# Expectation values for an harmonic oscillator

1. Jun 28, 2014

### carllacan

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.
3. 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?

2. Jun 28, 2014

### Orodruin

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

3. Jun 28, 2014

### carllacan

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?

4. Jun 28, 2014

### Orodruin

Staff Emeritus
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
$$\langle a\rangle = \langle a^\dagger\rangle^*$$

5. Jun 28, 2014

### bloby

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

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

Last edited: Jun 28, 2014
6. Jun 28, 2014

### carllacan

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

7. Jun 28, 2014

### Orodruin

Staff Emeritus
Is $\alpha$ necessarily real?

8. Jun 28, 2014

### carllacan

No. Why?

9. Jun 28, 2014

### Orodruin

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

10. Jun 28, 2014

### carllacan

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.

11. Jun 29, 2014

### 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:

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

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.

12. Jun 29, 2014

### carllacan

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
$\langle a \rangle = \alpha$,
$\langle a^{\dagger} \rangle = \langle a \rangle ^* = \alpha^*$,
$\langle a^2 \rangle = \langle \alpha \vert a^2 \vert \alpha \rangle = \langle \alpha \vert \alpha^*\alpha \vert \alpha \rangle = \alpha ^* \alpha = \vert \alpha\vert ^2$,
$\langle a_{\dagger}^2 \rangle = \langle a^2 \rangle ^* = \vert \alpha\vert ^2$,
$\langle a a_{\dagger} \rangle = \langle \alpha \vert a a_{\dagger} \vert\alpha \rangle = \alpha \langle \alpha \vert a_{\dagger} \vert\alpha \rangle = \vert \alpha\vert ^2$

13. Jun 29, 2014

### 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 )

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

14. Jun 29, 2014

### Orodruin

Staff Emeritus
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...)

15. Jun 29, 2014

### Oxvillian

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.

16. Jun 29, 2014

### 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 $\lambda ^* \lambda = \vert \lambda \vert ^2$

17. Jun 29, 2014

### Oxvillian

That would be the 5th one.

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