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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?

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What exactly is the question? –  DarenW Jan 29 '13 at 7:14
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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
    
The question was if my reasoning is sound :) –  Lagerbaer Jan 29 '13 at 16:21
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2 Answers

up vote 2 down vote accepted

We start with

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

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

and

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

respectively, so that

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

Here we have used that

$$\tag{5} [S_2,b]~=~x b^{\dagger}, \qquad [S_2,b^{\dagger}]~=~x b, $$

so that

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

Note that

$$\tag{7} [S_1,S_2] ~=~xS_1 .$$

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

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

where

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

is the generating function of Bernoulli numbers.

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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
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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.

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