14
$\begingroup$

My understanding so far is: a pure state is a basic state of a system, and a mixed state represents uncertainty about the system, i.e. the system is in one of a set of states with some (classical) probability. However, superpositions seem to be a kind of mix of states as well, so how do they fit into this picture?

For example, consider a fair coin flip. You can represent it as a mixed state of “heads” $\left|0\right>$ and “tails” $\left|1\right>$: $$ \rho_1 = \sum_j \frac{1}{2} \left|\psi_j\right> \left<\psi_j\right| = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

However, we can also use the superposition of “heads” and “tails”: specific state $\psi = \frac{1}{\sqrt{2}}\left( \left|0\right> + \left|1\right> \right)$ with density

$$ \rho_2 = \left|\psi\right> \left<\psi\right| = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$

If we measure in the computational basis, we will get the same result. What is the difference between a superposed and a mixed state?

$\endgroup$
8
$\begingroup$

No, a superposition of two different states is a completely different beast than a mixture of the same states. While it may appear from your example that $\rho_1$ and $\rho_2$ produce the same measurement outcomes (and that is indeed the case), as soon as you measure in a different basis they will give measurably different results.

A "superposition" like $\newcommand{\up}{|\!\!\uparrow\rangle}\newcommand{\down}{|\!\!\downarrow\rangle}|\psi\rangle=\frac{1}{\sqrt2}(\up+\down)$ is a pure state. This means that it is a completely characterised state. In other words, there is no amount of information that, added to its description, could make it "less undetermined". Note that every pure state can be written as superposition of other pure states. Writing a given state $|\psi\rangle$ as a superposition of other states is literally the same thing as writing a vector $\boldsymbol v$ in terms of some basis: you can always change the basis and find a different representation of $\boldsymbol v$.

This is in direct contrast to a mixed state like $\rho_1$ in your question. In the case of $\rho_1$, the probabilistic nature of the outcomes depends on our ignorance about the state itself. This means that, in principle, it is possible to acquire some additional information that will tell us whether $\rho_2$ is indeed in the state $\up$ or in the state $\down$.

A mixed state cannot, in general, be written as a pure state. This should be clear from the above physical intuition: mixed states represent our ignorance about a physical state, while pure states are completely defined states, which just so happen to still give probabilistic outcomes due to the way quantum mechanics work.

Indeed, there is a simple criterion to tell whether a given (generally mixed) state $\rho$ can be written as $|\psi\rangle\langle\psi|$ for some (pure) state $|\psi\rangle$: computing its purity. The purity of a state $\rho$ is defined as $\operatorname{Tr} \,(\rho^2)$, and it is a standard result that the purity of state is $1$ if and only if the state is pure (and lesser than $1$ otherwise).

$\endgroup$
9
$\begingroup$

The short answer is that there is more to quantum information than "uncertainty". This is because there is more than one way to measure a state; and that is because there is more than one basis in which, in principle, you can store and retrieve information. Superpositions allow you to express information in a different basis than the computational basis — but mixtures describe the presence of a probabilistic element, no matter which basis you use to look at the state.

The longer answer is as follows —

Measurement as you have described it is specifically measurement in the computational basis. This is often described just as "measurement" for the sake of brevity, and large subsets of the community think in terms of this being the primary way to measure things. But in many physical systems, it is possible to choose a measurement basis.

A vector space over $\mathbb C$ has more than one basis (even more than one orthonormal basis), and on a mathematical level there isn't much that makes one basis more special than another, aside from what is convenient for the mathematician to think about. The same is true in quantum mechanics: unless you specify some specific dynamics, there is no basis which is more special than the others. That means that the computational basis $$ \lvert 0 \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \qquad \lvert 1 \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$ is not fundamentally different physically from another basis such as $$ \lvert + \rangle = \tfrac{1}{\sqrt 2}\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \lvert - \rangle = \tfrac{1}{\sqrt 2}\begin{bmatrix} 1 \\ -1 \end{bmatrix},$$ which is also an orthonormal basis. That means that there should be a way to "measure" a state $\lvert \psi \rangle \in \mathbb C^2$ in such a way that the probabilities of the outcomes depend on projections onto these states $\lvert + \rangle$ and $\lvert - \rangle$.

In some physical systems, the way one performs this measurement is to literally take the same apparatus and tilt it so that it is aligned with the X axis instead of the Z axis. Mathematically, the way we do this is to consider the projectors $$ \Pi_+ = \lvert + \rangle\!\langle + \rvert = \tfrac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \qquad \Pi_- = \lvert - \rangle\!\langle - \rvert = \tfrac{1}{2}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$ and then to ask what the projections $\lvert \varphi_+ \rangle := \Pi_+ \lvert \psi \rangle$ and $\lvert \varphi_- \rangle := \Pi_- \lvert \psi \rangle$. The norm-squared of $\lvert \varphi_\pm \rangle$ determines the probability of "measuring $\lvert + \rangle$" and of "measuring $\lvert - \rangle$"; and normalising $\lvert \varphi_+ \rangle$ or $\lvert \varphi_- \rangle$ to have a norm of 1 yields the post-measurement state. (For a state on a single qubit, this will just be $\lvert + \rangle$ or $\lvert - \rangle$. More interesting post-measurement states may result if we consider multi-qubit states, and consider the projector $\Pi_+$ or $\Pi_-$ acting on one of many qubits.)

For density operators, one takes the state $\rho$ which you want to perform a measurement on, and consider $\rho_+ := \Pi_+ \rho \Pi_+$ and $\rho_- := \Pi_- \rho \Pi_-$. These operators may be sub-normalised in the same way that the states $\lvert \varphi_\pm \rangle$ might be, in the sense that they may have trace less than 1. The value of the trace of $\rho_\pm$ is the probability of obtaining the outcome $\lvert + \rangle$ or $\lvert - \rangle$ of the measurement; to renormalise, simply scale the projected operator to have trace 1.

Consider your state $\rho_2$ above. If you measure it with respect to the $\lvert \pm \rangle$ basis, what you will find is that $\rho_2 = \rho_{2,+} := \Pi_+ \rho_2 \Pi_+$. This means that projecting the operator with $\Pi_+$ does change the state, and that the probability of obtaining the outcome $\lvert + \rangle$ to the measurement is 1. If you do this instead with $\rho_1$, you will find a 50/50 chance of obtaining either $\lvert + \rangle$ or $\lvert - \rangle$. So the state $\rho_1$ is a mixed state, while $\rho_2 $ is not --- the difference being that $\rho_2$ has a definite outcome in a different measurement basis than the standard basis. You might say that $\rho_2$ stores a definite piece of information, albeit in a different basis than the computational basis.

More generally, a mixed state is one whose largest eigenvalue is less than 1, meaning that there is no basis in which you can measure it to get a definite outcome. Superpositions allow you to express information in a different basis than the computational basis; mixtures represent a degree of randomness about the state of the system you're considering, regardless of how you measure that system.

$\endgroup$
2
$\begingroup$

Along with glS' post:

A mixed state would be if you had a can of paint, but you weren't sure if it was blue or yellow. You know it is either one of the two, and once you pop the top and measure it, you'd know, but until you do it is in one of those two pure states. If you picked it up from a stack of cans where you knew there were equally many cans of blue paint as yellow, you would expect an equal chance of it being one or the other. 50% of the time it would be 100% yellow and 50% of the time it would be 100% blue.

A superposition is more like if you take half a can of blue and half a can of yellow and pour them together. You've now constructed a new pure state that is expressible as a combination of other pure states. If you test its 'blueness', it is about 50%. If you test its 'yellowness' it is about 50%. It is both yellow and blue at the same time. 100% of the time it is both 50% blue and 50% yellow.

If you measured the amount of blue and yellow in one stack of blue or yellow cans and then in another stack of green, you might be confused to see you have just as much blue and yellow in both stacks, but the difference is that the 'blueness' and 'yellowness' is in a mixed state in the later stack but is in a superposition in the latter.

$\endgroup$

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.