# A rather weird form of a coherent state

• Markus Kahn
In summary, the coherent state is usually defined as $$|\alpha \rangle =e^{-{|\alpha |^{2} \over 2}}\sum _{n=0}^{\infty }{\alpha ^{n} \over {\sqrt {n!}}}|n\rangle =e^{-{|\alpha |^{2} \over 2}}e^{\alpha {\hat {a}}^{\dagger }}|0\rangle,$$ and it can be pretty easily shown that ##\vert \alpha\rangle ## is indeed an eigenstate of ##a##.
Markus Kahn
Homework Statement
Show that the coherent state $$\vert \alpha\rangle = e^{\alpha p}\vert 0\rangle$$ is an eigenstate of the annihilation operator of the harmonic oscillator.
Relevant Equations
Hamiltonian of the harmonic oscillator: $$H = \frac{p^2}{2m}+ \frac{m\omega^2}{2}x^2.$$
As far as I know we can express the position and momentum operators in terms of ladder operators in the following way
{\begin{aligned}{ {x}}&={\sqrt {{\frac {\hbar }{2}}{\frac {1}{m\omega }}}}(a^{\dagger }+a)\\{{p}}&=i{\sqrt {{\frac {\hbar }{2}}m\omega }}(a^{\dagger }-a)~.\end{aligned}}.
With this the coherent state is usually defined as
$$|\alpha \rangle =e^{-{|\alpha |^{2} \over 2}}\sum _{n=0}^{\infty }{\alpha ^{n} \over {\sqrt {n!}}}|n\rangle =e^{-{|\alpha |^{2} \over 2}}e^{\alpha {\hat {a}}^{\dagger }}|0\rangle,$$
and it can be pretty easily shown that ##\vert \alpha\rangle ## is indeed an eigenstate of ##a##. But if we take the definition given in the exercise
$$\vert \alpha\rangle = e^{\alpha p}\vert 0\rangle$$
I have some doubts if this is really an eigenstate...
1. First of, ##p## has units of momentum, so how exactly can this be written in an exponential function? Do we assume that ##\alpha## has units of inverse momentum?
2. Secondly, how exactly does one show that this ##\vert \alpha\rangle## is an eigenstate of ##a##? My attempt was straight forward $$a \vert\alpha\rangle = a\sum\limits_{k=0}^\infty \frac{1}{k!}(\alpha p)^k \vert 0\rangle =\sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k ap^k \vert 0\rangle.$$ To continue form here on I would need to to rewrite ##ap^k## using the Binomial theorem, which resulted in $$ap^k = \Gamma^k a (a-a^\dagger)^k = \Gamma^k a \sum _{n=0}^{k}{k \choose n}a^{n}(-a^\dagger)^{k-n}.$$ Now the question arises what exactly ##a(a^\dagger)^{k-n}## is and I think it should be ##a(a^\dagger)^{k-n}=(k-n)(a^{\dagger})^{k-n-1}+(a^\dagger)^{k-n}a##, but I wasn't able to prove this yet.
At this point I stopped since I was starting to doubt my approach here... Is this the wrong way to show that ##e^{\alpha p}\vert 0\rangle## is an eigenstate of ##a##? If so, can somebody maybe help me out a bit in how to prove this here?

There are a few ways to do this, and your approach (including your idea of proving the formula $a(a^\dagger)^{k-n}=(k-n)(a^{\dagger})^{k-n-1}+(a^\dagger)^{k-n}a$, which is true) is one such way. To begin to prove that formula, try playing with small integer values of the exponent of $a^{\dagger}$ to find the pattern.

You're right that now $\alpha$ will need units. Since the eigenvalue of $a$ needs to be dimensionless, this means it won't simply be $\alpha$ anymore.

Markus Kahn

Now I was able to prove the statement: ##a(a^\dagger)^m = m(a^\dagger)^{m-1} +(a^\dagger)^ma## for ##m\in\mathbb{N}_0##.
Proof: For ##m=0## the statement is trivially true. Let us assume the statement is true for ##m=n##. We then have
\begin{align*}a(a^\dagger)^{n+1}&=aa^\dagger(a^\dagger)^n = (a^\dagger)^n+a^\dagger a (a^\dagger)^n= (a^\dagger)^n+a^\dagger \left(n(a^\dagger)^{n-1}+(a^\dagger)^na\right)\\&=(n+1)(a^\dagger)^n+(a^\dagger)^{n+1}a\end{align*}

After proving this I tried to continue but encountered a bit of an issue. What I want in the end is ##a\vert\alpha\rangle = \alpha\vert\alpha\rangle##, which is equivalent to
$$a\vert\alpha\rangle=\alpha \vert\alpha\rangle = \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^{k+1}p^k \vert 0\rangle.\tag{*}$$
If I now insert the above proof into my previous result I get
\begin{align*}a\vert\alpha\rangle& = \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k ap^k \vert 0\rangle\\ &= \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k \left[\Gamma^k a \sum _{n=0}^{k}{k \choose n}a^{n}(-a^\dagger)^{k-n}\right]\vert 0\rangle \\ &=\sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k \left[\Gamma^k \sum _{n=0}^{k}{k \choose n} a^n (-1)^{k-n}\left((k-n)(a^{\dagger})^{k-n-1}+(a^\dagger)^{k-n}a\right)\right]\vert 0\rangle .\end{align*}
The term with ##(a^\dagger)^{k-n}a## can be ignored since ##a\vert 0\rangle=0.## We therefore end up with
$$a\vert\alpha\rangle = \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k \left[\Gamma^k \sum _{n=0}^{k}{k \choose n} a^n (-1)^{k-n}(k-n)(a^{\dagger})^{k-n-1}\right]\vert 0\rangle.$$
If I compare this equation to eq. ##(*)## I need to show that
$$\sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^{k+1}p^k = \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k \left[\Gamma^k \sum _{n=0}^{k}{k \choose n} a^n (-1)^{k-n}(k-n)(a^{\dagger})^{k-n-1}\right].$$
After looking at this for a while I'm pretty sure I need to perfrom an index shift at some point, but the bigger problem for me right now is that I can't piece together the momentum operator ##p## on the righthand-side due to the Binomial coefficient. Can you maybe help me here?

Markus Kahn said:
What I want in the end is ##a\vert\alpha\rangle = \alpha\vert\alpha\rangle##
Since the operator ##a## is dimensionless, the eigenvalue will be dimensionless. So, the eigenvalue cannot be ##\alpha## since ##\alpha## has dimensions of inverse momentum.

$$a\vert\alpha\rangle = \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k ap^k \vert 0\rangle$$
Note that ##a p^k \vert 0 \rangle = [a, p^k] \vert 0 \rangle##. The commutator ##[a, p^k]## reduces to a fairly simple expression. First work out ##[a, p]##, ##[a, p^2]##, and ##[a, p^3]## to see the pattern.

Markus Kahn
Thanks for the help. Let ##\Gamma = i{\sqrt {{\frac {\hbar }{2}}{{m\omega }}}}## such that ##p=\Gamma (a^\dagger -a)##.
• ##[a,p]=\Gamma##
• ##[a,p^2]=2\Gamma^2 (a^\dagger-a)##
• ##[a,p^3]=\Gamma^3 (4{a^\dagger}^2-3a^2-3a^\dagger a -aa^\dagger)##
Now the last one is probably wrong, since there is no pattern I can recognize there (I checked it multiple times and always got this result, so I'm probably doing something systematically wrong)... If I had to guess based on the first two, I'd say the pattern is something along the lines of
$$[a,p^k]=k\Gamma^k (a^\dagger - a)^{k-1}= k\Gamma p^{k-1}$$
Proof. For ##k=0## the statement is trivially true. Assume the statement is true for ##k=n##. We then have
$$[a,p^{n+1}]=p^n[a,p]+[a,p^{n}]p = p^n\Gamma +n \Gamma p^n = (n+1)\Gamma p^n.$$
So it seems my guess was right.

From your input I'm now able to get
\begin{align*}a\vert\alpha\rangle &= \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k ap^k \vert 0\rangle=\sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k [a,p^k] \vert 0\rangle= \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k k\Gamma p^{k-1} \vert 0\rangle\\ &=\Gamma \sum\limits_{k=0}^\infty \frac{1}{(k-1)!}\alpha^k p^{k-1} \vert 0\rangle = \alpha\Gamma \sum\limits_{k=1}^\infty \frac{1}{k!}\alpha^k p^k \vert 0\rangle=\alpha\Gamma\vert \alpha\rangle, \end{align*}
where I'm not really sure about the last step, since I can't really justify why the first term in the sum is zero.

P.S. Am I correct in assuming that my first approach therefore couldn't produce the desired result.

Markus Kahn said:
$$[a,p^k]=k\Gamma^k (a^\dagger - a)^{k-1}= k\Gamma p^{k-1}$$
Yes

From your input I'm now able to get
\begin{align*}a\vert\alpha\rangle &= \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k ap^k \vert 0\rangle=\sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k [a,p^k] \vert 0\rangle= \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k k\Gamma p^{k-1} \vert 0\rangle\\ &=\Gamma \sum\limits_{k=0}^\infty \frac{1}{(k-1)!}\alpha^k p^{k-1} \vert 0\rangle = \alpha\Gamma \sum\limits_{k=1}^\infty \frac{1}{k!}\alpha^k p^k \vert 0\rangle=\alpha\Gamma\vert \alpha\rangle, \end{align*}
where I'm not really sure about the last step, since I can't really justify why the first term in the sum is zero.
Go to the first equality where you have
$$a\vert\alpha\rangle = \sum\limits_{k=0}^\infty \frac{1}{k!}\alpha^k ap^k \vert 0\rangle$$
You can see that k = 0 yields zero contribution.

P.S. Am I correct in assuming that my first approach therefore couldn't produce the desired result.
Your first approach might work. But it was looking messy.

Markus Kahn

## 1. What is a coherent state?

A coherent state is a quantum state of a system that has properties similar to a classical wave or oscillator. It is a state of matter where all the particles are in the same quantum state, resulting in a state that behaves like a classical wave.

## 2. How is a coherent state different from other quantum states?

A coherent state is different from other quantum states in that it exhibits both quantum and classical properties. It has a well-defined phase and amplitude, similar to a classical wave, but also follows the laws of quantum mechanics.

## 3. What makes "A rather weird form of a coherent state" unique?

The "rather weird form" of a coherent state refers to a state that is not described by a simple Gaussian distribution, but rather a more complex distribution. This form is unique because it exhibits non-classical properties, such as squeezing and entanglement, while still maintaining some classical characteristics.

## 4. What are some potential applications of "A rather weird form of a coherent state"?

One potential application of "A rather weird form of a coherent state" is in quantum computing, where these states could be used as qubits to perform complex calculations. Another potential application is in quantum communication, where these states could be used for secure information transfer.

## 5. Can "A rather weird form of a coherent state" be observed in nature?

Yes, "A rather weird form of a coherent state" has been observed in various physical systems, such as trapped ions, superconducting circuits, and optical systems. These states are created and manipulated in controlled laboratory environments, but they have also been observed in natural phenomena, such as in the behavior of certain types of crystals.

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