The difference between quantum superposition and statistical mixture

When learning about quantum computing, there are various new words that you start to hear and a lot of confusion arises between the classical “statistical(or probabilistic) mixture”, a “quantum superposition”, “mixed states”, “pure states” and “entangled states.

First of all, to clear any confusion, let’s just clarify that entanglement is a different beast. It describes the state of a multi-partite system. Here we focus on the description of independent simple (local) states.

A state can be separable and pure.
A state can be separable and mixed.
A state can be entangled and pure.
A state can be entangled and mixed.

So with that in mind, we will ignore entanglement for now as it is not necessary to understand what we are trying to explain in this blog post.

Mathematical hint to distinguish pure and mixed states:

Given the density matrix $ \rho $ of dimension $ d $ for a quantum state, how to tell if it is pure or mixed?

Simply multiply the matrix $ \rho $ by itself so that you get $ \rho^2 $, then compute the trace.

if $ tr(\rho^2)=1 \Rightarrow $ pure state
if $ \frac{\mathbb{I}}{d} <tr(\rho^2)<1 \Rightarrow $ mixed state
if $ tr(\rho^2) = \frac{\mathbb{I}}{d} \Rightarrow $ maximally mixed state [a particular case of mixed states.]

Now all that is left is to clear up the confusion between “mixed states”, “probabilistic mixtures” and “superposition”. For that, keep reading.

Superposition:

A superposition is a always pure state. (but not every pure state is a pure state!)

The physical description is that it can be represented by summing wave functions. The phenomenon of superposition is the result of quantum interference between the waves.

Simply put, a superposition looks like this, for example, $ \ket {\psi_s} = \frac{1}{\sqrt{2}} ( \ket{0} + \ket{1}) $ it is at both states at the same time. It is not that we don’t know which, there is no uncertainty here. We know for sure what the state is, it is just a state in-between.* The observation, however, when you measure it, is that you still get a probabilistic outcome (depending on the basis).

The difference in terms of density matrix representation:

Superposition has elements on the off-diagonal. These terms represent the presence of quantum interference in the system
Mathematically, if you use the hint given above for the (normalized) density matrix, you should find it is a pure state

*: This phrasing may depend a bit on the interpretation of quantum mechanics that you chose. See this answer on SE.

Mixture:

A mixed state Can only be described using density matrices (no wave functions). The terms “probabilistic”, “statistical” mixture, or “mixed state” all refer to the same thing.

A probabilistic mixture has nothing particularly quantum, it is just uncertainty. Just unknown. Like tossing a coin.

Physically, there is no quantum interference between the states, and therefore no (coherence) elements on the off-diagonal.

Example of a matrix that represents a chance of 0.5 for each possibility:

$$ \rho_m = \begin{bmatrix}
1/2 & 0 \\
0 & 1/2 \
\end{bmatrix} = 1/2 ( \ket{0}\bra{0} + \ket{1}\bra{1} ) $$

distinguishing both (with measurement):

I am going to skip a bit what a measurement exactly is, and for that, I will let you read the “measurement” section in density matrices on Wikipedia. Basically what we need to know here, is that measuring a state allows us to observe it and know exactly what it is, and it is mathematically achieved by multiplying a given measurement operator by the density matrix (the representation of our quantum state).

Aside from the hints given above to distinguish by squaring density operator and check the purity of the state, the two cases can be distinguished by a measurement.

Let $ \rho_s = \ket {\psi_s} \bra{\psi_s} = 1/2 ( \ket{0}\bra{0} + \ket{0}\bra{1} + \ket{1}\bra{0} + \ket{1}\bra{1} ) $

and $ \rho_m = 1/2 ( \ket{0}\bra{0} + \ket{1}\bra{1} ) $ as defined above.

Computational basis:

When you perform a measurement in the basis $ \{ \ket {0}, \ket{1} \} $ we will use as measurement operators $ A_{c0}= \ket{0}\bra{0} $ and $ A_{c1} = \ket{1}\bra{1}$, you won’t be able to tell the difference between both as both measurements will give the same outcome.

The math:

we know for $ \rho_s $ the probabilities are $ (\frac {1}{\sqrt{2}})^2 $ so $Pr(0) = 1/2 $ and $Pr(1) =1/2$
As for $ \rho_m , Pr(0) = Tr ( A_{c0} \rho_m) = 1/2 $ and $ Pr (1) =Tr ( A_{c1} \rho_m) = 1/2 $
Simply do the math between the matrices or check this on QC SE, it is a similar example.

So in both cases, we get similar probabilities, so the trick to distinguish a superposition from a mixture is to use a different measurement basis!

Hadamard basis:

But, When you measure in the Hadamard basis $ \{ \ket {+} , \ket{-} \} $, use as measurement operators $ A_h0= \ket{+}\bra{+} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \ \end{bmatrix} $ and $ A_h1= \ket{-}\bra{-}= \begin{bmatrix} 1/2 & -1/2 \\ -1/2 & 1/2 \ \end{bmatrix}$

The math:

With similar reasoning: For $ \rho_s $ :
$ Pr(0) = Tr ( A_{h0} \rho_s) = 1 $ and $ Pr(1) = Tr ( A_{h1} \rho_s) = 0 $
This makes perfect sense as the state we started with is indeed obtained by applying a Hadamard gate to $ \ket{0} $.

For $ \rho_m $ :

$ Pr(0) = Tr ( A_{h0} \rho_m) = Tr( \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \ \end{bmatrix} \begin{bmatrix} 1/2 & 0 \\ 0 & 1/2 \ \end{bmatrix} ) = 1/2 $ and $ Pr(1) = Tr ( A_{h1} \rho_m) = 1/2 $