I am currently running numerical simulations of a single qumode state acted upon by a parameterised unitary. The qumode state is realised as a Fock state with a fixed cutoff dimension $(d)$ and is represented by a column vector. The unitary is a universal operation in continuous variables and I determine it's truncated form by already using appropriate $d \times d$ approximations for the ladder operators in the truncated Fock basis.

Assuming this is indeed an appropriate way to define and use truncated states and operators numerically, my question is as follows -

  • Can the unitary be parameterised such that it's action on the $d-$dimensional Fock state outputs a state with support outside of the truncated Hilbert space and is this numerically verifiable?

My guess would be that given how I determine the truncated unitaries this should not be possible for any parametrisation as the unitary is (approximately) restricted to the the $d-$dimensional Fock space and as such the situation boils down to a qudit state being acted upon by this unitary. However, I'm not sure about whether this argument is foolproof.


1 Answer 1


Your guess is correct that this is not possible. A $d\times d$ matrix times an $d\times 1$ column vector gives you another $d\times 1$ column vector, so your truncated unitary cannot take the vector beyond the truncated space.

However, if you would like, you can indeed make a truncated unitary that does what you like. Consider your $d\times 1$ state vector to have the components $\psi_n=\langle n|\Psi\rangle$ for $n\in\{0,\cdots, d-1\}$ and for some arbitrary state $|\Psi\rangle$. The $d\times d$ truncated unitary has components $u_{m,n}=\langle m|U|n\rangle$ for $m,n\in\{0,\cdots, d-1\}$. When you act with the matrix on the vector, what you get is $$\psi_m^\prime=\sum_{n=0}^{d-1}u_{m,n}\psi_n=\sum_{n=0}^{d-1}\langle m|U|n\rangle\langle n|\Psi\rangle.$$ If there was no truncation error, we would have had $$\psi_m^\prime=\langle m| \Psi^\prime\rangle=\langle m| U|\Psi\rangle=\sum_{n=0}^\infty\langle m| U|n\rangle\langle n|\Psi\rangle,$$ where we used the resolution of identity $\mathbf{1}=\sum_{n=0}^\infty|n\rangle\langle n|$. The two expressions are the same if there is no truncation error on $|\Psi\rangle$; i.e., they are the same if $\langle n|\Psi\rangle=0$ for $n\geq d$.

But we also have to worry about the limit on $m$: the question asks for the output state to include information about the state's components outside of the original truncated zone, i.e., we would like to know $\langle m|\Psi^\prime\rangle$ for $m\geq d$. This is simply done by expanding $u$ to be a rectangular matrix. We keep the same definitions $u_{m,n}=\langle m|U|n\rangle$ with $n\in\{0,\cdots, d-1\}$ but now allow $m\in\{0,\cdots, N-1\}$ for $N> d$. The matrix $u$ now has size $N\times d$. This will let you know about the unitary's effect for taking the components of the state in the truncated basis and applying them outside of the truncated basis.

I note that errors will still occur in this situation, and it is possible that they are larger than you'd expect. We get more accuracy by including extra elements of $U$ and making $u$ be a rectangular matrix, but we still have neglected $n\geq d$ in the sum. This means we are still ignoring any contributions that should have come from parts of the original state that were truncated. It is okay to do that, but it is worth noting that these might have the largest contributions to the final state's parts outside of the truncated region. For example, if $U$ is just the identity, then any part of the original state that you truncate will not show up properly in the final state, even if you use a rectangular $U$.

You can think of the simplest case: take the state $\psi_0|0\rangle+\psi_1|1\rangle$ with column vector $\begin{pmatrix}\psi_0\\\psi_1\end{pmatrix}$ but truncate at $d=1$, then the truncated vector is $\begin{pmatrix}\psi_0\end{pmatrix}$, and $u=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ gets truncated to $u=\begin{pmatrix}1\end{pmatrix}$. In the truncated region, we find $U|\Psi\rangle=\begin{pmatrix}1\end{pmatrix}\begin{pmatrix}\psi_0\end{pmatrix}=\begin{pmatrix}\psi_0\end{pmatrix}$. If we extend $u$ to its rectangular form, we retain the components $u=\begin{pmatrix}1\\0\end{pmatrix}$. If we multiply the truncated state by this, we get $U|\Psi\rangle=\begin{pmatrix}1\\0\end{pmatrix}\begin{pmatrix}\psi_0\end{pmatrix}=\begin{pmatrix}\psi_0\\0\end{pmatrix}$. If we do the full non-truncated thing, we get $U|\Psi\rangle=\begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}\psi_0\\\psi_1\end{pmatrix}=\begin{pmatrix}\psi_0\\\psi_1\end{pmatrix}$. So you see that even including extra components of $U$ can trick you into thinking you've got a final state in a larger dimension, but will not necessarily give you the correct components for that state in the larger dimension. This is a radical example of course and it is true that extending $U$ to a rectangular matrix adds some accuracy to the final result, this example is just a warning.


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