# Uncertainty in eigenstate of $Q=\mu a+\nu a^\dagger$

Consider a particle in simple harmonic oscillator potential. Suppose the eigenstate of the operator

$$Q=\mu a+\nu a^\dagger$$ with eigenvalue $$\alpha$$ where $$\mu,\nu$$ and $$\alpha$$ are three complex numbers satisfying $$|\mu|^2-|\nu|^2=1$$. I'm trying to find uncertainty in this state $$|\alpha\rangle$$.

To find $$|\alpha\rangle$$ $$Q|\alpha\rangle =\alpha |\alpha\rangle$$ writing $$|\alpha\rangle =\sum_n c_n|n\rangle$$ and putting in above. I get $$\mu c_1=\alpha c_0,\ \ \ \&\ \ \ \ \mu c_{n+1} \sqrt{n+1}+\nu c_{n-1}\sqrt{n}=\alpha c_n,\ \ \ n=1,2,3,\ldots$$ I don't understand what to do now. Since it doesn't give an explicit solution for $$c_n$$. Please help me with this.

Edit:

\begin{align*} X&= \sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger) \\ P&= -i\frac{\sqrt{m\omega \hbar}}{2}(a-a^\dagger) \end{align*} Furthermore, $$Q=\mu a+\nu a^\dagger,\ \ \ \ Q^\dagger =\mu^*a^\dagger +\nu^* a$$ $$\rightsquigarrow a=\mu^* Q-\nu Q^\dagger,\ \ \ \ a^\dagger=\mu Q^\dagger -\nu^*Q$$ We can solve these for $$X$$ and $$P$$ in term of $$Q$$ and $$Q^\dagger$$. We have \begin{align*} X&=\sqrt{\frac{\hbar}{2m\omega}}[(\mu^*-\nu^*)Q+(\mu-\nu)Q^\dagger] \\ P&= -i\frac{\sqrt{m\omega \hbar}}{2}[(\mu^*+\nu*)Q-(\mu+\nu)Q^\dagger] \end{align*}

We would like to calculate $$\langle X\rangle$$ in state $$|\alpha\rangle$$. \begin{align*} \langle X\rangle &= \langle \alpha |X|\alpha\rangle \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left[(\mu^*-\nu^*)\langle \alpha|Q|\alpha\rangle +(\mu-\nu)\langle \alpha| Q^\dagger|\alpha\rangle \right] \\ &= \sqrt{\frac{\hbar}{2m\omega}}\left[\alpha(\mu^*-\nu^*)+\alpha^*(\mu-\nu)\right] \end{align*}

The operator $$X^2$$ given by

$$X^2=\frac{\hbar}{2m\omega}\left[(\mu^*-\nu^*)^2Q^2+(\mu-\nu)^2(Q^\dagger)^2+(\mu^*-\nu^*)(\mu-\nu)(QQ^\dagger+Q^\dagger Q)\right]$$

Now we can find the expectation value of the same in state $$|\alpha\rangle$$. We can explicitly see $$\langle \alpha |Q^\dagger Q|\alpha\rangle =\alpha^*\alpha\langle \alpha|\alpha\rangle =|\alpha|^2$$ Similarly, $$\langle \alpha |Q Q^\dagger|\alpha\rangle =\langle \alpha|(Q^\dagger Q)^\dagger|\alpha\rangle =\langle \alpha|Q^\dagger Q|\alpha\rangle^*= |\alpha|^2$$ Therefore, We have $$\langle \alpha |X^2|\alpha \rangle =\frac{\hbar}{2m\omega}\left[(\mu^*-\nu^*)^2(\alpha^2)+(\mu-\nu)^2(\alpha^*)^2+2|\alpha|^2(\mu^*-\nu^*)(\mu-\nu)\right]=\frac{\hbar}{2m\omega}\left[(\mu^*-\nu^*)\alpha+(\mu-\nu)\alpha^*\right]^2=\langle X\rangle ^2$$
But this means, $$\Delta X=0$$, How is it so?

• Oct 15 '21 at 17:28

You are working in the oscillator algebra $$[a,a^\dagger]=1$$ in a different basis, since $$[Q,Q^\dagger]=1 .$$ So you are inspecting its action on its normalized coherent states. $$\langle \alpha| Q|\alpha\rangle =\alpha~~~\leadsto \\ \langle \alpha| Q^2|\alpha\rangle -\langle \alpha| Q|\alpha\rangle ^2 = \alpha^2 -\alpha^2 =0.$$

The Fock space you should be working in has a vacuum annihilated by Q, not a, $$Q|0\rangle=0.$$

Edit in response to question edit

After the retargeting of your question, $$\langle QQ^\dagger+ Q^\dagger Q\rangle = \langle 2Q^\dagger Q + 1\rangle=2|\alpha|^2+1,$$ so, as you computed, $$\langle X^2\rangle = \langle X\rangle ^2 + \left ({\hbar\over 2m\omega} \right ) |\mu-\nu|^2.$$ As you might expect, for μ=ν, Χ collapses to 0, so small wonder its variance vanishes there.

(Irrelevant to this, but this might be as good a place as any to be exposed to Dirac's trick.)

• I added a little bit of calculation, and I'm getting $\Delta X$ to be zero in $|\alpha\rangle$ state, But the uncertainty shouldn't be zero, Where is the flaw? Oct 16 '21 at 5:07
• Commute Q and it’s hc. Your “similarly” expression is wrong. Oct 16 '21 at 5:32
• Got that, I can calculate that from commutation relation. Oct 16 '21 at 5:49
• Got it, thanks for the explanation Oct 20 '21 at 9:27