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Homework Statement:

Given the quantumoptics 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
$$X=\frac{a+a^*}{2}$$
$$P=\frac{aa^*}{2i}$$
$$a\alpha \rangle = \alpha \alpha \rangle$$
Relevant Equations:
 $$\alpha \rangle = \exp \Big(\frac{\alpha^2}{2}\Big) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} n \rangle$$
I've tried to square and compare ##\Delta X## and ##\Delta P## 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##
More can be found in Problem 1.1 Mandl & Shaw; I also attached the complete solution of it.
Thanks
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##
More can be found in Problem 1.1 Mandl & Shaw; I also attached the complete solution of it.
Thanks
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