Expectation values for an harmonic oscillator

[carllacan]

Homework Statement

Find the expectation values of x and p for the state
\vert \alpha \rangle = e^{-\frac{1}{2}\vert\alpha\vert^2}exp(\alpha a^{\dagger})\vert 0 \rangle, where $a$ is the destruction operator.

Homework Equations

Destruction and creation operators
$a=Ax+Bp$
$a^{\dagger}=Ax-Bp$
For some constants A (real) and B (imaginary) whose value is not important now.

The Attempt at a Solution

I've found a solution, but it is so simple it looks dumb. State the expectation value:
$\langle x \rangle = \langle 0 \vert e^{- \frac{1}{2}\vert\alpha \vert ^2} e^{\alpha^*a}\hat{x}e^{\alpha a^{\dagger}}e^{- \frac{1}{2}\vert\alpha \vert ^2} \vert 0\rangle =$

Now if we descompose the first exponential $e^{\alpha^*a}$ there will be an annihilation operator acting on the ground state, which is equal to 0 and therefore ANY expectation value for this state is zero...

That seems wrong. Where did I go wrong?

 

[Orodruin]

$a$ is never acting directly on a ground state ket. How do the creation and annihilation operators act on bra states?

 

[carllacan]

$a$ is never acting directly on a ground state ket. How do the creation and annihilation operators act on bra states?

I was not sure of this, but I still don't see the mistake. Look at what the expression for the expectation value I wrote. There is an exponential of the annihilation operator next to the ground state; if we decompose it... oh, wait.
Okay, I just realized that the first term of the series would not contain an $a$ operator, so is doesn't destroy it.

New try:

From the definition for $a$ I've found $x=\frac{a+a^{\dagger}}{2A}$ and $p=\frac{a-a^{\dagger}}{2B}$, and then its easy to find:

$\langle x \rangle = \langle 0 \vert e^{- \frac{1}{2}\vert\alpha \vert ^2} e^{\alpha^*a}\hat{x}e^{\alpha a^{\dagger}}e^{- \frac{1}{2}\vert\alpha \vert ^2} \vert 0\rangle = \frac{1}{2A} \langle 0 \vert e^{ -\frac{1}{2}\vert\alpha \vert ^2} e^{\alpha^*a}(a + a^{\dagger})e^{\alpha a^{\dagger}}e^{-\frac{1}{2}\vert \alpha \vert ^2} \vert 0\rangle$

Now this reduces to $\langle x \rangle = \frac{1}{2A} \left ( \langle a \rangle + \langle a^{\dagger} \rangle \right )$ and, operating similarly for the momentum operator $\langle p \rangle = \frac{1}{2B} \left ( \langle a \rangle - \langle a^{\dagger} \rangle \right )$.

Now, I think $\langle a \rangle = \langle a^{\dagger} \rangle$, is that right?

 

[Orodruin]

Neither operator in $\hat x$ act directly on the vacuum. They must first be commuted through the exponential expressions that contain operators.

By definition, we must have
$$
\langle a\rangle = \langle a^\dagger\rangle^*
$$

 

[bloby]

Yes it seems right at first look. So $<p>=0$. What about $<a>$?

[edit: oops yes $<>^*$]

 

[carllacan]

I know from a previous exercise that these states are eigenkets of $a$ with eigenvalue $\alpha$, so $\langle a \rangle = \alpha$ and $\langle x \rangle = \frac{\alpha}{A}$?

 

[Orodruin]

Is $\alpha$ necessarily real?

 

[carllacan]

Is $\alpha$ necessarily real?

No. Why?

 

[Orodruin]

If $\alpha$ is not real, is p zero?

 

[carllacan]

If $\alpha$ is not real, is p zero?

Oh, thank you. The actual solutions are
$\langle \hat{x} \rangle = \frac{Re(\alpha)}{A}$
$\langle \hat{p} \rangle = \frac{Im(\alpha)}{B}$
The momentum EV is actually real, because B is pure imaginary.

 

[Oxvillian]

Hi Carllacan - not sure if you figured this out or not - but anyway here's the cleanest way I can think of to do it:

\begin{eqnarray}<br /> \langle X \rangle &amp;=&amp; e^{-\vert \alpha \vert^2} \frac{1}{2A} \langle 0 \lvert e^{\alpha^*a}(a + a^{\dagger}) e^{\alpha a^{\dagger}} \lvert 0\rangle\\<br /> <br /> &amp;=&amp; \frac{1}{2A} e^{-\vert \alpha \vert^2} \left( \frac{\partial}{\partial \alpha} + \frac{\partial}{\partial \alpha^*}\right) \langle 0 \lvert e^{\alpha^*a} e^{\alpha a^{\dagger}} \lvert 0\rangle\\<br /> <br /> &amp;=&amp; \frac{1}{2A} e^{-\vert \alpha \vert^2} \left( \frac{\partial}{\partial \alpha} + \frac{\partial}{\partial \alpha^*}\right) e^{\alpha^* \alpha}\\<br /> <br /> &amp;=&amp; \frac{1}{2A} e^{-\vert \alpha \vert^2} \left( \alpha + \alpha^* \right) e^{\alpha^* \alpha}\\<br /> <br /> &amp;=&amp; \frac{\alpha + \alpha^*}{2A}\\<br /> <br /> &amp;=&amp; \frac{{\rm Re} \; \alpha}{A}<br /> <br /> \end{eqnarray}<br /> <br />
I'll let you do \langle P \rangle for yourself

Here's a couple of "for what it's worth" comments:

Expectation values for an harmonic oscillator

The state \lvert \alpha \rangle is called a "canonical coherent state" and uses the harmonic oscillator ground state as what's called its "fiducial vector". However note that there need not be a harmonic oscillator present - our particle could be in absolutely any potential.

The state \lvert \alpha \rangle has (the expectation values of) its position and momentum coded into the real and imaginary parts of \alpha. So we could rewrite it as simply \lvert p,q \rangle. Thus there's a nice one-to-one map from classical phase space onto canonical coherent states in the Hilbert space - it's as good as a map as the uncertainty principle allows.

 

[carllacan]

Here's a couple of "for what it's worth" comments:



The state \lvert \alpha \rangle is called a "canonical coherent state" and uses the harmonic oscillator ground state as what's called its "fiducial vector". However note that there need not be a harmonic oscillator present - our particle could be in absolutely any potential.

The state \lvert \alpha \rangle has (the expectation values of) its position and momentum coded into the real and imaginary parts of \alpha. So we could rewrite it as simply \lvert p,q \rangle. Thus there's a nice one-to-one map from classical phase space onto canonical coherent states in the Hilbert space - it's as good as a map as the uncertainty principle allows.

Thank you, I was wondering what was the point of having us work with these specific states.

Would you guys mind checking this?
We know $a\vert \alpha\rangle = \alpha \vert \alpha \rangle$, so
$\langle a \rangle = \alpha$,
$\langle a^{\dagger} \rangle = \langle a \rangle ^* = \alpha^*$,
$\langle a^2 \rangle = \langle \alpha \vert a^2 \vert \alpha \rangle = \langle \alpha \vert \alpha^*\alpha \vert \alpha \rangle = \alpha ^* \alpha = \vert \alpha\vert ^2$,
$\langle a_{\dagger}^2 \rangle = \langle a^2 \rangle ^* = \vert \alpha\vert ^2$,
$\langle a a_{\dagger} \rangle = \langle \alpha \vert a a_{\dagger} \vert\alpha \rangle = \alpha \langle \alpha \vert a_{\dagger} \vert\alpha \rangle = \vert \alpha\vert ^2$

 

[Oxvillian]

Well the notation \langle a\rangle looks a bit weird to me, because it suggests you're finding the expectation value of some observable - but a isn't an observable. I guess it's ok though.

I would say that:

(I'm changing \alpha to \lambda because with my eyesight I can't tell the difference between an \alpha and an a on the screen )

\begin{eqnarray}<br /> \langle \lambda \lvert a \lvert \lambda \rangle &amp;=&amp; \lambda\\<br /> \langle \lambda \lvert a^{\dagger} \lvert \lambda \rangle &amp;=&amp; \lambda^*\\<br /> \langle \lambda \lvert a^2 \lvert \lambda \rangle &amp;=&amp; \lambda^2\\<br /> \langle \lambda \lvert (a^{\dagger})^2 \lvert \lambda \rangle &amp;=&amp; (\lambda^*)^2\\<br /> \langle \lambda \lvert a^{\dagger}a \lvert \lambda \rangle &amp;=&amp; \lambda^*\lambda\\<br /> \end{eqnarray}
So I disagree with you about some of them at least.

 

[Orodruin]

I agree with Oxvillian's results. I would however not write the coherent state as $|p,q\rangle$ without some additional hint to it being a coherent state. Otherwise there is a risk of interpreting it as an eigenstate of the position and momentum operators (there are no simultaneous eigenstates, but anyway...)

 

[Oxvillian]

I agree with Oxvillian's results. I would however not write the coherent state as $|p,q\rangle$ without some additional hint to it being a coherent state. Otherwise there is a risk of interpreting it as an eigenstate of the position and momentum operators (there are no simultaneous eigenstates, but anyway...)

Indeed, \lvert p,q \rangle looks like a total violation of page 1 of the quantum mechanics rulebook

You just have to remember that the labels p and q are just parameters, not eigenvalues.

Anyway when you get past the notation, there's a lot to be said for working in the \lvert p,q \rangle basis instead of using \vert p \rangle and/or \vert q \rangle. The coherent states are nice and smooth rather than delta-function spikey. They lead, for instance, to much nicer path integrals.

 

[carllacan]

About the third one: doesn't $a\vert \lambda \rangle = \lambda \vert \lambda \rangle$ imply $\langle \lambda\vert a^{\dagger} = \langle \lambda \vert \lambda^*$?

And therefore the result is $\lambda ^* \lambda = \vert \lambda \vert ^2$

 

[Oxvillian]

That would be the 5th one.

The 3rd one is
\langle \lambda \lvert aa \lvert \lambda \rangle = \lambda \langle \lambda \lvert a \lvert \lambda \rangle = \lambda^2 \langle \lambda \lvert \lambda \rangle = \lambda^2