# Quantum Optics Question Involving Coherent States

Given the quantum-optics coherent states

$$|\alpha \rangle = \exp \Big(-\frac{|\alpha|^2}{2}\Big) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle$$

Show that

$$\langle (\Delta X)^2 \rangle_{\alpha} = \langle (\Delta P)^2 \rangle_{\alpha} = \frac 1 4$$

Where

$$|n \rangle$$ are the photon number states

$$X=\frac{a+a^*}{2}$$

$$P=\frac{a-a^*}{2i}$$

$$a|\alpha \rangle = \alpha |\alpha \rangle$$

My attempt:

$$|\alpha \rangle = \exp \Big(-\frac{|\alpha|^2}{2}\Big) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle$$

I've tried to square $$\Delta X$$ and $$\Delta P$$ compare and but they are not equal

I have to say I am pretty lost here and a hint would be appreciated.

I have studied coherent states and I know how to proof some properties related to it.

For instance, I see how to proof that the state is normalized:

$$\langle \alpha|\alpha \rangle = \exp(-|\alpha|^2) \Big(\sum_{m=0}^{\infty} \langle m| \frac{\alpha^{*m}}{\sqrt{m!}}\Big) \Big(\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n!}}|n\rangle \Big)$$

Based on $$\langle m | n \rangle = \delta_{mn}$$ we indeed get $$\langle \alpha | \alpha \rangle = 1$$

Hint: What is $$\langle\alpha |a| \alpha \rangle$$? (you can find it from the fact that the coherent state is an eigenstate of $$a$$)
Then, what is $$\langle\alpha |a^\dagger| \alpha \rangle$$? What are the values of $$\langle\alpha| a^\dagger a| \alpha \rangle$$, $$\langle\alpha|a a^\dagger| \alpha \rangle$$, $$\langle\alpha| {a^\dagger} ^2| \alpha \rangle$$ and $$\langle\alpha| a^2| \alpha \rangle$$?
Now, you can find $$\langle\alpha |X| \alpha \rangle$$, $$\langle\alpha |P| \alpha \rangle$$, $$\langle\alpha |X^2| \alpha \rangle$$, $$\langle\alpha |P^2| \alpha \rangle$$. (you have already written expressions of $$X$$ and $$P$$ in terms of $$a$$ and $$a^\dagger$$).
What are the expectation values of $$\Delta X$$ and $$\Delta P$$ in terms of these quantities?