Quantum Optics Question Involving Coherent States

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  • #1
JD_PM
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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 https://ia800108.us.archive.org/32/items/FranzMandlGrahamShawQuantumFieldTheoryWiley2010/Franz%20Mandl%2C%20Graham%20Shaw-Quantum%20Field%20Theory-Wiley%20%282010%29.pdf; I also attached the complete solution of it.

Thanks :smile:
 

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Answers and Replies

  • #2
strangerep
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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
Well? It's rather difficult for anyone to help you if you don't show your work.

Btw, you should probably include the definition of ##\langle (\Delta X)^2 \rangle_{\alpha}## in your "relevant equations" section.
 
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  • #3
JD_PM
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Well? It's rather difficult for anyone to help you if you don't show your work.

Of course, my bad.

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

PS: I did not forget about the series expansion of LT thread; I appreciate your help :smile:
 
  • #4
dRic2
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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.
 
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  • #5
strangerep
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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?
 
  • #6
vanhees71
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Also, if ##a## is a usual annihilation operator and ##a^*=a^{\dagger}## the corresponding creation operator you have ##[a,a^{\dagger}]=1## and thus
$$(a+a^*)=a^2 + (a^*)^2 + a a^* + a^* a=a^2 + (a^*)^2 + 2 a a^*-1.$$
 
  • #7
JD_PM
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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 :rolleyes:) to get

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

Is this approach OK?
 
  • #8
JD_PM
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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.
 
  • #9
JD_PM
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Well? It's rather difficult for anyone to help you if you don't show your work.

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##
 
  • #10
JD_PM
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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##
 
  • #11
JD_PM
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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...
 
  • #12
strangerep
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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##.

You don't need to use integrals here. Just use annihilation/creation operators and bra-ket notation. (You've already written in post #1 how X and P are expressed in terms of annihilation and creation operators.)

Btw, AFAICT, Griffiths does not even mention "coherent state" in his "Intro to QM" textbook. [Correction: Griffiths mentions coherent states in later editions of his textbook -- I was previously looking at the 1st edition. Thanks ProxyMelon for mentioning this.]

Sakurai and Ballentine do a bit on coherent states. Mandel & Wolf is the "bible" of Quantum Optics, albeit rather more advanced.
 
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  • #13
ProxyMelon
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Griffiths has an extensive exercise about coherent states (page 129 in the second edition).
The problem with Griffiths, in my opinion, is that it overuses position representation, for example in a problem of this kind you are clearly supposed to work in braket notation and not do a single integral basically.

What you could to do work this problem out is:
1) Before you even begin, do the math and write $$x,x^2,x^3,...,p,p^2,p^3,...$$ in terms of $$a,a^\dagger$$ and N $$N=a^\dagger a , N|n\rangle = n|n\rangle$$
The main focus should be then to get fluent with how the operators work on a state n, so
2) Evaluate $$\langle n|\hat{o}|n\rangle = \langle o \rangle_n$$
on a generic state n (and with all the operators you want), you should be able to prove that the uncertainty principle must hold for any state, note, you should be able to do this calculation by knowing how the state n is written in terms of the ground state, and not doing a single integral apart from maybe $$\langle 0|\hat{o}|0\rangle = \langle o \rangle_0 = \int \psi_0^* O \psi_0 dx$$

Notice that by exploiting orthonormality, a lot of terms in your operators are going to not be relevant in an expected value (because raising and lowering operators will create a scalar product between different eigenstates), this could save you tons of useless calculations if you ever need to evaluate operators raised to higher powers

at this point you should be easily capable of evaluating your EV in a coherent state, you just have to use the added properties you get (pretty much being eigenvectors of the lowering operator).

This seems a lot of work, but is in fact quite an easy task, it's just a bit tricky to get used to work with lowering and raising operators
 
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  • #14
JD_PM
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Thank you ProxyMelon for suggesting that exercise! I've worked it out and now things are much more clear!

1) Before you even begin, do the math and write $$x,x^2,x^3,...,p,p^2,p^3,...$$ in terms of $$a,a^\dagger$$ and N $$N=a^\dagger a , N|n\rangle = n|n\rangle$$
The main focus should be then to get fluent with how the operators work on a state n, so

$$x=\sqrt{\frac{\hbar}{2m \omega}}(a_{+} + a_{-})$$

$$x^2=\frac{\hbar}{2m\omega}\Big(a_{+}^2 + a_{-}^2 + a_{+}a_{-} + a_{-}a_{+} \Big)=\frac{\hbar}{2m\omega}\Big(a_{+}^2 + a_{-}^2 + 2a_{+}a_{-} + 1 \Big)$$

$$p=i\sqrt{\frac{\hbar m \omega}{2}}(a_{+} - a_{-})$$

$$p^2=-\frac{\hbar m\omega}{2}\Big(a_{+}^2 + a_{-}^2 - a_{+}a_{-} - a_{-}a_{+} \Big)=-\frac{\hbar m\omega}{2}\Big(a_{+}^2 + a_{-}^2 - 2a_{+}a_{-} -1 \Big)$$

Where I've used

$$[a_{-}, a_{+}]=1$$

OK. I think all we need is ##a_{-}|\alpha \rangle = \alpha |\alpha \rangle## (after doing the suggested exercise, I think the person who wrote the thread's statement made a typo writing ##a|\alpha \rangle = \alpha |\alpha \rangle##).

Here we go; let's compute ##\langle x \rangle##

$$\langle x \rangle = \langle \alpha | x \alpha \rangle = \sqrt{\frac{\hbar}{2m \omega}} \langle \alpha | (a_{+} + a_{-}) \alpha \rangle=\sqrt{\frac{\hbar}{2m \omega}} \Big( \langle \alpha |a_{+} \alpha \rangle + \langle \alpha |a_{-} \alpha \rangle \Big) = \sqrt{\frac{\hbar}{2m \omega}} \Big( \langle a_{-} \alpha |\alpha \rangle + \langle \alpha |a_{-} \alpha \rangle \Big)$$ $$=\sqrt{\frac{\hbar}{2m \omega}}(\alpha^*+\alpha)$$

Where I've used

$$a^* = a^{\dagger}$$

$$\langle \alpha | a_{+} \alpha \rangle = \langle a_{+}^{\dagger} \alpha | \alpha \rangle = \langle a_{-} \alpha | \alpha \rangle$$

$$\langle \alpha | a_{-} = \langle \alpha | \alpha^*$$

OK. We now compute ##\langle x^2 \rangle##

$$\langle x^2 \rangle = \langle \alpha | x^2 \alpha \rangle = \frac{\hbar}{2m\omega}\Big(\langle a_{-}^2 \alpha | \alpha \rangle + 2\langle \alpha | a_{+} a_{-} \alpha \rangle +\langle \alpha | \alpha \rangle + \langle \alpha | a_{-}^2 \alpha \rangle \Big)$$ $$=\frac{\hbar}{2m\omega} \Big( 1 + (\alpha^* + \alpha)^2 \Big)$$

Where we noticed that ##\langle \alpha | a_{+} a_{-} \alpha \rangle = \langle a_{-} \alpha | a_{-} \alpha \rangle##

We now compute ##\langle p \rangle##

$$\langle p \rangle = \langle \alpha | p \alpha \rangle =i\sqrt{\frac{\hbar m \omega}{2}} \langle \alpha | (a_{+} - a_{-}) \alpha \rangle= i\sqrt{\frac{\hbar m \omega}{2}} \Big( \langle \alpha |a_{+} \alpha \rangle - \langle \alpha |a_{-} \alpha \rangle \Big) = i\sqrt{\frac{\hbar m \omega}{2}} \Big( \langle a_{-} \alpha |\alpha \rangle - \langle \alpha |a_{-} \alpha \rangle \Big)$$ $$=-i\sqrt{\frac{\hbar m \omega}{2}}(-\alpha^*+\alpha)$$

We now compute ##\langle p^2 \rangle##

$$\langle p^2 \rangle = \langle \alpha | p^2 \alpha \rangle = -\frac{\hbar m\omega}{2}\Big(\langle a_{-}^2 \alpha | \alpha \rangle - 2\langle \alpha | a_{+} a_{-} \alpha \rangle - \langle \alpha | \alpha \rangle + \langle \alpha | a_{-}^2 \alpha \rangle \Big)$$ $$=\frac{\hbar m\omega}{2} \Big( 1 - ( \alpha -\alpha^*)^2 \Big)$$

We only have to compute ##\sigma_x## and ##\sigma_p## and compare them.

$$\sigma_x^2 := \langle x^2 \rangle - \langle x \rangle^2 = \frac{\hbar}{2 m \omega}$$

$$\sigma_p^2 := \langle p^2 \rangle - \langle p \rangle^2 = \frac{\hbar m \omega}{2}$$

Thus I get

$$\sigma_x^2 = \frac{1}{m^2 \omega^2} \sigma_p^2$$

Instead of the provided solution

$$\sigma_x^2 = \sigma_p^2 = \frac 1 4$$

Mmm what am I missing?

PS: It turns out that I get the same result that Griffiths' sections a), b). I guess there's a key difference between both exercise I do not see right now (may be there was no typo after all?!). I attach Griffiths' sections a), b) below

CoherentStatesGriffiths.jpeg
 
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  • #15
ProxyMelon
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Well, realize that no matter what ##\sigma_x\sigma_p\geq\frac{\hbar}{2}## must hold, so your provided solution is clearly "wrong".

Probably it is not even wrong, you are just using ##\hbar = 1##, could this be?

Regardless, your results net you the correct value.
 
  • #16
JD_PM
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Probably it is not even wrong, you are just using ##\hbar = 1##, could this be?

Yes. Actually after some thinking, I'd say that what they really meant is

$$\sigma_x^2 \sigma_p^2 = \frac 1 4$$

Setting ##\hbar =1##.

And I'd say their provided solution ##\sigma_x^2 = \frac 1 4 = \sigma_p^2## is simply wrong.
 

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