# Quantum Optics Question Involving Coherent States

## Homework Statement:

Given the quantum-optics coherent states

$$|\alpha \rangle = \exp \Big(\frac{|\alpha|^2}{2}\Big) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle$$

Show that

$$\langle (\Delta X)^2 \rangle_{\alpha} = \langle (\Delta P)^2 \rangle_{\alpha} = \frac 1 4$$

Where

##|n \rangle## are the photon number states

$$X=\frac{a+a^*}{2}$$

$$P=\frac{a-a^*}{2i}$$

$$a|\alpha \rangle = \alpha |\alpha \rangle$$

## Relevant Equations:

$$|\alpha \rangle = \exp \Big(\frac{|\alpha|^2}{2}\Big) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle$$
I've tried to square and compare ##\Delta X## and ##\Delta P## but they are not equal

I have to say I am pretty lost here and a hint would be appreciated.

I have studied coherent states and I know how to proof some properties related to it.

For instance, I see how to proof that the state is normalized:

$$\langle \alpha|\alpha \rangle = \exp(-|\alpha|^2) \Big(\sum_{m=0}^{\infty} \langle m| \frac{\alpha^{*m}}{\sqrt{m!}}\Big) \Big(\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n!}}|n\rangle \Big)$$

Based on ##\langle m | n \rangle = \delta_{mn}## we indeed get ##\langle \alpha | \alpha \rangle = 1##

More can be found in Problem 1.1 Mandl & Shaw; I also attached the complete solution of it.

Thanks

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strangerep
Homework Statement:: Show that $$\langle (\Delta X)^2 \rangle_{\alpha} = \langle (\Delta P)^2 \rangle_{\alpha} = \frac 1 4$$ [...] I've tried to square and compare ##\Delta X## and ##\Delta P## but they are not equal

Btw, you should probably include the definition of ##\langle (\Delta X)^2 \rangle_{\alpha}## in your "relevant equations" section.

JD_PM, DrClaude and vanhees71

$$\Big( \frac{a+a^*}{2} \Big)^2 = \frac 1 4 (a^2+(a^*)^2+2aa^*) \neq \frac 1 4 (-a^2-(a^*)^2+2aa^*)=\Big( \frac{a-a^*}{2i} \Big)^2$$

dRic2
Gold Member
Didn't you just compared ##X^2## and ##P^2## instead of ##\Delta X^2## and ##\Delta P^2##?

Btw, I don't know Quantum Optics but there is something that caught my attention. That "normalization coefficient" is a Gaussian times some other stuff. There is a theorem called "uncertainty principle" which states that (https://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle)
##\left( \int (x-x_0)^2 f(x) \right) \left( \int ( \eta- \eta_0)^2 \hat f( \eta ) \right) \geq \frac 1 {16 \pi^2}##
You can easily check that the equal sign holds only for a Gaussian since the fourier transform of a Gaussian is still a Gaussian. Moreover you know that momentum operator is linked with the fourier transform of the position operator. Maybe you can avoid all the calculation somehow by using this theorem... that would be fun.

Last edited:
JD_PM
strangerep
Of course, my bad.$$\Big( \frac{a+a^*}{2} \Big)^2 = \frac 1 4 (a^2+(a^*)^2+2aa^*) \neq \frac 1 4 (-a^2-(a^*)^2+2aa^*)=\Big( \frac{a-a^*}{2i} \Big)^2$$
As @dRic2 has noted, you have only calculated squares of the operators. You have not calculated their respective variances in the coherent state ##|\alpha\rangle##.

So,... do what I suggested in the "Btw" sentence in my post #2.

(If you don't know what "variance" means in this context, then review the concepts of "expectation value" and "uncertainty" in ordinary QM.)

Which Quantum Optics and/or QM textbook(s) are you working from?

JD_PM
vanhees71
Gold Member
2019 Award
Also, if ##a## is a usual annihilation operator and ##a^*=a^{\dagger}## the corresponding creation operator you have ##[a,a^{\dagger}]=1## and thus
$$(a+a^*)=a^2 + (a^*)^2 + a a^* + a^* a=a^2 + (a^*)^2 + 2 a a^*-1.$$

JD_PM
OK let me try to set up the problem first

I am used to solve problems like the following (see for instance 1.14 of Griffiths' QM book, first edition).

Given a wave function (i.e. state of system) ##\psi##.

a) Compute the expectation value of ##x, x^2, p, p^2##

b) Find its variance (i.e. ##\langle (\Delta X)^2 \rangle_{\psi}## and ##\langle (\Delta P)^2 \rangle_{\psi}##).

We proceed as follows

a)

$$\langle x \rangle = \int_{-\infty}^{\infty} x |\psi|^2 dx$$

$$\langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 |\psi|^2 dx$$

$$\langle p \rangle = -i \hbar \int_{-\infty}^{\infty} \Big( \psi^* \frac{\partial \psi}{\partial x}\Big)dx$$

$$\langle p^2 \rangle = -i \hbar \int_{-\infty}^{\infty} \Big( \psi^* \frac{\partial \psi}{\partial x}\Big)dx$$

b)

$$\langle (\Delta X)^2 \rangle_{\psi}=\sigma_x^2 = \langle x^2 \rangle - \langle x \rangle^2, \ \ \ \ \langle (\Delta P)^2 \rangle_{\psi}=\sigma_p^2 = \langle p^2 \rangle - \langle p \rangle^2$$

OK.

I'd go for an analogous approach in our problem; I'd first compute the expectation values

$$\langle x \rangle = \int_{-\infty}^{\infty} \frac{a+a^*}{2} |\alpha|^2 dx$$

$$\langle x^2 \rangle = \int_{-\infty}^{\infty} \Big(\frac{a+a^*}{2}\Big)^2 |\alpha|^2 dx$$

$$\langle p \rangle = -i \hbar \int_{-\infty}^{\infty} \Big( \alpha^* \frac{\partial \alpha}{\partial x}\Big)dx$$

$$\langle p^2 \rangle = -i \hbar \int_{-\infty}^{\infty} \Big( \alpha^* \frac{\partial \alpha}{\partial x}\Big)^2dx$$

And then I'd go for

$$\langle (\Delta X)^2 \rangle_{\alpha}=\sigma_x^2 = \langle x^2 \rangle - \langle x \rangle^2, \ \ \ \ \langle (\Delta P)^2 \rangle_{\alpha}=\sigma_p^2 = \langle p^2 \rangle - \langle p \rangle^2$$

And then we indeed expect (well may be not the best verb choice here but anyway ) to get

$$\langle (\Delta X)^2 \rangle_{\alpha}=\langle (\Delta P)^2 \rangle_{\alpha}=\frac 1 4$$

Is this approach OK?

So,... do what I suggested in the "Btw" sentence in my post #2.
But I cannot edit my #1 :(

Which Quantum Optics and/or QM textbook(s) are you working from?
I like Griffiths but may be I should go for another book to complement; I had a brief look at Sakurai's and looks nice.

I've seen (never read) that one of the most brilliant mathematicians of the 20th century, John Von Neumann, also has a book on QM.

I see you recommend Ballentine; I've never had a copy of it though.

Btw, you should probably include the definition of ##\langle (\Delta X)^2 \rangle_{\alpha}## in your "relevant equations" section.
I think that its definition is as follows

$$\langle (\Delta X)^2 \rangle_{\alpha} := \langle X^2 \rangle - \langle X \rangle^2$$

From here on the idea is to compute ##\langle X \rangle## and ##\langle X^2 \rangle##

Is this approach OK?
Well, I better try to compute the given integrals and see if the approach makes sense.

Let us focus on ##\langle X \rangle##

$$\langle X \rangle = \int_{-\infty}^{\infty} X |\alpha|^2 dX=0$$

Because the definite integral of an odd function is zero

OK. Now let's tackle ##\langle X^2 \rangle ##

$$\langle X^2 \rangle = 2\int_{0}^{\infty} X^2 |\alpha|^2 dX = 2\int_{0}^{\infty} \Big( \Big(\frac{a+a^*}{2}\Big)^2 \exp(-|\alpha|^2) \sum_{n=0}^{\infty} \frac{\alpha^{*n}}{\sqrt{n!}} \sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n!}} \Big)dX$$

This looks like I am overcomplicating myself again, so any help would be appreciated.

I've recently been thinking that the right way to go with this integral may be a change of variables so that we end up integrating over ##a## instead of ##X##; this thought is enforced by the fact that I am hinted to use ##a|\alpha\rangle = \alpha|\alpha\rangle##

To compute ##\langle P \rangle## and ##\langle P^2 \rangle## I am using the following formulas

$$\langle P \rangle = -i \hbar \int_{-\infty}^{\infty} \Big( \alpha^* \frac{\partial \alpha}{\partial X}\Big)dX$$

$$\langle P^2 \rangle = -i \hbar \int_{-\infty}^{\infty} \Big( \alpha^* \frac{\partial \alpha}{\partial X}\Big)^2dX$$

The issue here is that the state ##|\alpha \rangle## does not explicitly depend on ##X##, which means that ##\frac{\partial \alpha}{\partial X}=0##. So I guess this approach is incorrect...

strangerep
The expectation value of any observable ##Z## in the state ##|\alpha\rangle## is $$\langle \alpha| Z |\alpha\rangle ~,$$which is sometimes abbreviated as ##\langle Z \rangle_\alpha##.