How to prepare mixed states on a quantum computer?

Question

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 $\sigma \neq \rho$.

Answer 1

I don't know if that's still useful but I've been asking this to myself recently and I've found a simple answer.
If you want to prepare the mixed state $$\rho = \frac{1}{d}\sum_{i}^{d} |i\rangle\langle i|$$ you can start by preparing the maximally entangled pure state $$|\varphi\rangle = \frac{1}{\sqrt{d}}\sum_{i}^{d} |i\rangle|i\rangle$$
The density matrix of $|\varphi\rangle$ would be $$|\varphi\rangle\langle \varphi| = \frac{1}{d}\sum_{i}^{d} \sum_{j}^{d}|i\rangle|i\rangle\langle{j}|\langle{j}| = \sum_{i}^{d} \sum_{j}^{d}|i\rangle\langle{j}|\otimes|i\rangle\langle{j}|$$. Tracing out the second qubit would result in: $$Tr_2[\frac{1}{d}\sum_{i}^{d} \sum_{j}^{d}|i\rangle\langle{j}|\otimes|i\rangle\langle{j}|] =$$ $$\frac{1}{d}\sum_{i}^{d} \sum_{j}^{d}|i\rangle\langle{j}|\cdot Tr[|i\rangle\langle{j}|] = $$ $$\frac{1}{d}\sum_{i}^{d} \sum_{j}^{d}|i\rangle\langle{j}|\delta_{ij} = \frac{1}{d}\sum_{i}^{d} |i\rangle\langle i|=\rho$$

Moreover, instead of using $|\varphi\rangle = \frac{1}{\sqrt{d}}\sum_{i}^{d} |i\rangle|i\rangle$ you could start with any state of the form $$|\varphi\rangle = \frac{1}{\sqrt{d}}\sum_{i}^{d} |u_i\rangle|u_i\rangle$$
where {$u_i$} is an orthonormal basis for $H^{\otimes d}$ and you would still have $$Tr_2[|\varphi\rangle\langle \varphi|] = Tr_1 [|\varphi\rangle\langle \varphi|] = \frac{\mathbb{I}}{d} = \frac{1}{d}\sum_{i}^{d} |i\rangle\langle i|$$

Answer 2

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?

Answer 3

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.