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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?

Cosmas Zachos
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Himanshu
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1 Answers1

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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.)

Cosmas Zachos
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