# Coherent states for Klein-Gordon field

• CharlieCW
In summary: Hmmm... Since I know the result, I know that the generalization of the commutator is correct. So, I think I'll just have to trust it.
CharlieCW

## Homework Statement

Show that the coherent state ##|c\rangle=exp(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})|0\rangle## is an eigenstate of the anhiquilation operator ##a_{\vec{p}}##. Express it in terms of the states of type ##|\vec{p}_1...\vec{p}_N\rangle##

## Homework Equations

$$a^{\dagger}_{\vec{p}_i}|\vec{p}_1...\vec{p}_N\rangle=|\vec{p}_1...\vec{p}_i...\vec{p}_N\rangle$$

$$a_{\vec{p}}|\vec{p}_1...\vec{p}_i...\vec{p}_N\rangle=|\vec{p}_1...\vec{p}_N\rangle$$

$$e^{A}=1+A+\frac{A^2}{2!}+...$$

## The Attempt at a Solution

I tried applying directly the operator:

$$a_{\vec{p}}|c\rangle=a_{\vec{p}}exp(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})|0\rangle$$

Since the other operator is in an exponential, the only thing I could think about is expanding the exponential:

$$a_{\vec{p}}|c\rangle=a_{\vec{p}}(1+\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}}+\frac{1}{2!}(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})^2+...)|0\rangle$$

However, I don't know if I can just introduce the anhiquilation operator randomly into the integral, since it depends on ##\vec{p}## and the integral is an integral on ##d^3p##. Moreover, I have an integral squared which I'm not sure how to reduce. Could you give some advice?

CharlieCW said:

## Homework Statement

Show that the coherent state ##|c\rangle=exp(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})|0\rangle## is an eigenstate of the anhiquilation operator ##a_{\vec{p}}##. Express it in terms of the states of type ##|\vec{p}_1...\vec{p}_N\rangle##

## Homework Equations

$$a^{\dagger}_{\vec{p}_i}|\vec{p}_1...\vec{p}_N\rangle=|\vec{p}_1...\vec{p}_i...\vec{p}_N\rangle$$

$$a_{\vec{p}}|\vec{p}_1...\vec{p}_i...\vec{p}_N\rangle=|\vec{p}_1...\vec{p}_N\rangle$$

$$e^{A}=1+A+\frac{A^2}{2!}+...$$

## The Attempt at a Solution

I tried applying directly the operator:

$$a_{\vec{p}}|c\rangle=a_{\vec{p}}exp(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})|0\rangle$$

Since the other operator is in an exponential, the only thing I could think about is expanding the exponential:

$$a_{\vec{p}}|c\rangle=a_{\vec{p}}(1+\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}}+\frac{1}{2!}(\int \frac{d^3p}{(2\pi)^3}c(\vec{p})a^{\dagger}_{\vec{p}})^2+...)|0\rangle$$

However, I don't know if I can just introduce the anhiquilation operator randomly into the integral, since it depends on ##\vec{p}## and the integral is an integral on ##d^3p##. Moreover, I have an integral squared which I'm not sure how to reduce. Could you give some advice?
You should use different labels for the annihilation operator acting on the coherent state and the ones in the exponential. For example, use
$$a_{\vec{p}'}|c\rangle$$
I hope it is clear: I put a prime on the three-vector.

Now apply this on your expansion of the exponential. Do you know the commutator of ##a_{\vec{p}'}## and ##a^\dagger_{\vec{p}}##?

For the square term, you have to think of it as $$\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3} \frac{d^3p_2}{(2\pi)^3} c(\vec{p_1}) c(\vec{p_2}) a^{\dagger}_{\vec{p_1}}a^{\dagger}_{\vec{p_2}}$$

Now pass the ##a_{\vec{p}'}## through this expression, using commutators. After doing this, you will probably see how it generalizes to all the powers.

Thank you. I see, then the idea was to take differently the labels, so I'm saving this general idea for future problems. Indeed, I calculated the commutator for the operators (for simplicity I'll omit the arrows on the ##\vec{p_i}## so it's easier to see and write):

$$[a_{p'},a^{\dagger}_{p}]=(2\pi)^3\delta(p'-p)$$

Which can be found using the result of each operator on the general (bosonic) particle states given by:

$$a^{\dagger}_p|p_1,...,p_n\rangle=\sqrt{2E_p}|p,p_1,...,p_n\rangle$$

$$a_p|p_1,...p_n\rangle=\sum^{n}_{r=1}(2\pi)^3\delta(p-p_r)|p_1,...\boldsymbol{p_r},...p_n\rangle$$

Where the ##\boldsymbol{p_r}## denotes eliminating the state of particle with momentum ##p_r## (I couldn't find a way to cross the text here, though).

Moving on, expanding the exponential we have:

$$|c\rangle=(1+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p_1)a_{p_1}^{\dagger}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^2}c(p_1)c(p_2)a_{p_1}^{\dagger}a_{p_2}^{\dagger}+...)|0\rangle$$

$$a_{p'}|c\rangle=(a_{p'}+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p_1)a_{p'}a_{p_1}^{\dagger}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^2}c(p_1)c(p_2)a_{p'}a_{p_1}^{\dagger}a_{p_2}^{\dagger}+...)|0\rangle$$

The first term obviously vanishes since ##a_{p'}|0\rangle=0##, and the other commutators are:

$$[a_{p'},a_{p_1}]=(2\pi)^3\delta(p'-p_1)$$

$$[a_{p'},a_{p_1}a_{p_2}]=-a_{p_1}[a_{p'},a_{p_2}]-[a_{p'},a_{p_1}]a_{p_2}=(2\pi)^3(\delta(p'-p_1)a_{p_2}+\delta(p'-p_2)a_{p_1})$$

In general, we can generalize the commutator (I just put here the result, I did the proof in my notes) as:

$$[a_{p'},a_{p_1}...a_{p_n}]=(2\pi)^3\sum_{i=1}^{n} \delta(p'-p_i)\prod_{j\neq i}^{n} a_{p_j}$$

(Edit: typing error, it was actually ##a_{p_j}## not ##p_j##)

Substituting into the expression before:

$$a_{p'}|c\rangle=(\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p_1)(a_{p_1}^{\dagger}a_{p'}+(2\pi)^3(p'-p_1))+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^2}c(p_1)c(p_2)((a_{p_1}^{\dagger}a_{p_2}^{\dagger}a_{p'}+(2\pi)^3(\delta(p'-p_1)a^{\dagger}_{p_2}+\delta(p'-p_2)a^{\dagger}_{p_1}))+...)|0\rangle$$

Again all the terms which have ##a_{p'}## on the right side vanish since ##a_{p'}|0\rangle=0##. Moreover, the delta terms "kill" one integral:

$$a_{p'}|c\rangle=(\frac{c(p')}{1!}+\frac{c(p')}{2!}\int \frac{d^3p_1}{(2\pi)^3}c(p_1)a^{\dagger}_{p_1}+\frac{c(p')}{2!}\int \frac{d^3p_2}{(2\pi)^3}c(p_2)a^{\dagger}_{p_2}+...)|0\rangle=c(p')(\frac{1}{1!}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}c(p_1)a^{\dagger}_{p_1}+\frac{1}{2!}\int \frac{d^3p_2}{(2\pi)^3}c(p_2)a^{\dagger}_{p_2}+...)|0\rangle$$

Which kind of resembles the expression of the coherent state except for the fact that now the creation operators have different labels.

My doubt now is, if I could take all the creation operators with the same momentum ##p##, that is, make ##a^{\dagger}_{p_1}=a^{\dagger}_{p_2}=...a^{\dagger}_{p}##. I ask this because, if I do this, I actually get:

$$a_{p'}|c\rangle=c(p')(\frac{1}{0!}+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p)a^{\dagger}_{p}+\frac{1}{2!}\int \frac{d^3p}{(2\pi)^3}\frac{d^3p}{(2\pi)^3}c(p)c(p)a^{\dagger}_{p}a^{\dagger}_{p}+...)|0\rangle=c(p')|c\rangle$$

Which is the result I wanted. However, I'm not sure if this is a valid step, do you think I can do this?

Last edited:
CharlieCW said:
Thank you. I see, then the idea was to take differently the labels, so I'm saving this general idea for future problems. Indeed, I calculated the commutator for the operators (for simplicity I'll omit the arrows on the ##\vec{p_i}## so it's easier to see and write):

$$[a_{p'},a^{\dagger}_{p}]=(2\pi)^3\delta(p'-p)$$

Which can be found using the result of each operator on the general (bosonic) particle states given by:

$$a^{\dagger}_p|p_1,...,p_n\rangle=\sqrt{2E_p}|p,p_1,...,p_n\rangle$$

$$a_p|p_1,...p_n\rangle=\sum^{n}_{r=1}(2\pi)^3\delta(p-p_r)|p_1,...\boldsymbol{p_r},...p_n\rangle$$

Where the ##\boldsymbol{p_r}## denotes eliminating the state of particle with momentum ##p_r## (I couldn't find a way to cross the text here, though).

Moving on, expanding the exponential we have:

$$|c\rangle=(1+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p_1)a_{p_1}^{\dagger}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^2}c(p_1)c(p_2)a_{p_1}^{\dagger}a_{p_2}^{\dagger}+...)|0\rangle$$

$$a_{p'}|c\rangle=(a_{p'}+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p_1)a_{p'}a_{p_1}^{\dagger}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^2}c(p_1)c(p_2)a_{p'}a_{p_1}^{\dagger}a_{p_2}^{\dagger}+...)|0\rangle$$

The first term obviously vanishes since ##a_{p'}|0\rangle=0##, and the other commutators are:

$$[a_{p'},a_{p_1}]=(2\pi)^3\delta(p'-p_1)$$

$$[a_{p'},a_{p_1}a_{p_2}]=-a_{p_1}[a_{p'},a_{p_2}]-[a_{p'},a_{p_1}]a_{p_2}=(2\pi)^3(\delta(p'-p_1)a_{p_2}+\delta(p'-p_2)a_{p_1})$$

In general, we can generalize the commutator (I just put here the result, I did the proof in my notes) as:

$$[a_{p'},a_{p_1}...a_{p_n}]=(2\pi)^3\sum_{i=1}^{n} \delta(p'-p_i)\prod_{j\neq i}^{n} p_j$$
I guess you meant ##a_{p_j}## at the extreme right.

Substituting into the expression before:

$$a_{p'}|c\rangle=(\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p_1)(a_{p_1}^{\dagger}a_{p'}+(2\pi)^3(p'-p_1))+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^2}c(p_1)c(p_2)((a_{p_1}^{\dagger}a_{p_2}^{\dagger}a_{p'}+(2\pi)^3(\delta(p'-p_1)a^{\dagger}_{p_2}+\delta(p'-p_2)a^{\dagger}_{p_1}))+...)|0\rangle$$

Again all the terms which have ##a_{p'}## on the right side vanish since ##a_{p'}|0\rangle=0##. Moreover, the delta terms "kill" one integral:

$$a_{p'}|c\rangle=(\frac{c(p')}{1!}+\frac{c(p')}{2!}\int \frac{d^3p_1}{(2\pi)^3}c(p_1)a^{\dagger}_{p_1}+\frac{c(p')}{2!}\int \frac{d^3p_2}{(2\pi)^3}c(p_2)a^{\dagger}_{p_2}+...)|0\rangle=c(p')(\frac{1}{1!}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}c(p_1)a^{\dagger}_{p_1}+\frac{1}{2!}\int \frac{d^3p_2}{(2\pi)^3}c(p_2)a^{\dagger}_{p_2}+...)|0\rangle$$

Which kind of resembles the expression of the coherent state except for the fact that now the creation operators have different labels.
Note that the two last terms (with factors of ##1/2!##) are identical so they can be added (the variable of integration is a dummy variable).

My doubt now is, if I could take all the creation operators with the same momentum ##p##, that is, make ##a^{\dagger}_{p_1}=a^{\dagger}_{p_2}=...a^{\dagger}_{p}##. I ask this because, if I do this, I actually get:

$$a_{p'}|c\rangle=c(p')(\frac{1}{0!}+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p)a^{\dagger}_{p}+\frac{1}{2!}\int \frac{d^3p}{(2\pi)^3}\frac{d^3p}{(2\pi)^3}c(p)c(p)a^{\dagger}_{p}a^{\dagger}_{p}+...)|0\rangle=c(p')|c\rangle$$

Which is the result I wanted. However, I'm not sure if this is a valid step, do you think I can do this?
This is NOT the result you want. The result you want is

$$a_{p'}|c\rangle=c(p')(\frac{1}{0!}+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p)a^{\dagger}_{p}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}c(p_1)c(p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_2}+...)|0\rangle=c(p')|c\rangle$$You are in the right direction, you are basically there.

nrqed said:
I guess you meant ##a_{p_j}## at the extreme right.

Indeed, typing error. I edited the correct expression in the previous message, it should be:

$$[a_{p'},a_{p_1}...a_{p_n}]=(2\pi)^3\sum_{i=1}^{n} \delta(p'-p_i)\prod_{j\neq i}^{n} a_{p_j}$$

nrqed said:
Note that the two last terms (with factors of ##1/2!##) are identical so they can be added (the variable of integration is a dummy variable).

Thanks, you're right, since I'm performing an integration on the variables, I'm free to add them as they're dummy variables; this was the step I was missing. Therefore, adding these (I change 1! to 0! just for convenience):

$$a_{p'}|c\rangle=c(p')(\frac{1}{0!}+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p)a^{\dagger}_{p}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}c(p_1)c(p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_2}|0\rangle+...)|0\rangle=c(p)|0\rangle$$

Where the last term was obtained by analyzing the cubic term of the original expansion and using the commutator written above:

$$\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_1)c(p_2)c(p_3)a_{p'}a^{\dagger}_{p_1}a^{\dagger}_{p_2}a^{\dagger}_{p_3}|0\rangle=\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_1)c(p_2)c(p_3)(a^{\dagger}_{p_1}a^{\dagger}_{p_2}a^{\dagger}_{p_3}a_{p'}+(2\pi)^3\delta(p'-p_1)a^{\dagger}_{p_2}a^{\dagger}_{p_3}+(2\pi)^3\delta(p'-p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_3}+(2\pi)^3\delta(p'-p_3)a^{\dagger}_{p_1}a^{\dagger}_{p_2})|0\rangle$$

Again, the term ##a_{p'}|0\rangle## vanishes and the Dirac deltas cancel one integral, so we get:

$$c(p')(\frac{1}{3!}\int \frac{d^3p_2}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_2)c(p_3)a^{\dagger}_{p_2}a^{\dagger}_{p_3}|0\rangle+\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_1)c(p_3)a^{\dagger}_{p_1}a^{\dagger}_{p_3}|0\rangle+\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}c(p_1)c(p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_2}|0\rangle)=\frac{c(p')}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}c(p_1)c(p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_2}|0\rangle$$

Which again can be added since the variables are just dummy variables. It is easy to see this pattern continues, so we can generalise the result and thus the proof is complete.

Thanks, the problem was solved and I gained some interesting insight in how to treat exponentials which will prove useful later on.

CharlieCW said:
Indeed, typing error. I edited the correct expression in the previous message, it should be:

$$[a_{p'},a_{p_1}...a_{p_n}]=(2\pi)^3\sum_{i=1}^{n} \delta(p'-p_i)\prod_{j\neq i}^{n} a_{p_j}$$
Thanks, you're right, since I'm performing an integration on the variables, I'm free to add them as they're dummy variables; this was the step I was missing. Therefore, adding these (I change 1! to 0! just for convenience):

$$a_{p'}|c\rangle=c(p')(\frac{1}{0!}+\frac{1}{1!}\int \frac{d^3p}{(2\pi)^3}c(p)a^{\dagger}_{p}+\frac{1}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}c(p_1)c(p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_2}|0\rangle+...)|0\rangle=c(p)|0\rangle$$

Where the last term was obtained by analyzing the cubic term of the original expansion and using the commutator written above:

$$\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_1)c(p_2)c(p_3)a_{p'}a^{\dagger}_{p_1}a^{\dagger}_{p_2}a^{\dagger}_{p_3}|0\rangle=\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_1)c(p_2)c(p_3)(a^{\dagger}_{p_1}a^{\dagger}_{p_2}a^{\dagger}_{p_3}a_{p'}+(2\pi)^3\delta(p'-p_1)a^{\dagger}_{p_2}a^{\dagger}_{p_3}+(2\pi)^3\delta(p'-p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_3}+(2\pi)^3\delta(p'-p_3)a^{\dagger}_{p_1}a^{\dagger}_{p_2})|0\rangle$$

Again, the term ##a_{p'}|0\rangle## vanishes and the Dirac deltas cancel one integral, so we get:

$$c(p')(\frac{1}{3!}\int \frac{d^3p_2}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_2)c(p_3)a^{\dagger}_{p_2}a^{\dagger}_{p_3}|0\rangle+\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_3}{(2\pi)^3}c(p_1)c(p_3)a^{\dagger}_{p_1}a^{\dagger}_{p_3}|0\rangle+\frac{1}{3!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}c(p_1)c(p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_2}|0\rangle)=\frac{c(p')}{2!}\int \frac{d^3p_1}{(2\pi)^3}\frac{d^3p_2}{(2\pi)^3}c(p_1)c(p_2)a^{\dagger}_{p_1}a^{\dagger}_{p_2}|0\rangle$$

Which again can be added since the variables are just dummy variables. It is easy to see this pattern continues, so we can generalise the result and thus the proof is complete.

Thanks, the problem was solved and I gained some interesting insight in how to treat exponentials which will prove useful later on.
Good job!

## 1. What are coherent states for Klein-Gordon field?

Coherent states for Klein-Gordon field are quantum states that describe a system in a state of minimal uncertainty. They are also known as "squeezed states" as they have a narrower uncertainty in one variable while the uncertainty in the conjugate variable is increased.

## 2. How are coherent states for Klein-Gordon field different from other quantum states?

Coherent states for Klein-Gordon field are different from other quantum states in that they have a well-defined classical limit. This means that in the classical limit, the expectation values of position and momentum for a coherent state match those of a classical system.

## 3. What are the applications of coherent states for Klein-Gordon field?

Coherent states for Klein-Gordon field have applications in quantum optics, quantum information, and quantum computing. They are also used in the study of Bose-Einstein condensates and in the description of quantum states of light.

## 4. How are coherent states for Klein-Gordon field constructed?

Coherent states for Klein-Gordon field are constructed by applying a displacement operator to the vacuum state of the system. This displacement operator shifts the vacuum state in phase space to create a state with minimal uncertainty.

## 5. Can coherent states for Klein-Gordon field be experimentally observed?

Yes, coherent states for Klein-Gordon field have been experimentally observed in various systems, including trapped ions, superconducting circuits, and optical systems. These experiments confirm the theoretical predictions of these states and their properties.

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