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?

  • 1
    $\begingroup$ For the general case, I think you should take a look at Baker-Campbell-Hausdorff formula, it canbe useful. $\endgroup$
    – Martino
    Jan 29, 2013 at 11:11

2 Answers 2


We start with the bosonic commutation relation

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

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


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

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

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


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

is the generating function of Bernoulli numbers.

  • $\begingroup$ 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... $\endgroup$
    – Lagerbaer
    Jan 30, 2013 at 2:06
  • $\begingroup$ The discrepancy is caused by the fact that the shift and the Bogoliubov transformation do not commute. $\endgroup$
    – Qmechanic
    Jan 30, 2013 at 12:11
  • $\begingroup$ 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. $\endgroup$
    – Lagerbaer
    Jan 30, 2013 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.


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