# Harmonic oscillator coherent state expectation values

I'm looking to calculate the expected values of a coherent state (of a harmonic oscillator) evolving in time. I know that the $x$ and $p$ expectation values are as in classical motion, but I'm wondering about $x^2$ and $p^2$.

Let's say I'm starting with the coherent state $| b \rangle$, with $b \in \mathbb{R}$, so the wavefunction is the ground state displaced by $bx_0\sqrt{2}$:

$$\psi_b (x) = \psi_0(x-bx_0\sqrt{2})$$

Or similarly the Wigner function will be

$$W_b(x,p) = W_0(x-bx_0\sqrt{2},p)$$

Now I know the expected values of $x$ and $p$ are classical:

$$\langle x(t) \rangle = bx_0\sqrt{2}\cos(-\omega t)$$ $$\langle p(t) \rangle = bp_0\sqrt{2}\sin(-\omega t)$$

But what about $\langle x^2(t) \rangle$ and $\langle p^2(t) \rangle$ and ?

• The energy is conserved... Jan 11, 2015 at 21:26
• Of course, I must've been tired writing this. Still, what about the position and momentum squared. Jan 11, 2015 at 21:34
• Use the expansions into ladder operators (e.g. $x\sim (a+a^{\dagger})$), and then the fact that coherent states are right eigenstates of the annihilation operator, and left eigenstates of the creation operator. Jan 12, 2015 at 0:49
• What are left and right eigenstates? Jan 12, 2015 at 1:05
• See mathworld for info on eigenvectors. If you look up the Wiki page on coherent states you will find info about their relationship with the ladder operators. Jan 12, 2015 at 1:19

Let $\alpha \in {\Bbb C}$, and let $\vert{n}\rangle$ be the harmonic oscillator state with energy $(n+\textstyle\frac{1}{2})\hbar\omega$. At $t=0$, the coherent state $\vert {\alpha(0)}\rangle$ is defined by $$\vert{\alpha(0)}\rangle= e^{-\vert \alpha \vert^2/2}\,\left( \sum_{n=0}^{\infty} \displaystyle{\alpha^n\over \sqrt{n!}}\,\vert{n}\rangle\right) \tag{1}$$

What is $\vert{\alpha(t)}\rangle$, the coherent state at time $t$? Start with (1). Since $\left\vert n\right\rangle$ is an eigenstate of the harmonic oscillator hamiltonian $\hat{H}=\left( \hat a^{\dagger }\hat a+\frac{1}{2}\right) \hbar \omega$ with eigenvalue $\left( n+\frac{1}{2}\right) \hbar \omega ,$ the time evolution of $\left\vert n\right\rangle$ is simply $\left\vert n(t)\right\rangle =e^{-i(n+\frac{1}{2})\omega t}\left\vert n\right\rangle$ and thus $$\left\vert \alpha (t)\right\rangle =e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{% 1}{2})\omega t}\left\vert n\right\rangle \right) .$$ It is easy to show that $\left\vert \alpha (t)\right\rangle$ is normalized.

Now we first need to show that $a\vert{\alpha(t)}\rangle=\alpha e^{i\hbar \omega t}\vert{\alpha(t)}\rangle$. Recall that $\hat{a}\left\vert n\right\rangle =\sqrt{n}\left\vert n-1\right\rangle .$ \ Then, since $\hat{a}$ is linear, \begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}% e^{-i(n+\frac{1}{2})\omega t}\hat{a}\left\vert n\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{1}{2})\omega t}\sqrt{n}\left\vert n-1\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{\left( n-1\right) !}}e^{-i(n+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) , \\ &=&\alpha e^{-i\omega t}e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n-1}}{\sqrt{\left( n-1\right) !}}% e^{-i(n-1+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) .\quad \end{eqnarray} The sum properly starts at $n=1$ since the $n=0$ term does not exist. Thus, setting $m=n-1,$ we can rewrite this sum in terms of $m,$ with $m$ starting at $m=0.$ Hence \begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&\alpha e^{-i\omega t}\left[ e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{m=0}^{\infty }\frac{% \alpha ^{m}}{\sqrt{m!}}e^{-i(m+\frac{1}{2})\omega t}\left\vert m\right\rangle \right) \right] \\ &=&\alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle . \end{eqnarray} A useful secondary result, which follows immediately from above, is \begin{eqnarray} \left[ \hat{a}\left\vert \alpha (t)\right\rangle \right] ^{\dagger } &=&\left\langle \alpha (t)\right\vert \hat{a}^{\dagger } \\ &=&\left[ \alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle \right] ^{\dagger }=\alpha ^{\ast }e^{i\omega t}\left\langle \alpha (t)\right\vert \end{eqnarray}

Now $\langle \hat p(t) \rangle$ and $\langle \hat x(t)\rangle$ for $\vert{\alpha(t)}\rangle$. Starting from the definitions $$\hat{a} =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}+\frac{i}{% m\omega }\hat{p}\right) , \qquad \hat{a}^{\dagger } =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}-% \frac{i}{m\omega }\hat{p}\right) ,$$ we have $$\hat{x} =\sqrt{\frac{\hbar }{2m\omega }}\left( \hat{a}^{\dagger }+% \hat{a}\right) , \qquad \hat{p} =i\sqrt{\frac{m\omega \hbar }{2}}\left( \hat{a}^{\dagger }-% \hat{a}\right) ,$$ and thus \begin{eqnarray} \left\langle x(t)\right\rangle &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle +\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right]\, , \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] , \end{eqnarray} which is real, as expected. We can clean this up by writing $\alpha =\left\vert \alpha \right\vert e^{i\theta }$ to obtain% $$\left\langle x(t)\right\rangle =\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left( \omega t-\theta \right) \tag{2}$$

Likewise, \begin{eqnarray} \left\langle p(t)\right\rangle &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle -\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right] \\ &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&-\sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left( \omega t-\theta \right) \tag{3} \end{eqnarray} which is again real.

In your specific case you are starting with a coherent state for which, at $t=0$, we have $$\langle x(0)\rangle= b\sqrt{2}x_0\, ,\qquad \langle p(0)\rangle=0$$ so this implies from (2) and (3) evaluated at $t=0$ that $$b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left(\theta \right)\, , \qquad 0= \sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left(\theta \right)$$ Comparing with your initial conditions gives $\theta=0$ and $b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }} \alpha$ with $\alpha$ real.

Finally, $\hat{x}^{2}$ and $\hat{p}^{2}.$ From $\hat{x}$ and $\hat{p},$ we find \begin{eqnarray} \hat{x}^{2} &=&\frac{\hbar }{2m\omega }\left( \hat{a}^{\dagger }+ \hat{a}\right) ^{2}=\frac{\hbar }{2m\omega }\left( \left( \hat{a} ^{\dagger }\right) ^{2}+\hat{a}^{\dagger }\hat{a}+\hat{a}\hat{a} ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&\frac{\hbar }{2m\omega }\left( \left( \hat{a}^{\dagger }\right) ^{2}+2 \hat{a}^{\dagger }\hat{a}+1+\left( \hat{a}\right) ^{2}\right) , \\ \hat{p}^{2} &=&-\frac{m\omega \hbar }{2}\left( \hat{a}-\hat{a} ^{\dagger }\right) ^{2}=-\frac{m\omega \hbar }{2}\left( \left( \hat{a} ^{\dagger }\right) ^{2}-\hat{a}^{\dagger }\hat{a}-\hat{a}\hat{a}% ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&-\frac{m\omega \hbar }{2}\left( \left( \hat{a}^{\dagger }\right) ^{2}-2% \hat{a}^{\dagger }\hat{a}-1+\left( \hat{a}\right) ^{2}\right) , \end{eqnarray} where $$\hat{a}\hat{a}^{\dagger }=\hat{a}\hat{a}^{\dagger }-\hat{a}% ^{\dagger }\hat{a}+\hat{a}^{\dagger }\hat{a}=\left[ \hat{a},% \hat{a}^{\dagger }\right] +\hat{a}^{\dagger }\hat{a}=1+\hat{a}% ^{\dagger }\hat{a}$$ has been used. Thus, \begin{eqnarray} \left\langle x^{2}(t)\right\rangle &=&\frac{\hbar }{2m\omega }\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle +2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad +1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}+2\alpha ^{\ast }\alpha +1+\left( \alpha e^{-i\omega t}\right) ^{2}% \right]\, ,\\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right) ^{2}+1\right] , \\ &=&\frac{\hbar }{2m\omega }\left[ 4\left\vert \alpha \right\vert ^{2}\cos ^{2}\left( \omega t-\theta \right) +1\right] . \\ \left\langle p^{2}(t)\right\rangle &=&-\frac{m\omega \hbar }{2}\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle -2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad -1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}-2\alpha ^{\ast }\alpha -1+\left( \alpha e^{-i\omega t}\right) ^{2}\right]\, ,\\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right) ^{2}-1\right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ -4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) -1\right]\, ,\\ &=&\frac{m\omega \hbar }{2}\left[ 4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) +1% \right] . \end{eqnarray}