We suppose for simplicity we have a 1D oscillator, but this is a question about the general CCR in oscillators, second quantization, quantum field theory etc.

We know coherent states form a non-orthonormal overcomplete basis. We know they satisfy $A |\alpha>=\alpha |\alpha>$, where $|\alpha>=e^{-\frac{|\alpha|^2}{2}} e^{\alpha A^+} \Psi_0$

How can we calculate a generic scalar product between two different coherent states with eigenvalues $\alpha$ and $\beta$?

$$ e^{-\frac{|\alpha|^2+|\beta|^2}{2}}\Psi_0 e^{\beta A} e^{\alpha A^+} \Psi_0 $$

  • 1
    $\begingroup$ Start with a Taylor expansion, i.e., $\mbox{exp}(\alpha a^{\dagger})|0\rangle = \sum_n \frac{\alpha^n}{n!}(a^{\dagger})^n|0\rangle$ (and remember that number states are orthogonal!) $\endgroup$
    – wsc
    Feb 24, 2011 at 0:36
  • 1
    $\begingroup$ @wsc So I'll obtain $e^{-\frac{|\alpha|^2+|\beta|^2}{2}}e^{\hat\beta\alpha}$ ? $\endgroup$
    – Boy S
    Feb 24, 2011 at 0:54
  • $\begingroup$ if by that hat you mean the conjugate of $\beta$, yes... Now, can you think of a more suggestive way to write it? $\endgroup$
    – wsc
    Feb 24, 2011 at 1:26
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    $\begingroup$ actually that was rather vague, and it's difficult unless you know what i'm looking for. The point is you can also write it as $\mbox{exp}(-|\alpha-\beta|^2)$, which allows you to interpret the space of coherent states a bit more graphically, IMO. $\endgroup$
    – wsc
    Feb 24, 2011 at 1:46
  • 1
    $\begingroup$ Dear wsc, it is not true that $-(|a|^2+|b|^2)/2 +b^* a = -|a-b|^2$. First of all, a factor of two is missing. Second of all, the left-hand side has $b^* a$ instead of its real part, $(b^* a + a^* b)/2$, which appears on the right hand side. Because those quadratic expressions appear in the exponent, their difference - which is a purely imaginary number - only changes the phase of the result. So your latest geometric result is OK up to a wrong phase of the inner product. Just to be sure, Boy's answer 3 comments higher is exactly right. $\endgroup$ Feb 24, 2011 at 7:12

2 Answers 2


The computation can be done alternatively in the Bargmann representation of the harmonic oscillator. In the following, the required inner product evaluation will be described in this representation. In my opinion, this method is computationally superior as well as many other advantages. This method is based on the isomorphism between the Hilbert spaces $L^2(\mathbb{R})$ and the Bargmann space of analytical functions on $\mathbb{C}$ with respect to the inner product

$$(f,g) = \frac{1}{2\pi}\int_{\mathbb{C}} f(z) \overline{g(z)} \exp(-z\bar{z})dz d\bar{z}$$

(The isomorphism is given explicitely by means of the Bargmann transform)

In the Bargmann representation, the creation operator is representaed by the multiplication by $z$ and the anihilation operator derivative with respect to z and the vacuum state by the constant unit function (and, by the way, the energy eigenfunctions of the harmonic oscillator by the monomials $z^n$ - up to a normalization). Thus the $\alpha$ coherent state is represented by:

$$\psi_{\alpha}(z) = \exp\left(-\frac{|\alpha|^2}{2}\right) \exp(\bar{\alpha}z) $$

and the inner product is therefore given by:

$$(\psi_{\beta},\psi_{\alpha}) = \frac{\exp\left(-\frac{\left(|\alpha|^2+|\beta|^2\right)}{2}\right)}{2\pi}\int_{\mathbb{C}} \exp(\bar{\beta}z) \exp(\alpha\bar{z}) \exp(-z\bar{z}) dz d\bar{z} = \exp\left(-\frac{\left(|\alpha|^2+|\beta|^2\right)}{2}\right) \exp(\bar{\beta} \alpha)$$

The integral is easily evaluated by coordinate translation and square completion.

This example is a prototype of quantization in Kahler polarization.


Use the Baker-Campbell-Hausdorff identities. They are your best friends for this. Two links, for example:

Wikipedia has a page, that has an example under the heading "Application in Quantum Mechanics" that is pretty much what you need.


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