# Coherent states - scalar product [closed]

$$\newcommand\norm[1]{\lVert#1\rVert}$$ $$\newcommand\ket[1]{|#1\rangle}$$ $$\newcommand\mean[1]{\langle #1\rangle}$$ $$\newcommand\braket[2]{\langle #1|#2\rangle}$$ $$\newcommand\ketbra[2]{|#1\rangle\langle #2|}$$

One shows that the wave-function of the coherent states of a system can be written in terms of a Gaussian function, as follows $$$$\label{general wave-function} \psi(x) = \frac{\exp(-\frac{i}{2\pi}\mean{X}\mean{P})}{\left(\pi l^2\right)^{1/4}}\exp(-\frac{\left(x-\mean{X}\right)^2}{2l^2}+\frac{i\mean{P}x}{\hbar})$$$$ where we have $$l=\sqrt{2}\Delta X$$.

Independently, one shows that $$\psi(x) = \braket{x}{\psi}$$ and that $$\psi(x)$$ is indeed a coherent state, in the sense that $$\ket{\psi}=\ket{\alpha}$$ for any $$\alpha$$ in the complex plane.

One might wonder what $$\braket{0}{\alpha}$$ could possibly mean. Using the creation and annihilation operators introduced to solve the Quantum Harmonic Oscillator, one shows that the solution to this question is given by $$\braket{0}{\alpha} = e^{-\frac{1}{2}\norm{\alpha}^2}$$.

My question concerns the proof that $$\braket{0}{\alpha}=e^{-\frac{1}{2}\norm{\alpha}^2}$$, using the wave-function introduced above. Here is how I tried to do it.

I start by using the well-known relation \begin{align} \mathbb{I} &= \int_{-\infty}^{\infty}dx \; \ketbra{x}{x}, \end{align} which I shall refer to as the completing relationship.

\begin{align} \braket{x}{\alpha} &= \int_{-\infty}^{\infty}dx \; \braket{0}{x}\braket{x}{\alpha} \end{align}

Let us note that $$\braket{0}{x}$$ is the fundamental state of the Harmonic Oscillator, which is an established function (just plug-in $$x=0$$ in the first equation). One finds oneself having to deal with a Gaussian integral, which one can readily resolve as

\begin{align} \braket{0}{\alpha} &= \exp(-\frac{i}{2\hbar}\mean{X}\mean{P}-\frac{\mean{X}^2}{2l^2}+\frac{\alpha^2}{2}) = \exp(-\frac{\alpha\mean{X}}{\sqrt{2}l}+\frac{\norm{\alpha}^2}{2}) \end{align} where I used the fact that $$\alpha = \mean{a} = \frac{1}{\sqrt{2}}\left(\frac{\mean{X}}{l}+\frac{il\mean{P}}{\hbar}\right)$$.

I don't know how I can go further, without previously knowing the expected result. Can anyone help?

• There is a relationship between $\alpha$ and $\langle X\rangle$ and $\langle P\rangle$ which can be derived using the fact that $\alpha$ is an eigenvalue of the lowering operator and that the lowering operator can be written in term so the operators $\hat{X}$ and $\hat{P}$. Use that! Commented Nov 17, 2021 at 16:27
• @march : I updated the question following your advice. Thanks a lot! Commented Nov 17, 2021 at 16:33
• Could you check my solution below? @march Commented Jan 2, 2022 at 16:14

$$\renewcommand\inf{\infty}$$ $$\renewcommand\ket[1]{|#1\rangle}$$ $$\renewcommand\ketbra[2]{|#1\rangle\langle#2|}$$ $$\renewcommand\braket[2]{\langle #1|#2 \rangle}$$ $$\renewcommand\mean[1]{\langle #1 \rangle}$$ $$\renewcommand\expval[1]{\mean{#1}}$$ $$\renewcommand\h{\hbar}$$ $$\renewcommand\norm[1]{||#1||}$$

I have found two methods to answer this. The most general method uses the wave-function associated with a coherent state. It is given by $$$$\braket{x}{\psi} = \psi(x) = \frac{\exp(-\frac{i}{2\pi}\mean{X}\mean{P})}{\left(\pi l^2\right)^{1/4}}\exp(-\frac{\left(x-\mean{X}\right)^2}{2l^2}+\frac{i\mean{P}x}{\h})$$$$ where $$l = 2\Delta X$$.

Resolution using a gaussian integral

In the X-representation of Quantum Mechanics, one has the closure relation $$$$\int_{-\inf}^{+\inf}dx \; \ketbra{x}{x}$$$$

Using this property, one was find a solution to $$\ketbra{0}{\alpha}$$. Indeed, $$$$\ketbra{0}{\alpha} = \int_{-\inf}^{+\inf} dx \; \braket{0}{x}\braket{x}{\alpha}$$$$ The term $$\braket{x}{0}$$ is known as the fundamental state of a coherent state. The second ($$\braket{x}{\alpha}$$) is the wave-function given here-above. Doing the math, one finds \begin{align*} \braket{0}{\alpha} &= \int_{-\infty}^{+\infty}dx \; \left(\frac{\exp(-\frac{x^2}{2l^2})}{\left(\pi l^2\right)^{1/4}}\right)\left[\frac{\exp(-\frac{i}{2\pi}\expval{X}\expval{P})}{\left(\pi l^2\right)^{1/4}}\exp(-\frac{\left(x-\expval{X}\right)^2}{2l^2}+\frac{i\expval{P}x}{\hbar})\right]\\ &= \frac{\exp(-\frac{i}{2\pi}\expval{X}\expval{P}-\frac{\expval{X}^2}{2l^2})}{l\sqrt{\pi}}\int_{-\infty}^{+\infty}dx \; \exp(-\left(\frac{x}{l}\right)^2+x\underbrace{\left(\frac{\expval{X}}{l^2}+\frac{i\expval{P}}{\h}\right)}_{\frac{\sqrt{2}\alpha}{l}}) \end{align*}

We then have a Gaussian integral, which can be solved explicitely into: \begin{align*} \braket{0}{\alpha} &= \frac{\exp(-\frac{i}{2\pi}\expval{X}\expval{P}-\frac{\expval{X}^2}{2l^2})}{l\sqrt{\pi}}\sqrt{\pi}l\exp(\frac{\alpha^2}{2}) = e^{-\frac{\norm{\alpha}^2}{2}} \end{align*}

Resolution using properties of coherent states

For any complex number $$\alpha$$, one has that $$\ket{\alpha} = e^{-\norm{\alpha}^2/2}\sum_n \frac{\alpha^n}{\sqrt{n!}}\ket{n}$$. Using this property, the proof requested is trivial: $$$$\braket{0}{\alpha} = e^{-\frac{1}{2}\norm{\alpha}^2}\sum_{n\geq 0} \frac{\alpha^n}{\sqrt{n!}}\ketbra{0}{n} = e^{-\frac{1}{2}\norm{\alpha}^2}$$$$ where one uses the fact that $$\braket{0}{n} = 0$$ for all $$n\neq 0$$.