Expectation value in coherent state

[spaghetti3451]

Homework Statement

In a coherent state $|\alpha\rangle$, letting $P(n)$ denote the probability of finding $n^{\text{th}}$ harmomic oscillator state. Show that

$$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)=|\alpha|^{2}}$$

Homework Equations

The Attempt at a Solution

A coherent state $\alpha\rangle$ is defined as $D(\alpha)|\alpha\rangle = \exp(\alpha a^{\dagger}-\alpha^{*}a)|\alpha\rangle$ so that

$$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)= \sum\limits_{n}n\ |\langle n|\alpha\rangle|^{2}}$$

Therefore, it is prudent to evaluate $|\alpha\rangle$ in terms of $|n\rangle$ as follows:

$\displaystyle{|\alpha\rangle = \exp\ (\alpha a^{\dagger}-\alpha^{*}a)|0\rangle}$

$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \exp\ (\alpha a^{\dagger})|0\rangle}$

$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{n'!}(a^{\dagger})^{n'}|0\rangle}\bigg)$

$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}|n'\rangle}\bigg)$

$\displaystyle{\langle n|\alpha\rangle = \langle n|\left(\exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}\right) |n'\rangle}.$

Am I correct so far?

 

[strangerep]

$\displaystyle{|\alpha\rangle = \exp\ (\alpha a^{\dagger}-\alpha^{*}a)|0\rangle}$

$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \exp\ (\alpha a^{\dagger})|0\rangle}$

OK, so you're using the Zassenhaus formula in this step, but note that the formula can be written in 2 ways. I.e., you could put $\exp(-\alpha^*a)$ on the right and change the sign of the $|\alpha|^2$ exponent. Then, what is $$\exp(-\alpha^*a)|0\rangle $$ ?

 

[spaghetti3451]

OK, so you're using the Zassenhaus formula in this step, but note that the formula can be written in 2 ways. I.e., you could put $\exp(-\alpha^*a)$ on the right and change the sign of the $|\alpha|^2$ exponent. Then, what is $$\exp(-\alpha^*a)|0\rangle $$ ?

$\displaystyle{\exp(-\alpha^*a)|0\rangle=|0\rangle}$, but then, using Baker-Campbell-Hausdorff, I get

$\displaystyle{\exp(\alpha a^{\dagger}-\alpha^{*}a)}|0\rangle = \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) \exp\left(\frac{1}{2}[\alpha a^{\dagger},-\alpha^{*}a]\right)|0\rangle$

$= \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) \exp\left(-\frac{1}{2}|\alpha|^{2}[a^{\dagger},a]\right)|0\rangle$

$= \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) \exp\left(\frac{1}{2}|\alpha|^{2}\right)|0\rangle$

$= \exp\left(\frac{1}{2}|\alpha|^{2}\right) \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) |0\rangle$

$= \exp\left(\frac{1}{2}|\alpha|^{2}\right) \exp(\alpha a^{\dagger}) |0\rangle,$

when, in fact, I should have

$= \exp\left(-\frac{1}{2}|\alpha|^{2}\right) \exp(\alpha a^{\dagger}) |0\rangle.$

What am I doing wrong here?

 

[strangerep]

$\displaystyle{\exp(\alpha a^{\dagger}-\alpha^{*}a)}|0\rangle = \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) \exp\left(\frac{1}{2}[\alpha a^{\dagger},-\alpha^{*}a]\right)|0\rangle$

Check the Zassenhaus formula. Shouldn't the 3rd exponent have a minus sign?

(BTW, you can shorten your latex a bit by using just double-dollar, instead of double-hash and \displaystyle.}

 

[spaghetti3451]

Check the Zassenhaus formula. Shouldn't the 3rd exponent have a minus sign?

(BTW, you can shorten your latex a bit by using just double-dollar, instead of double-hash and \displaystyle.}

Why use the Zassenhaus formula and not the BCH formula?

 

[strangerep]

Why use the Zassenhaus formula and not the BCH formula?

Look up the "Zassenhaus formula" on Wikipedia. It is simply a special case of BCH.

 

[spaghetti3451]

But then the BCH formula says that

$$\exp(A+B)=\exp(A)\exp(B)\exp\left(\frac{1}{2}[A,B]\right)$$

and the Zassenhaus formula says that

$$\exp(t(X+Y))=\exp(tX)\exp(tY)\exp\left(-\frac{t^{2}}{2}[X,Y]\right)$$.

I don't see how the Zassenhaus formula is a special case of the BCH formula, if $A=tX$ and $B=tY$.

 

[strangerep]

But then the BCH formula says that
$$\exp(A+B)=\exp(A)\exp(B)\exp\left(\frac{1}{2}[A,B]\right)$$

Where are you getting your BCH formula from? There's another form of BCH which looks like this:
$$e^A e^B ~=~ e^{A+B+[A,B]/2 ~+~ \dots} ~.$$

 

[spaghetti3451]

Where are you getting your BCH formula from? There's another form of BCH which looks like this:
$$e^A e^B ~=~ e^{A+B+[A,B]/2 ~+~ \dots} ~.$$

Sorry, my bad!

How do you prove the Zassenhaus formula from the BCH formula?

 

[spaghetti3451]

Wait. I don't need to see the proof.

But, I am not really sure why the Zassenhaus formula can be used in this case. After all, we can't be sure if $t=\alpha$ or if $t=-\alpha^{*}$ in https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula#The_Zassenhaus_formula.

 

[strangerep]

Wait. I don't need to see the proof.

Actually, I think you do, at least for the current simple case. In fact that's trivial because we're only dealing here with the case where $[A,B]$ is a c-number (meaning it commutes with everything). So you can factor it out and transfer it across to the other side of the equation.

But, I am not really sure why the Zassenhaus formula can be used in this case. After all, we can't be sure if $t=\alpha$ or if $t=-\alpha^{*}$ in https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula#The_Zassenhaus_formula.

The $t$ is a red herring. Just think in terms of A and B (i.e., put $t=1$ in those Wiki formulas). Then put $A = \alpha a^\dagger$, etc.

 

[spaghetti3451]

Ah, I see! So, I have

$\displaystyle{|\alpha\rangle = \exp\left(-\frac{1}{2}|\alpha|^{2}\right)\exp\left(\alpha a^{\dagger}\right)\exp\left(-\alpha^{*}a\right)|0\rangle}$

$\displaystyle{|\alpha\rangle = \exp\left(-\frac{1}{2}|\alpha|^{2}\right)\exp\left(\alpha a^{\dagger}\right)|0\rangle}$

$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{n'!}(a^{\dagger})^{n'}|0\rangle}\bigg)$

$\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}|n'\rangle}\bigg)$

$\displaystyle{\langle n|\alpha\rangle = \langle n|\left(\exp (-|\alpha|^{2}/2) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}} |n'\rangle\right)}$

$\displaystyle{= \left(\exp (-|\alpha|^{2}/2) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}} \langle n|n'\rangle\right)}$

$\displaystyle{= \left(\exp (-|\alpha|^{2}/2) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}} \delta_{nn'}\right)}$

$\displaystyle{= \left(\exp (-|\alpha|^{2}/2) \frac{\alpha^{n}}{\sqrt{n!}}\right)},$

which is a Poisson distribution with mean $|\alpha|^{2}$ and $n$ events in each interval so that

$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)}$

$\displaystyle{= \sum\limits_{n}n \left(\exp (-|\alpha|^{2}/2) \frac{\alpha^{n}}{\sqrt{n!}}\right)\left(\exp(-|\alpha|^{2}/2)\frac{{\alpha^{*}}^{n}}{\sqrt{n!}}\right)}$

$\displaystyle{= \sum\limits_{n}n \exp (-|\alpha|^{2}) \frac{|\alpha|^{2n}}{n!}}$

is the mean squared of the Poisson distribution and is equal to $|\alpha|^{2}$.

What do you think?

 

[strangerep]

It's looking a lot better. Unfortunately, I have to dash out now. I'll try to check it more carefully later.