Is there a way to disentangle the exponential of the sum of the number, the annihilation and the creation operator? For example,

$$e^{\alpha N + \beta a + \gamma a^\dagger } = e^{G a^\dagger}e^{A N}e^{B a}$$

where $G$, $A$, and $B$ are each functions of possibly all three parameters $\alpha$, $\beta$, and $\gamma$.

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    $\begingroup$ Did you try the Baker-Cambell-Haussdorff formula? $\endgroup$ Jan 5, 2021 at 12:40
  • $\begingroup$ And what do you mean precisely by "disentangle"? What do you want to end up with - that will tell you want you need to do. $\endgroup$ Jan 5, 2021 at 12:58
  • $\begingroup$ i mean precisely whats written in the equation: go from an exponential of a sum of operators to a product of exponentials of operators. $\endgroup$
    – oweydd
    Jan 5, 2021 at 13:16

2 Answers 2


Not an answer, but an extended comment on your basically sound approach, since the comment format does not permit such extended comments. The group involved is the oscillator group, and the 3d rep you found is a faithful one, so any group relation for it will also hold for the abstract group in general, so, all representations! I will call your central element C of your answer Z, and it can filter out of all expressions, commuting with everything.

The generic statement supported by Lie's theorem is that the product of all group elements will close to an exponential of some linear combination of all generators in the Lie algebra, so, then, $$ 𝑒^{πœƒZ} 𝑒^{πΊπ‘Ž^†} 𝑒^{𝐴𝑁}𝑒^{π΅π‘Ž}=𝑒^{πœ™'Z+𝛼𝑁+π›½π‘Ž+π›Ύπ‘Ž^†}. $$ However, since Z commutes with everything, we can invert the first factor of the l.h.s. to the right, and incorporate it into a new parameter $\phi'-\theta=\phi$, so that $$ 𝑒^{πΊπ‘Ž^†} 𝑒^{𝐴𝑁}𝑒^{π΅π‘Ž}=𝑒^{πœ™Z+𝛼𝑁+π›½π‘Ž+π›Ύπ‘Ž^†}, \tag{*} $$ where the parameters $\phi,\alpha,\beta, \gamma$ are guaranteed to be functions of $G,A,B$.

Now, by the nilpotency of the first three generators, and the diagonally of the fourth, the l.h.side trivially evaluates to $$ e^{-A/2} \begin{bmatrix}e^A & G & BG\\0 &1 &B\\0 &0 &e^A\end{bmatrix}, $$ with determinant $e^{A/2}$.

This must equal $$ \exp \begin{bmatrix} \alpha/2 & \gamma & -\phi\\0 &-\alpha/2 &\beta\\0 &0 &\alpha/2\end{bmatrix}. $$ Its determinant is $e^{\alpha/2}$ by the identity $e^{\operatorname{Tr} M} = \det e^M$.

Now, to second order in its parameters, it expands to $$ \begin{bmatrix}1+ \alpha/2 +\alpha^2/8& \gamma & -\phi-\phi\alpha/2+\beta\gamma/2\\0 &1-\alpha/2 +\alpha^2/8&\beta\\0 &0 &1+\alpha/2+\alpha^2/8\end{bmatrix}. $$

Comparing with the above l.h.side dictates, to second order, $$A=\alpha, \qquad B=\beta e^{\alpha/2}, \qquad G=\gamma e^{\alpha/2},$$ but then you realize the upper-rightmost entry is mismatched, and requires a non-vanishing $\phi$, $$ BGe^{-A/2}= \beta\gamma e^{\alpha/2}= \beta\gamma/2 -\phi(1+\alpha/2), $$ to pick up the slack. One had to go to second order to see this, as your need at least one commutation $[a,a^\dagger]$ to produce the central element.

So, then, $\phi$ is actually essential in your amended expression (*): this is not a degree of freedom that could be omitted. Apologies (with Pascal) for lacking the time to make the comment shorter.

  • $\begingroup$ thanks for your answer. However, could you explain why it does not matter that $[a, a^\dagger]=I$ but when you use the sigma matrix representation $[\sigma_m, \sigma_p]=-\sigma_3$? $\endgroup$
    – oweydd
    Jan 5, 2021 at 10:51
  • $\begingroup$ Also, the fact that the bottom-right hand matrix element on the RHS of the equation is equal to one presumably means that the two sides can't be matched for general $\alpha$, $\beta$, $\gamma$? But perhaps other orderings of the RHS side might have solutions for general $\alpha$, $\beta$, $\gamma$? $\endgroup$
    – oweydd
    Jan 5, 2021 at 11:52
  • $\begingroup$ Thanks again for the elaboration on your comment. Can i ask specifically what theorem is it that allows the product of group elements to be written as an exponential of a linear combination of generators. Is it okay for the coefficients to be complex as long as the exponential of the generator multiplied by the coefficient is in the group? $\endgroup$
    – oweydd
    Jan 7, 2021 at 15:35
  • $\begingroup$ There is latitude of terminology, but, loosely, the idea is that the Lie algebra generates the group, so all exponentials of linear combinations of generators are in the group; since products of group elements are in the group, near the identity, any group element must be the exponential of some linear combination of generators (Lie algebra elements), as here. Yes, the coefficients ("angles") of generators may be chosen complex. $\endgroup$ Jan 7, 2021 at 15:47
  • $\begingroup$ so you can do this for any lie group? what if we're not near the identity? I realise ive asked a lot of questions and this isnt really the platform for getting a lesson in lie group theory. If you have a reference which explains how the "Lie group–Lie algebra correspondence" allows us to write any group element as a product of exponentials that would be really appreciated. $\endgroup$
    – oweydd
    Jan 7, 2021 at 17:46

I think i have found a method using the answers to these two questions:


How do disentangling and reordering of exponential operators work?

We can map the following matrices to the ladder operators:

$a^\dagger\equiv A=\left[\matrix{0 & 1 & 0\\0 &0 &0\\0 &0 &0}\right]$, $a \equiv B=\left[\matrix{0 & 0 & 0\\0 &0 &1\\0 &0 &0}\right]$, $I\equiv C=\left[\matrix{0 & 0 & -1\\0 &0 &0\\0 &0 &0}\right]$, $N\equiv D= \frac12\left[\matrix{1 & 0 & 0\\0 &-1 &0\\0 &0 &1}\right]$

The matrices A,B,C,D satisfy the commutation relations of the ladder operators. Then evaluate left hand side and right hand side using these matrices and match coefficients. It seems to work, but I'd like some confirmation this is the right approach as i have no experience with lie algebras.

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    $\begingroup$ Congratulations, the algebra works (unlike my deleted one), and the Heisenberg subalgebra is the standard one, up to a few redefinitions. But this algebra has 4, not 3 elements, so there should be four terms in the exponent of the l.h.side. In your place, I'd transpose the r.h.side with the augmented l.h.side, guaranteed to work by Lie's theorem. Do you want an easy demonstration that the augmentation of the unified exponential is necessary? $\endgroup$ Jan 5, 2021 at 17:04
  • $\begingroup$ Thanks Cosmas, but it was only after seeing your answer that i could see what to do. I should have mentioned in my answer - I included an identity term on both sides of the equation. What do you mean by transposing the RHS with the augmented LHS? Yes an explanation would be great thank you! $\endgroup$
    – oweydd
    Jan 6, 2021 at 10:21
  • $\begingroup$ What I mean is the multiplication of 3 group elements must equal the generic group element by Lie s theorem, and that is the exponential of a linear combination of all Four algebra elements, including the center C. Without it, the determinants won’t match. $\endgroup$ Jan 6, 2021 at 11:31

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