12
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ Possible duplicate of What's the difference between a pure and mixed quantum state? $\endgroup$ – Mithrandir24601 Mar 29 '18 at 21:00
  • $\begingroup$ As per gls's answer, a 'superposition state' is a pure state, so your question title as it currently stands reads as "What is the relationship between [pure] and mixed states?" which compares with the other question "What's the difference between a pure and mixed quantum state?" If you edit the question to make it clearer that you're not asking the same as the other question, I'll gladly retract my vote (if you want to know how purity relates to superposition, you could make that more of a point in the question?) $\endgroup$ – Mithrandir24601 Mar 29 '18 at 21:14
  • $\begingroup$ @Mithrandir24601 Superpositions being pure might be clear to you, but to a layman like me it's a bit less obvious. I see where I went wrong now, but "superposition vs mixed state" very much was my question. I'll try to edit, but I doubt I can make it different enough from the other question, so feel free to close this. $\endgroup$ – Norrius Mar 29 '18 at 21:26
  • $\begingroup$ @Mithrandir24601 All right, with the knowledge from the answers, I rephrased it a bit. Would you say it is clearer now? $\endgroup$ – Norrius Mar 29 '18 at 22:18
  • $\begingroup$ It's a bit clearer, yes, but I'm afraid it still looks like a duplicate to me :/ I hope you at least got a satisfactory answer! $\endgroup$ – Mithrandir24601 Mar 30 '18 at 9:18
5
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ You say that |0>+|1> is a pure state. Are there no mixed superpositions? I think I understand the difference between pure and mixed, I'm asking how superpositions fit into this. $\endgroup$ – Norrius Mar 29 '18 at 20:32
  • $\begingroup$ @Norrius "mixture" and "superposition" are two separate concepts. A mixed state is always given by a mixture of different states. Indeed, a mixed state is defined as a convex sum (mixture) of other states. If a mixed state cannot be written as (convex) sum of other (pure) states, than it is not a mixed state but a pure state. Finally, a superposition of pure states is always, by definition, a pure state $\endgroup$ – glS Mar 29 '18 at 20:35
  • $\begingroup$ Okay. Can you take a superposition of mixed states? Is that a meaningful operation? $\endgroup$ – Norrius Mar 29 '18 at 20:39
  • $\begingroup$ @Norrius no it is not. Note that you cannot "take superpositions" of different states in practice. You can evolve a certain state into another state which can be described as a superposition of certain states. You can apply the same evolution to a mixed state, and what you will get is a mixture of two states, each one of which is described as a superposition in a given basis. This will still be a mixed state though. $\endgroup$ – glS Mar 29 '18 at 20:43
  • $\begingroup$ Can you see if I understand it right? A superposition is just a way of writing a state through other states, I could write |0> = |+> + |-> if I wanted, effectively stating that the ground state is a superposition. And this is independent of purity (so the last sentence in my post is right). $\endgroup$ – Norrius Mar 29 '18 at 20:58
6
$\begingroup$

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.

$\endgroup$

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.