Getting a strange result here. I am studying the Quantum Harmonic oscillator and dealing with coherent states.

I know that I can expand the position operator as $x = \sqrt{ \frac{\hbar}{2 m \omega} } (a^{\dagger}+ a)$, where $a$ and $a^{\dagger}$ are the usual ladder operators, satisfying $[a,a^{\dagger}]=\mathbb{I}$.

For some complex number $\lambda \in \mathbb{C}$, a coherent state is a normalized ket such that $a \left| \lambda\right> = \lambda\left|\lambda\right>$. This implies that $a^{\dagger}\left|\lambda\right> = \lambda^{\ast}\left|\lambda\right>$.

Here is my issue:

\begin{align} \left< \lambda | x^{2} | \lambda \right> &= \frac{\hbar}{2m\omega} \left< \lambda | \left( a^{\dagger} a^{\dagger} + a^{\dagger} a + a a^{\dagger} + a a \right) | \lambda \right> \\&= \frac{\hbar}{2m\omega} \left( (\lambda^{\ast})^{2} + 2|\lambda|^{2} + \lambda^{2} \right) \end{align}

HOWEVER, I can use the identity $[a,a^{\dagger}]=\mathbb{I}$, to say that $aa^{\dagger} = \mathbb{I} + a^{\dagger} a$, which means that equivalently, I can write

\begin{align} \left< \lambda | x^{2} | \lambda \right> &= \frac{\hbar}{2m\omega} \left< \lambda | \left( a^{\dagger} a^{\dagger} + 2 a^{\dagger} a + \mathbb{I} + a a \right) | \lambda \right> \\ &= \frac{\hbar}{2m\omega} \left( (\lambda^{\ast})^{2} + 2 |\lambda|^{2} + 1 + \lambda^{2} \right) \end{align}

How is it possible that I am getting two different answers here? The second one is supposedly the right one, but how? I don't understand the source of the discrepancy - I am very confused.


The coherent state is only an eigenstate of the annihilation operator, not of the creation operator. Because the lowest mode is not created.

in other words, $a\left|\lambda\right> = \lambda\left|\lambda\right>$ implies $\left<\lambda\right| a^\dagger = \left<\lambda\right| \lambda^*$

Thus, you cannot directly calculate $\left<\lambda|aa^\dagger|\lambda\right>$, and have to use your second form.

  • $\begingroup$ Thank you! This makes perfect sense, so I can only apply $a^{\dagger}$ to $<\lambda|$. Great $\endgroup$ – Greg.Paul Oct 10 '16 at 11:34
  • 3
    $\begingroup$ Use \rangle and \langle: $\langle \text{physics} | \text{happiness} \rangle = \delta(\text{not} - \text{possible})$. $\endgroup$ – QuantumBrick Oct 10 '16 at 11:36

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