Let's say I have a quantum harmonic oscillator $H = \omega a^\dagger a$, where $a^\dagger$ is the raising operator and $a$ is the lowering operator and $H |n\rangle = \omega n |n\rangle$.

Now assume that we add some kind of shift so that we get the following Hamiltonian (up to a constant)

$$H = \omega a^\dagger a + \omega (a + a^\dagger) = \omega (a+1)^\dagger (a+1)- \omega $$

Is it possible to express the eigenstates of this shifted harmonic oscillator with respect to the old eigenstates?

It should be possible by using a coherent state I guess, because a coherent state can be seen as kind of a 'shifted' number state. Do you have any ideas/experiences on how to do this?

The background of the question is that I want do diagonalize a similar hamiltonian of two coupled systems, where the coupling is of the order of $\omega$.


2 Answers 2


The key concept to look for is displaced number states. These are, quite simply, the number states $|n⟩$, moved by the displacement operator $$D(\alpha)=\exp\left[\alpha a^\dagger-\alpha^*a\right]$$ to some point $\alpha=x+ip$ on the complex phase space. The ground state of a harmonic oscillator which has been displaced to a real $\alpha=x$ is, as you note, the coherent state $$|\alpha⟩=D(\alpha)|0⟩.$$ The rest of the eigenstates are, similarly, $$|\alpha,n⟩=D(\alpha)|n⟩.$$

The simplest way to prove this is to use the commutation/displacement relations between the displacement operator and the ladder operators, $$aD(\alpha)=D(\alpha)(a+\alpha)\quad\text{and}\quad a^\dagger D(\alpha)=D(\alpha)(a^\dagger+\alpha^*).$$ You can then transform the eigenvalue equation $a^\dagger a|n⟩=n|n⟩$ into $$ \begin{align} (a^\dagger-\alpha^*)(a-\alpha)|\alpha,n⟩ &=\left(D(\alpha)a^\dagger D(\alpha)^\dagger\right) \left(D(\alpha)aD(\alpha)^\dagger\right) D(\alpha)|n⟩ \\ &= D(\alpha)a^\dagger a|n⟩=nD(\alpha)|n⟩ \\&=n~|\alpha,n⟩, \end{align} $$ or alternatively $$ \left[a^\dagger a-(\alpha a^\dagger +\alpha^*a)\right]~|\alpha,n⟩ =(n-|\alpha|^*)~|\alpha, n⟩. $$

Now, as all states in Hilbert space, the displaced number states can be written in the basis of number states, as $|\alpha,n⟩=\sum_{m=0}^\infty c_m(\alpha)|m⟩$. The coefficients in this expansion are simply the matrix elements of the displacement operator in the number basis:


This can be calculated in a couple of ways, which either look messy or seem pulled out of thin air, but what matters is ultimately the result. They come out as

$$ ⟨m|D(\alpha)|n⟩=\sqrt{\frac{n!}{m!}}\alpha^{m-n}e^{-\tfrac12|\alpha|^2}L_n^{(m-n)}(|\alpha|^2)\quad\text{when }m\geq n, \tag 1 $$ where $L_n^{(m-n)}$ is a Laguerre polynomial.

The standard reference in the literature for this matrix element is Appendix B of

Ordered Expansions in Boson Amplitude Operators. K. E. Cahill and R. J. Glauber. Phys. Rev. 177 no. 5, 1857-1881 (1969).

It's worth noting that most of the literature simply quotes the result $(1)$ and attributes it to Cahill and Glauber, but the most papers cite both the paper above and a second back-to-back paper, also by Cahill and Glauber, which does not contain the result. So be careful and show that you've read what you cite!

My bachelor's thesis,

E. Pisanty. Estados coherentes generalizados y estructura analítica del operador de aniquilación [pdf] (Generalized coherent states and the analytical structure of the annihilation operator). Tesis de licenciatura, Universidad Nacional Autónoma de México, 2011.

was on this topic and it contains an alternative calculation, though it's all in Spanish (un?)fortunately.


I) Right, the operator

$$\tag{1} \hat{A}~\equiv~ \hat{a}-\alpha{\bf 1}, \qquad \alpha\in \mathbb{C},$$

satisfies the same commutation relations

$$\tag{2} [\hat{A},\hat{A}^{\dagger}] ~=~{\bf 1}$$


$$\tag{3} [\hat{a},\hat{a}^{\dagger}] ~=~{\bf 1}.$$

(In OP's example the complex number $\alpha=-1$.)

II) Define number operator

$$\tag{4} \hat{N}~:=~\hat{A}^{\dagger}\hat{A}$$

in analogy with

$$\tag{5} \hat{n}~:=~\hat{a}^{\dagger}\hat{a}.$$

III) The only non-trivial issue is the corresponding Fock space. The usual normalized Fock states read

$$\tag{6} | n \rangle~:=~\frac{1}{\sqrt{n!}} (\hat{a}^{\dagger})^n|0 \rangle, \qquad \hat{a}|0 \rangle~=~0, \qquad \langle 0|0 \rangle~=~1 .$$

The usual un-normalized coherent states reads

$$\tag{7} | z )~:=~e^{z\hat{a}^{\dagger}}|0 \rangle, \qquad \hat{a}| z )~=~z| z ), \qquad (w |z ) ~=~ e^{\bar{w}z}. $$

Define new normalized vacuum state

$$\tag{8} | 0 ] := e^{-\frac{|\alpha|^2}{2}}| \alpha ),\qquad \hat{A}|0 ]~=~0 , \qquad [ 0|0 ]~=~1 . $$

Define new normalized Fock states

$$\tag{9} | N ]~:=~\frac{1}{\sqrt{N!}} (\hat{A}^{\dagger})^N | 0 ]. $$

  • 1
    $\begingroup$ Thank you for your answer. I am in particular interested in the coefficients of $|N\rangle = \sum_{n=0}^\infty c_n |n\rangle$. Do know if there is an explicit expression of the coefficients $c_n$? $\endgroup$
    – physicsGuy
    Aug 6, 2014 at 8:01

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