1. In this problem, we will derive the expression for the coherent state for bosons. This state is an eigenstate of the boson annihilation operator, and so satisfies \(\)\bhat\ket{z} = z\ket{z}.\(\) It must also satisfy that \(\bra{z}\ket{z} = 1\). Using these two properties, derive an expression for the boson coherent state in terms of the basis \(\ket{n}\) signifying \(n\) particles in a particular state.
    As a first step, we will write \(\)\ket{z} = \sum_{n=0}^\infty a_{nz}\ket{n}.\(\) Our goal is to calculate \(a_{nz}\) such that \(\ket{z}\) satisfies the above two conditions. Our first step is to recast the two conditions into constraints on the coefficients \(a_{nz}\). Using the eigenstate condition, we find that $$\begin{align} \bhat\ket{z} ={}& \bhat\sum_{n=0}^\infty a_{nz}\ket{n} \\ ={}& \sum_{n=0}^\infty a_{nz}\sqrt{n}\ket{n-1} \\ ={}& \sum_{n=1}^\infty a_{nz}\sqrt{n}\ket{n-1} \\ z\sum_{n=0}^\infty a_{nz}\ket{n} ={}& \sum_{n=0}^\infty a_{n+1,z}\sqrt{n+1}\ket{n} \end{align}$$ and by matching coefficients, we find that \(\)za_{nz} = \sqrt{n+1}a_{n+1,z}\(\) which gives us a recursion relation capturing the first constraint on our coefficients: \(\)\frac{a_{n+1,z}}{a_{nz}} = \frac{z}{\sqrt{n+1}}.\(\) The second constraint of normalization simply yields that \(\)\sum_{n=0}^\infty |a_{nz}|^2 = 1.\(\) Equipped with these constraints, we start by solving the recursion relation, which gives \(\)a_{nz} = \frac{z^n}{\sqrt{n!}}a_{0z}.\(\) Now plugging this into the normalization condition, we find that \(\)\sum_{n=0}^\infty \frac{|z|^{2n}}{n!}|a_{0z}|^2 = 1 = e^{|z|^2}\(\) and find that \(\)a_{0z} = e^{-|z|^2/2}.\(\) Lastly, we write $$\begin{align} \ket{z} ={}& e^{-|z|^2/2} \sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}}\ket{n} \\ ={}& e^{-|z|^2/2}\sum_{n=0}^\infty \frac{z^n}{n!} \sqrt{n!}\ket{n} \\ ={}& e^{-|z|^2/2}\sum_{n=0}^\infty \frac{z^n}{n!} (\bhat^\dagger)^n \ket{0} \\ ={}& e^{-|z|^2/2} e^{z\bhat^\dagger}\ket{0} \end{align}$$