1
$\begingroup$

For some state $|\psi\rangle$ it is possible to recover the uncertainty principle using the fact that $$\left|(\hat{\sigma_{Q}}-i\lambda\hat{\sigma_{P}})|\psi\rangle\right|^{2}\geq0,$$where$$\hat{\sigma}_{Q}=\hat{Q}-\langle\hat{Q}\rangle,\hat{\sigma}_{P}=\hat{P}-\langle\hat{P}\rangle$$ and $\hat{Q}$ and $\hat{P}$ represent canonically conjugate operators, $[\hat{Q},\hat{P}]=i\hbar$. The minimum uncertainty state is the one which saturates the inequality, such that $$\left|(\hat{\sigma_{Q}}-i\lambda\hat{\sigma_{P}})|\psi\rangle\right|^{2}=0,$$ This means that the miminimum uncertainty state can be obtained by by solving the the differential equation resulting from the above expression in $\hat{Q}$-space: $$\left(q-\langle\hat{Q}\rangle+\lambda\hbar\frac{d}{dq}-i\lambda\langle\hat{P}\rangle\right)\psi(q)=0 $$ This differential equation has the general, normalised solution: \begin{equation} \psi(q)=\left(\frac{1}{2\pi(\Delta\hat{Q})^{2}}\right)^{\frac{1}{4}}\exp\left(-\frac{(q-q_{0})^{2}}{4(\Delta\hat{Q})^{2}}+\frac{i}{\hbar}qp_{0}\right) \end{equation} wherer $(q_{0},p_{0})=(\langle\hat{Q}\rangle,\langle\hat{P}\rangle)$. Integration to obtain the Wigner function using the formula, we obtain $$W_{\psi}(q,p)=\frac{1}{2\pi\Delta\hat{Q}\Delta\hat{P}}\exp\left(-\frac{(q-q_{0})^{2}}{2(\Delta\hat{Q})^{2}}\right)\exp\left(-\frac{(p-p_{0})^{2}}{2(\Delta\hat{P})^{2}}\right)$$ This represents the Wigner function of a squeezed coherent state, $$|\alpha,z\rangle=\hat{D}(\alpha)\hat{S}(z)|0\rangle$$ where $\hat{D}(\alpha)$ and $\hat{S}(z)$ represent the displacement and squeezing operators respectively. The displacement operator displaces the center of the gaussian in phase space to $(q_{0},p_{0})$and the squeeze operator has changed the value of $\Delta Q$and $\Delta P$ ( such that it takes a squeezed shape in dimensionless phase space). Crucially however, this is only the case if $z$ is real. I have seen that, generally, the argument of the squeeze operator may be complex, and in these scenarios,the Wigner function will squeeze relative to an axis in phase space which depends on the argument of the complex number $z$. However I don't fully understand why this happens, and my aim was to derive the effect by obtaining the Wigner function from the minimum uncertainty state wavefunction. I have concluded that the minimum uncertainty state obtained using the method above is not in fact the most general minimum uncertainty wavefunction, since there are the minimum uncertainty states produced by the squeeze operator with a complex arguments. So, my question is, is it possible to derive the most general wavefunction for a minimum uncertainty state? If so, how?

$\endgroup$
4
  • $\begingroup$ your "problem" is that you're implicitly working in the position representation since you have $\hat q\mapsto q$ and $\hat p\mapsto -i \hbar \frac{d}{dp}$. "Rotating" in $qp$ space amounts to working with "quasi-quadratures" of the form $q\pm i p$ (up to scale factors). $\endgroup$ Commented Oct 25, 2022 at 20:13
  • $\begingroup$ Thanks for your response. Do you know any good treatments of this rotation effect? stuggling to find much online $\endgroup$ Commented Oct 25, 2022 at 20:22
  • 1
    $\begingroup$ Presumably you are familiar with the FFT of Condon? $\endgroup$ Commented Oct 26, 2022 at 14:03
  • $\begingroup$ I'll have a read, thanks $\endgroup$ Commented Oct 26, 2022 at 20:49

1 Answer 1

2
$\begingroup$

You don't get anything "new" by rotating in the $xp$ plane. Simply define \begin{align} X&=x\cos\theta -p \sin\theta \, ,\\ P&= x\sin\theta +x \sin\theta \, . \end{align} Since $[X,P]=[x,p]$, the action of $\hat X$ in the X-basis is multiplication by $X$: $\hat X\psi(X)=x\Psi(x)$ whereas $P\mapsto -i\hbar \frac{d}{dX}$. The original eigenvalue problem in terms of $x,p$ is now simply one in terms of $X,P$.

$\endgroup$
3
  • $\begingroup$ I see, if there a name for this kind of transformation of canonically conjugate basis? I’d like to do some more reading $\endgroup$ Commented Oct 26, 2022 at 9:17
  • $\begingroup$ I don't think there's any special. The result follows from the Stone- vonNeumann theorem. You're just doing a canonical transformation in phase space, and because it's linear there's no ordering issue. $\endgroup$ Commented Oct 26, 2022 at 15:35
  • $\begingroup$ ah i see, thanks. $\endgroup$ Commented Oct 26, 2022 at 20:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.