# Displacement Operator definition

1. Aug 13, 2012

### McLaren Rulez

Hi,

Could someone explain how the following two definitions of the displacement operator are equal? The first is the standard one

1) $e^{\alpha a^{\dagger}-\alpha*a}$

2) $e^{\frac{-|\alpha^{2}|}{2}}e^{\alpha a^{\dagger}}$

What is the reasoning behind not using the second equation to define the operator that acts on a vacuum state to produce a coherent state? Or if it is fine to use definition 2), then can someone help me show that the two definitions are equivalent?

Thank you

2. Aug 13, 2012

### strangerep

They're not equal as operators, but if you act on the vacuum with them, e.g.,
$$e^{\alpha a^{\dagger}-\bar\alpha a} \, |0\rangle$$
you should be able to use one of the BCH theorems (Baker-Campbell-Hausdorff) -- or is it Zassenhaus? -- to convert the exponential into a product of exponentials, where the rightmost one involves only the annihilation operator, and hence leaves the vacuum unchanged.

http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula

The advantage of version (1) is that its formally unitary. Both are useful in different circumstances. Try Mandel & Wolf for lots more on this.

3. Aug 13, 2012

### McLaren Rulez

Using the BCH equations, the article arrived at $e^{\frac{|\alpha^{2}|}{2}}e^{\alpha a^{\dagger}}e^{\alpha^{*} a}$ for the displacement operator. Now, I expand the last exponential term in powers of a.

But if the rightmost one only involves the annihilation operator, then we get zero. Isn't it true that $a\left|0\right\rangle = 0$, not $\left|0\right\rangle$?

4. Aug 13, 2012

### Bill_K

It's an exponential whose exponent involves only a. And e0 = 1.

5. Aug 14, 2012

### McLaren Rulez

Oh of course! I forgot the identity term in the Taylor expansion. Thank you very much Bill_K and strangerep