5
$\begingroup$

I am a little bit confused by density matrix notation in quantum algorithms. While I am pretty confident with working with pure states, I never had the need to work with algorithm using density matrices. I am aware we can have/create a quantum register proportional to a density matrix by tracing out some qubits on a bigger states, but I don't have any intuition to how this can be done in practice on a quantum computer.

It is simple with just an Hadamard gate (for n being a power of 2) to prepare the following state:

$$ |\psi\rangle = \frac{1}{\sqrt{n}} \sum_i^n |i\rangle $$ The density matrix version of this state is: $$\sigma= \sum_{i,y}^{n,n} |i\rangle\langle y|$$

But instead, I would like to know how to prepare a quantum register in the following state: $$\rho = \frac{1}{n}\sum_{i}^{n} |i\rangle\langle i|$$

Unfortunately, I have no intuition how I can think this state in a quantum register, as I am too used to work with pure state. The density matrix should incorporate our (classical) ignorance about a quantum system, but why should I ignore the outcome of discarding (i.e. measuring) a bigger state that gives me the totally mixed state on a quantum computer?

Rephrased in other words my question is: what is the pure state state of $\rho$? We know it must exist, because density matrices of pure states have the property that $\rho^2 = \rho$ page 20 of preskill's lecture notes. Intuitively, is $\psi$, but it is not, as $\sigma \neq \rho$.

| improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ But the $\rho$ that you’ve given does not satisfy $\rho^2=\rho$. $\endgroup$ – DaftWullie Feb 19 at 6:42
4
$\begingroup$

Easiest way to prepare a mixed state is to decompose it into a sum of pure states that are easy to construct, and then classically make a random selection.

Sure, it's now in some pure state, but from the point of view of someone who doesn't know which that is, it's in a mixed state.

As far as I can tell, there's no good reason ever to work with mixed states that aren't Bell pair halves or something else entangled. Pure states are simpler, and a mixed state is essentially just a piece of a pure state. If you don't care about the thing the pure state's entangled with, why not just make a random selection?

| improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I rephrased better the question, now it should be clearer. $\endgroup$ – leimer0rozzh Feb 19 at 3:53
2
$\begingroup$

Given an arbitrary state $\rho$ in a space $H_A$, you can always find a pure state $\newcommand{\tr}{\operatorname{Tr}}\newcommand{\ket}[1]{|#1\rangle}\newcommand{\ketbra}[1]{|#1\rangle\!\langle #1|}\ket\psi$ on some $H_A\otimes H_B$ such that $\rho=\tr_B(\ketbra\psi)$. Any such $\ket\psi$ is called a purification of $\rho$. If the eigendecomposition of your $\rho$ reads $\rho=\sum_k p_k\ketbra{\psi_k}$, any pure of the form $$ \ket\psi = \sum_k \sqrt{p_k} \ket{\psi_k}\otimes\ket{u_k}, $$ for any set of orthonormal vectors $\ket{u_k}$, is a viable purification.

To actually generate experimentally such a $\rho$, two straightforward ways are

  1. Actually run the experiment using the different $\ket{\psi_k}$ as input, rather than $\rho$. You can then mix the experimental outcomes according to the weights $p_k$. This will give you identical answers as if you used $\rho$.
  2. Use a purification $\ket\psi$ as input for the experiment, but only operate and measure a part of the system (what we denoted with $H_A$ above).

The density matrix should incorporate our (classical) ignorance about a quantum system, but why should I ignore the outcome of discarding (i.e. measuring) a bigger state that gives me the totally mixed state on a quantum computer?

This depends on why you want to use a non-pure state to begin with.

| improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.