In the most simple example I can think of, we have:

  • a linear operator $A$, which is also a 2 x 2 matrix.

  • a vector $|v_i⟩$, which can be considered a 2 x 1 matrix.

If we see an example of an operation such as $A|v_i⟩$, which is an operation on the vector. Are we, in all cases, to consider this equivalent to a simple matrix multiplication?

Are there any other considerations to be aware of? This is fascinating stuff, thanks for considering this simple problem.

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    $\begingroup$ A dot product returns a scalar (a single number). An operator $A$ acting on a ket (vector) $|v_i\rangle$ returns another vector $A|v_i\rangle$. $\endgroup$ – Mark S Sep 5 '19 at 17:05
  • $\begingroup$ thanks and fixed; "dot product" subbed for general "matrix multiplication" $\endgroup$ – neutrino Sep 5 '19 at 17:07

The question is a bit vague (in other equivalent representations of QM applying operators can have nothing to do with linear algebra) so I will answer with respect to the specific example given of $\mathbf{A} | v \rangle $.

If you've chosen to represent your states and operators as vectors in finite-dimensional Hilbert space, then you can describe the corresponding as matrices and vectors in terms of the elements of those matrices, eg:

$$ \mathbf{A} := A_{ij}\\ |v\rangle := v_i \\ $$

with $i,j = 1, \dots, 2^n$ then yes, $\mathbf{A} | v \rangle $ will produce a new column vector $c_j = \sum_i A_{ij} v_i$ and the previous answer said as much. However, there's a more interesting an efficient way to do this! Suppose we know that $\mathbf{A}$ operates on only a size-k subset of the qubits that $|v\rangle$ is defined over and that we redefine how we index our operator and state so that they're tensors over two-dimensional axes:

$$ \mathbf{A} := A_{i_1,i_2,\cdots, i_{2k}}\\ |v\rangle := v_{i_1,\cdots,i_n} \\ $$

with $i_m \in \{0, 1\}$ and $2k \leq n$. This is certainly allowed because $\mathbf{A}$ and $|v\rangle$ each have the same number of elements, just arranged in different ways to make computation easier. In certain cases, this allows us to calculate the action of $\mathbf{A}$ more efficiently. For example if $\mathbf{A}$ is a permutation matrix, then computing $\mathbf{A} | v \rangle $ is equivalent to shuffling some of the dimensions of $|v\rangle$ and requires no multiplication at all$^\dagger$. I give a worked example of this kind of operation at the end of this post.

Finally, I'll note that simulations of Clifford circuits (wherein $\mathbf{A} \in \{H, CNOT, S\}$ for example and $|v\rangle$ is generated by a similar such circuit) uses what is essentially a lookup table to compute $\mathbf{A} | v \rangle$. This method also uses minimal linear algebra which is what allows simulations of those kinds of circuits to be very efficient.

Worked example

Say $\mathbf{A}$ is $X^{(j)}$, the Pauli-X operator on the j-th qubit. The procedure to compute $X^{(j)}|v\rangle$ is as follows:

  1. Initialize $\phi := v_{\cdots i_{j-1} ,0 ,i_{j+1} \cdots}$
  2. $v_{\cdots i_{j-1} ,0, i_{j+1} \cdots} \rightarrow v_{\cdots i_{j-1} ,1 ,i_{j+1} \cdots}$ ("swap" the amplitudes of $v$ representing $|0\rangle_j$ with the amplitudes representing $|1\rangle_j$)
  3. $v_{\cdots i_{j-1} ,1, i_{j+1} \cdots} \rightarrow \phi$ (do another "swap", but this time use the amplitudes that we saved in step (1) that would have been overwritten otherwise)

$^\dagger$ In practice, this process has a lot of memory overhead and requires an exponential number of read/write operations. But often that's still better than doing matrix multiplication which can require polynomial-of-exponential resources to do the same thing.

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  • $\begingroup$ can you give an example of what you mean by "in other equivalent representations of QM applying operators can have nothing to do with linear algebra"? $\endgroup$ – glS Sep 9 '19 at 17:55
  • $\begingroup$ The example I had in mind was the Schrodinger equation $\partial_t \psi = H \psi$ in which operators satisfy the canonical commutation relation $[x, p] = 0$. Then the equivalence follows from the Stone-von Neumann theorem. $\endgroup$ – forky40 Sep 9 '19 at 18:52

I think all unitary operations can be thought of as matrix multiplication, but something like a measurement is non-unitary, and looks like a partial trace over the density matrix, so there is a bit more going on there.

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