- #1

Markus Kahn

- 112

- 14

- 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...

$${\begin{aligned}{ {x}}&={\sqrt {{\frac {\hbar }{2}}{\frac {1}{m\omega }}}}(a^{\dagger }+a)\\{{p}}&=i{\sqrt {{\frac {\hbar }{2}}m\omega }}(a^{\dagger }-a)~.\end{aligned}}.$$

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...

- 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?
- 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.