# Constructing the exponential form of a unitary operator

I think I've got this figured out but wanted to make sure I'm doing this right.

Working with operators that satisfy bosonic commutation relations $$[b,b^\dagger] = 1,$$ I define a very general unitary transformation on them: $$\gamma^\dagger = ub^\dagger + vb + w,$$ where the coefficients $$u$$, $$v$$, and $$w$$ are real. My goal is to find the unitary operator $$U$$ written in the form $$e^S$$ where $$S$$ is anti-Hermitian, so that $$\gamma^\dagger = e^S b^\dagger e^{-S}.$$

Here's my attempt at this: The constant $$w$$ is straightforward using the expansion of $$e^{S} A e^{-S}$$ in terms of the commutator, i.e. $$e^{S} A e^{-S} = A + [S,A] + \frac{1}{2}[S,[S,A]] + \dots$$ and we see immediately that setting $$S = w(b-b^\dagger)$$ satisfies this. The more interesting part comes next:

The requirement that the transformation is unitary is equivalent to demanding that $$[\gamma, \gamma^\dagger] = 1$$ as well. This leads to the requirement $$u^2 - v^2 = 1$$, which in turn means that we could write these coefficients as $$u = \cosh(x), \quad v = \sinh(x)$$ for some parameter $$x$$. The infinitesimal transformation, $$x = \epsilon \ll 1$$, would then read, up to first order in $$\epsilon$$: $$\gamma^\dagger = b^\dagger + \epsilon b.$$ Comparing this with the commutator-expansion above, we now just need to find an anti-Hermitian operator whose commutator with $$b^\dagger$$ is $$b$$. We don't even worry about the higher orders of the expansion since those are second order in $$\epsilon$$, but here they even vanish exactly: We simply set $$S = \frac{\epsilon}{2}(b^2 - (b^\dagger)^2)$$ since the commutator $$[b^2, b^\dagger] = 2b$$, and $$[b,b]=0$$.

I think then that I'm done: For the full transformation, I just set $$S = w(b-b^\dagger) + \frac{x}{2}\left(b^2 - b^{\dagger, 2}\right), \quad \cosh(x) = u.$$

This should be true if it's true that in order to reach "angle" $$x$$ with our "rotation", we just have to rotate $$N$$ times with "angle" $$x/N$$, and hence in the limit $$N \rightarrow \infty$$ we can use the infinitesimal generator.

I know this argument works for rotations in space, where instead of $$\cosh$$ and $$\sinh$$ we'd be working with $$\cos$$ and $$\sin$$, but I'm sure this also works for the hyperbolical functions.

EDIT: The final question, now, is: Is the above reasoning sound?

• For the general case, I think you should take a look at Baker-Campbell-Hausdorff formula, it canbe useful. – Bzazz Jan 29 '13 at 11:11

$$[b,b^{\dagger}]~=~\mathbb{1},\tag{1}$$

and two numbers $$w$$ and $$x$$. The shift and Bogoliubov transformation are encoded by $$S_1 ~:=~ w(b-b^{\dagger}),\tag{2}$$

and

$$S_2 ~:=~ \frac{x}{2}(b^2 - (b^{\dagger})^2),\tag{3}$$

respectively, so that

$$\gamma ~:=~\cosh(x)b + \sinh(x)b^\dagger + w\mathbb{1} ~\stackrel{(6)}{=}~ e^{S_2}(b+w\mathbb{1})e^{-S_2}$$ $$~\stackrel{(1)+(2)}{=}~ e^{S_2}e^{S_1} b e^{-S_1}e^{-S_2}~\stackrel{(8)}{=}~ e^{S_3} b e^{-S_3}. \tag{4}$$

Here we have used that

$$[S_2,b]~\stackrel{(1)+(3)}{=}~x b^{\dagger}, \qquad [S_2,b^{\dagger}]~\stackrel{(1)+(3)}{=}~x b, \tag{5}$$

so that

$$e^{S_2} b e^{-S_2}~=~e^{[S_2,\cdot]}b ~=~\cosh([S_2,\cdot])b + \sinh([S_2,\cdot])b$$ $$~\stackrel{(5)}{=}~\cosh(x)b + \sinh(x)b^{\dagger}.\tag{6}$$

Note that

$$[S_1,S_2] ~\stackrel{(1)+(2)+(3)}{=}~xS_1 .\tag{7}$$

Therefore $$S_3$$ in eq. (4) is given by the Baker-Campbell-Hausdorff formula

$$S_3~:=~\ln(e^{S_2}e^{S_1})~=:~{\rm BCH}(S_2,S_1) ~\stackrel{(7)}{=}~S_2+B(x)S_1 ~=~\underline{\underline{S_2+\frac{x}{e^x-1}S_1}}, \tag{8}$$

where

$$B(x)~:=~\frac{x}{e^x-1}\tag{9}$$

is the generating function of Bernoulli numbers.

• Hm, now I wonder that since your $S_3$ and my $S$ are not the same, who of us made a mistake and where it is. Using the expansion of $e^S A e^{-S}$ in terms of the commutators, I still think I'm right... – Lagerbaer Jan 30 '13 at 2:06
• The discrepancy is caused by the fact that the shift and the Bogoliubov transformation do not commute. – Qmechanic Jan 30 '13 at 12:11
• That makes sense. I guess it works for the infinitesimal part, but then if I have to chain them together it wouldn't work anymore. I like your solution; it's cleaner to first start out with separate transformations for the translation and the rotation. – Lagerbaer Jan 30 '13 at 16:31

Yes, your derivation works. Note that trigonometric and hyperbolic functions are very similar; in fact, if you define them in terms of exponential function, they differ just by a factor of $i$ in the exponential (and, in case of $\sin$ and $\sinh$ in the denominator).

Moreover, the transformation you defined is very well known in quantum optics – it is squeezing (this is where the terms $u\hat{b}^\dagger+v\hat{b}$ come from) and a (real) displacement $w$. You can find a lot in many quantum optics textbooks about them, e.g., in books by Walls and Milburn, or Scully and Zubairy.