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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 ?

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  • $\begingroup$ The energy is conserved... $\endgroup$ – Fabian Jan 11 '15 at 21:26
  • $\begingroup$ Of course, I must've been tired writing this. Still, what about the position and momentum squared. $\endgroup$ – Spine Feast Jan 11 '15 at 21:34
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    $\begingroup$ 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. $\endgroup$ – Mark Mitchison Jan 12 '15 at 0:49
  • $\begingroup$ What are left and right eigenstates? $\endgroup$ – Spine Feast Jan 12 '15 at 1:05
  • $\begingroup$ 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. $\endgroup$ – Mark Mitchison Jan 12 '15 at 1:19
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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 \begin{equation} \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) . \end{equation} 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% \begin{equation} \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} \end{equation}

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 \begin{equation} \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} \end{equation} 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}

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