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle \frac{{(-itw)}^n}{n!}}{\widehat{a}}^{+}$

$=z{e}^{-itw}{\widehat{a}}^{+}$

$|\psi (t)⟩$ $=c(z)e^{z{e}^{-itw}{\widehat{a}}^{+}}{e}^{-\frac{it}{\mathrm{\hslash }}\widehat{H}}|0⟩$

$=c(z)e^{-\frac{itw}{2}}{e}^{z{e}^{-itw}{\widehat{a}}^{+}}|0⟩$

$=e^{-\frac{itw}{2}}|z{e}^{-itw}⟩$

Just as a reminder we are now working in the eigenbasis of $\widehat{z}$ as $z{e}^{-itw}$ is no longer an integer and hence not in the eigenbasis of $\widehat{k}(\widehat{H}=\mathrm{\hslash }w\widehat{k})$.

d) In equation form, the question is really asking for ${\psi }_z(x)=⟨x|z⟩$.

$\widehat{a}|z⟩$ $=z|z⟩$

$\Rightarrow ⟨x|\widehat{a}|z⟩$ $=z⟨x|z⟩$

$\iff {\displaystyle \frac{1}{\sqrt{2m\mathrm{\hslash }w}}}⟨x|mw\widehat{x}+i\widehat{p}|z⟩$ $=z{\psi }_z(x)$

$\iff {\displaystyle \frac{1}{\sqrt{2m\mathrm{\hslash }w}}}(mwx{\psi }_z(x)+\mathrm{\hslash }{\psi }_z^{\prime }(x))$ $=z{\psi }_z(x)$

$\Rightarrow \sqrt{{\displaystyle \frac{\mathrm{\hslash }}{2mw}}}{\psi }_z^{\prime }(x)$ $=\left(z-\sqrt{{\displaystyle \frac{mw}{2\mathrm{\hslash }}}}x\right){\psi }_z(x)$

$\Rightarrow \sqrt{{\displaystyle \frac{\mathrm{\hslash }}{2mw}}}{\displaystyle \frac{{\psi }_z^{\prime }(x)}{{\psi }_z(x)}}$ $=z-\sqrt{{\displaystyle \frac{mw}{2\mathrm{\hslash }}}}x$

$\iff {\displaystyle \frac{d}{dx}}\ln ({\psi }_z(x))$ $=\sqrt{{\displaystyle \frac{2mw}{\mathrm{\hslash }}}}{\displaystyle \frac{d}{dx}}\left(zx-{\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{mw}{2\mathrm{\hslash }}}}{x}^2\right)$

$\Rightarrow \ln ({\psi }_z(x))$ $=\sqrt{{\displaystyle \frac{2mw}{\mathrm{\hslash }}}}\left(zx-{\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{mw}{2\mathrm{\hslash }}}}{x}^2\right)+\lambda$

$\Rightarrow {\psi }_z(x)$ $=d{e}^{\sqrt{\frac{2mw}{\mathrm{\hslash }}}\left(zx-\frac{1}{2}\sqrt{\frac{mw}{2\mathrm{\hslash }}}{x}^2\right)}$

$=d\exp \left(\sqrt{{\displaystyle \frac{2mw}{\mathrm{\hslash }}}}zx\right)\exp \left(-{\displaystyle \frac{mw}{2\mathrm{\hslash }}}{x}^2\right)$

$=d\exp \left(-{\displaystyle \frac{mw}{2\mathrm{\hslash }}}\left(x^2-\sqrt{{\displaystyle \frac{8\mathrm{\hslash }}{mw}}}zx\right)\right)$

$=d\exp \left(-{\displaystyle \frac{mw}{2\mathrm{\hslash }}}\left({\left(x-\sqrt{{\displaystyle \frac{2\mathrm{\hslash }}{mw}}}z\right)}^2-{\displaystyle \frac{2\mathrm{\hslash }}{mw}}{z}^2\right)\right)$

$=d\exp (z^2)\exp \left(-{\displaystyle \frac{mw}{2\mathrm{\hslash }}}{\left(x-\sqrt{{\displaystyle \frac{2\mathrm{\hslash }}{mw}}}z\right)}^2\right)$

${|{\psi }_z(x)|}^2$ $={|d|}^2\exp \left(\sqrt{{\displaystyle \frac{2mw}{\mathrm{\hslash }}}}(z+z^{*})x\right)\exp \left(-{\displaystyle \frac{mw}{\mathrm{\hslash }}}{x}^2\right)$

$={|d|}^2\exp \left(2\sqrt{{\displaystyle \frac{2mw}{\mathrm{\hslash }}}}\mathrm{\Re }(z)x\right)\exp \left(-{\displaystyle \frac{mw}{\mathrm{\hslash }}}{x}^2\right)$

$={|d|}^2\exp \left(-{\displaystyle \frac{mw}{\mathrm{\hslash }}}\left(-2\sqrt{{\displaystyle \frac{2\mathrm{\hslash }}{mw}}}\mathrm{\Re }(z)x+x^2\right)\right)$

$={|d|}^2\exp \left(-{\displaystyle \frac{mw}{\mathrm{\hslash }}}\left({\left(x-\sqrt{{\displaystyle \frac{2\mathrm{\hslash }}{mw}}}\mathrm{\Re }(z)\right)}^2-{\displaystyle \frac{2\mathrm{\hslash }}{mw}}{(\mathrm{\Re }(z))}^2\right)\right)$

$={|d|}^2\exp \left(2{(\mathrm{\Re }(z))}^2\right)\exp \left(-{\displaystyle \frac{mw}{\mathrm{\hslash }}}{\left(x-\sqrt{{\displaystyle \frac{2\mathrm{\hslash }}{mw}}}\mathrm{\Re }(z)\right)}^2\right)$

However, we must remember that our $|z⟩$ is really $e^{-\frac{itw}{2}}|e^{-itw}{z}_0⟩$. Hence,

$\mathrm{\Re }(z)=\mathrm{\Re }(z_0)\cos (wt)+\mathrm{\Im }(z_0)\sin (wt)$

This shows that the time evolution corresponding to the wavefunction is a Gaussian which does not change form but oscillates with frequency $w$.