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Coherent states of Quantum harmonic oscillator .

The Hamiltonian of Quantum harmonic oscillator is $H=(a^+ a+\frac{1}{2})\hbar \omega$,$a=\sqrt{\frac{m \omega}{2 \hbar}}(\hat{x}+\frac{i \hat{p}}{m \omega})$,$\hat{N}=a^+ a$

The Coherent states are defined as eigenstates of $a$,we mark it $$a|\lambda \rangle=\lambda|\lambda \rangle$$ In $N$-representation,we can show that $$|\lambda \rangle=\sum_n c_n |n\rangle ,c_n=\frac{\lambda ^n}{\sqrt{n!}}e^{-\frac{|\lambda|^2}{2}}$$

my question:

  1. can we give the exact value of $\lambda$?

  2. In N-representation,the matrix representation of $a $ is \begin{equation} \left( \begin{matrix} 0&\sqrt{1}&0&0&0&\cdots\\ 0& 0&\sqrt{2}&0&0&\cdots \\ 0& 0&0&\sqrt{3}&0&\cdots \\ 0& 0&0&0&\sqrt{4}&\cdots \\ 0& 0&0&0&0&... \\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{matrix} \right) \end{equation} I want to calculate the eigenvalues of it. But all eigenvalues are $0$. Is it the reason that on the finite dimension?

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  1. With the number-basis coefficients as you have (correctly) defined them, $|\lambda⟩$ is an eigenstate of $a$ for all complex-valued $\lambda\in\mathbb C$. This can sound weird, but remember that the annihilation operator is not self-adjoint and it is not normal, so the spectral properties can come from a much wider set of possibilities than, say, a compact self-adjoint operator.

  2. In the $N$ representation, the annihilation operator takes the form of the infinite matrix you give. If you truncate it, you get an interesting and related object, but it is no longer the same object.

    In particular, the truncation to $N$ photons or less turns the annihilation into a single Jordan block with eigenvalue $\lambda=0$; this then means that no other eigenvalues will be present. This behaviour is perfectly normal and it is entirely due to the basis truncation.

Shameless plug: my undergraduate thesis has a deeper exploration of these themes - if you can read physics in Spanish ;-).

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  • $\begingroup$ could I think like this ?$\langle \lambda _1 |\lambda_2\rangle =e^{-\frac{|\lambda_1|^2+|\lambda_2|^2-2\lambda_{1}^{*}\lambda_2}{2}}$,not$0$,it can't give the constraint of the $\lambda$ $\endgroup$ – Young1997 Jan 10 '18 at 13:48
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A good way to think of $\lambda$ is to consider the coherent state as a displaced harmonic oscillator ground state. Writing $\lambda=\lambda_r+i\lambda_i$ with $\lambda_{r,i}$ the real and imaginary parts of $\lambda$ respectively, $$ \langle\lambda \vert \hat x\vert \lambda \rangle=\sqrt{\frac{2\hbar}{m\omega}}\lambda_r\, ,\qquad \langle\lambda \vert \hat p\vert \lambda \rangle=\sqrt{ 2m\hbar\omega}\lambda_i\, . $$ Clearly the possible displacements in the $(x,p)$ plane are unbounded so there is no restriction on the possible values of $\lambda_r$ and $\lambda_i$.

Since the harmonic oscillator ground state is a Gaussian shape, the probability density in $x$ is concentrated not about the origin but about $\lambda_r$, and the probability in $p$ is concentrated about $\lambda_i$. This is made evident by looking at the Wigner function of a coherent state. Thus the truncation of basis of harmonic oscillator states $\{\vert n\rangle\}$ is usually done on physical grounds after a value of $n$ sufficiently large so as to capture "most" of the probability density.

Note that the coherent state is not an eigenstate of the harmonic oscillator hamiltonian so it's probability density will be time-dependent.

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