A rather weird form of a coherent state

[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?

 

[king vitamin]

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]

Thank you for the reply!

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?

 

[TSny]

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.

 

[TSny]

$$[a,p^k]=k\Gamma^k (a^\dagger - a)^{k-1}= k\Gamma p^{k-1}$$

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.