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I'm trying to compute (numerically) the matrices of some simple quantum optical operations, which in principle are unitary. However, in my case they are unitary in an infinite-dimensional space, so I have to truncate them. The result is not necessarily unitary anymore, but if all the entries are correct up to the size that I choose, I'm happy.

So I compute the generator, truncate it to the size of my liking and then I exponentiate it, right? Nope. It doesn't work that way: the entries can be actually very wrong. In some cases they are almost correct, in some other cases they are all messed up.

Example 1: the beam splitter $\exp[i\theta(a^\dagger b + ab^\dagger)]$.

  1. compute $a$ and $a^\dagger$ up to dimension (say) $m$.
  2. multiply them with the kronecker product
  3. exponentiate

result: the entries are almost right, except for the last row and column of both spaces as in this figure (for $m=4$):

enter image description here

The only correct parts are the ones in the white spaces. In this case the solution is to truncate $a$ and $a^\dagger$ to size $m+1$ and then throw away the wrong rows/columns.

Example 2: the single-mode squeezer $\exp[\frac{1}{2}(z^*a^2-z{a^\dagger}^2)]$

This is all a mess: as I increase the size of $a$, the entries of the final result (which are correctly placed in a "checkerboard pattern") seem to converge to their correct values, but in order to have the first (say) 4x4 block somewhat correct I have to truncate $a$ to $m\approx 50$ and then truncate the result of the exponentiation to a 4x4 size.

Am I doing this the wrong way? Eventually I would like to produce the matrices of rather non-linear operations, where the $a$ and $a^\dagger$ operators are raised to large powers, how do I know if I'm doing it right?

UPDATE: In the first case (the beamsplitter) the unitary is in $SU(2)$, which is compact and admits finite-dimensional irreps. So I can exponentiate them individually and from those I can build the truncated unitary 😁

In the second case (the squeezer) the unitary is in $SU(1,1)$ which is non-compact and in fact the Casimir operator has two infinite-dimensional eigenspaces: one corresponding to even and one to odd Fock states. Also for the two-mode squeezer the eigenspaces of the Casimir are infinite-dimensional (although countably infinite). So I can't use the multiplet method in this case.

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  • $\begingroup$ It might be too naive, but if I had to do something like that I would try to diagonalize the generator $M=U D U^{-1}$ and then $exp(M)=U exp(D) U^{-1}$. Maybe it helps ? Keep me posted, I'm interested in the answer. $\endgroup$ – Adam Mar 8 '17 at 7:28
  • $\begingroup$ This is precisely what the matrix exponential is doing behind the scenes. $\endgroup$ – Ziofil Mar 8 '17 at 15:44
  • $\begingroup$ No. Have a look at what Matlab does for instance. $\endgroup$ – Adam Mar 8 '17 at 15:50
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    $\begingroup$ @Ziofil For Example 1, you may want to observe that $$\left[ a^\dagger b + a b^\dagger, \; a^\dagger a + b^\dagger b \right] = 0$$which tells you that the matrix exponential has very neat eigensubspaces, spanned by states with given total number of $a$ and $b$ quanta. For Example 2 and the squeezer op things are more complicated. You may want to take advantage of its underlying su(1,1) decomposition and its own action on $a$ and $a^\dagger$. $\endgroup$ – udrv Mar 9 '17 at 17:59
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    $\begingroup$ Is there something like Jordan normal form so the matrix multiplications doesnt mix so much in infinite dimensional cases? $\endgroup$ – Emil Mar 9 '17 at 20:58
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As a practical matter, this is something where you expect the exponentiation to converge as the size of the truncated input grows. So, without knowing anything about the problem in question, I would

  1. find the desired size of the answer matrix (say, $n\times n$),
  2. truncate the input matrix at twice that size ($2n \times 2n$) and calculate the exponential,
  3. calculate the exponential again at one larger ($[2n+1]\times[2n+1]$),
  4. see if the $n\times n$ matrix has converged to the desired numerical accuracy, if it has, terminate, if not, double again, etc.

Ideally, as @Adam mentioned in a comment, you would find the eigenvectors and eigenvalues of the operator/matrix in question to do the calculation. Examining that approach is instructive in figuring out whether the above algorithm can converge. If the expression for the original operator is: $$A_{ij} = \sum_{k} V_{ki}^\star \lambda_k V_{kj},$$ then the exponential is: $$[\exp(A)]_{ij} = \sum_{k} V_{ki}^\star \mathrm{e}^{\lambda_k} V_{kj}.$$

Convergence of the second expression requires that the eigenvectors do not mix the states too widely, otherwise the sum will diverge. For example, if the eigenvalues grow linearly with $k$ then the eigenvectors have to fall faster than exponentially in $i-k$. In the ideal case, each eigenvector will have a finite number of components, guaranteeing convergence, but that may not always be the case.

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I found the answer, at least for some of the cases that I thought were intractable, including my examples. The main tool is the disentangling theorem for the relevant group.

Example 1: the transformation is in $SU(2)$, so we need the $SU(2)$ disentangling theorem: $$ \exp(z J_+-z^* J_-) = \exp(\tau_+ J_+)\exp(\tau_ 0J_0)\exp(\tau_-J_-) $$ Here $\{J_0,J_\pm\}$ satisfy the $su(2)$ algebra relations: $[J_\pm,J_0]=\mp J_\pm,\ [J_-,J_+]=-2J_0$, and if $z=re^{i\phi}$, then $\tau_\pm=\pm e^{\pm i\phi}\tan(r),\tau_0=2\log\sec(r)$. If we apply this to the beamsplitter transformation in the form $U(\theta)=\exp[\theta(a^\dagger b-ab^\dagger)]$, we obtain $$ U(\theta) = \exp(\tan(\theta)a^\dagger b)\exp[\log\sec(\theta)(a^\dagger a-b^\dagger b)]\exp(-\tan(\theta)a b^\dagger) $$ which has no truncation issues.

Example 2: the transformation is in $SU(1,1)$, so we need the $SU(1,1)$ disentangling theorem: $$ \exp(z K_+-z^* K_-) = \exp(\sigma_+ K_+)\exp(\sigma_0K_0)\exp(\sigma_-K_-) $$ Here $\{K_0,K_\pm\}$ satisfy the $su(1,1)$ algebra relations: $[K_\pm,K_0]=\mp K_\pm,\ [K_-,K_+]=2K_0$, and if $z=re^{i\phi}$, then $\sigma_\pm=\pm e^{\pm i\phi}\tanh(r),\sigma_0=-2\log\cosh(r)$. If we apply this to the squeezing operator $S(z)=\exp[z \frac{{a^\dagger}^2}{2}-z^* \frac{{a}^2}{2}]$, we obtain $$ S(z) = \exp\left(e^{i\phi}\tanh(|z|)\frac{{a^\dagger}^2}{2}\right)\exp\left(-\log\cosh(|z|)(a^\dagger a+\frac{1}{2})\right)\exp\left(-e^{-i\phi}\tanh(|z|)\frac{{a}^2}{2}\right) $$ which has no truncation issues.

General case: in general it might be impossible to have a suitable disentangling theorem, but results like this might help approximating it (see eq. (30)-(35) for the two examples above).

I hope this will help those who will stumble upon my same issue.

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    $\begingroup$ Do you understand why you don't have the same issues with this form of S(z) ? Is each new exponential matrix better behaved, or is it the combination of the three that compensate each other truncation errors ? $\endgroup$ – Adam Mar 14 '17 at 16:23
  • $\begingroup$ It's the fact (I suppose) that in each exponential the generator is either diagonal ($K_0$) or upper-/lower- diagonal ($K_\pm$), so each exponential is either diagonal or upper-/lower- triangular. $\endgroup$ – Ziofil Mar 14 '17 at 17:05
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If a matrix $A$ is not diagonal or in Jordan normal form, then you have to truncate the exponential $B=e^A$ after computing it, and not just truncate the matrix $A$ and then compute $B$. This is because the matrix elements $A_{nm}$, $n,m>N$, that you neglect by truncating $A$ to $N\times N$, are contributing to the matrix elements $B_{nm}$, $n,m\leq N$, as you can see by expanding $B=I+A+\frac{1}{2}A^2+\cdots$. The contributions to $B_{nm}$, $n,m\leq N$, from $A^2$ etc. that you neglect by truncating $A$ to $N\times N$, are $\displaystyle \sum_{k=N+1}^{\infty} A_{nk}A_{km}$, etc., which in general are not zero. What you can do is to try to compute explicitly the matrix elements $\langle n|B|m\rangle$ in the Fock basis, up to $N$. However this is not practical and can be done in limited cases, e.g. for the displacement operator $D(\alpha)=\text{exp}(\alpha \hat{a}^{\dagger}-\alpha^* \hat{a})$: $$\langle n|D(\alpha)|m\rangle=(n!/m!)^{1/2} \alpha^{m-n} e^{-|\alpha|^2/2} L_n^{(n-m)}(|\alpha|^2),$$ where $L$ are associated Laguerre polynomials (see Glauber). Unfortunately for the squeeze operator $S(\xi)$, $\xi=r e^{i\theta}$, you can easily compute only the first column and row, by using $$\langle 2n| S(\xi)|0\rangle=\frac{(-\tanh r)^n}{\sqrt{\cosh r}}\frac{\sqrt{(2n)!}}{2^n n!}e^{i n \theta},$$ $$\langle 2n+1| S(\xi)|0\rangle=0$$ for $n=0,..., N$.

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