# What state is a qubit system in if it is measured by someone, but we do not know the measurement?

Suppose there is a qubit Q in the state

$$\lvert+\rangle = \frac{1}{\sqrt2}(\lvert 0\rangle+ \lvert 1\rangle)$$

Now somebody makes a computational basis measurement on Q, but we do not learn the result of the measurement. After the measurement, what state do we assign to Q?

This question came up in one of my quantum information theory courses and the only answer I can think of is that it remains in the same state (why would it be anything else?).

I have no idea how to tackle it otherwise. Any help or suggestions would be greatly appreciated. Thanks!

• Do you know about density matrices? – zeldredge Oct 19 '15 at 13:59

The state of the qubit does not depend on what you or anyone else knows about it. However, the state of a qubit typically does change if it is measured. Suppose the qubit Q and measurement device M start in the state $$\tfrac{1}{\sqrt{2}}(|0\rangle_Q+|1\rangle_Q)|0\rangle_M,$$ and that the measurement interaction has the form $|0\rangle_Q|0\rangle_M\to|0\rangle_Q|0\rangle_M,|1\rangle_Q|0\rangle_M\to|1\rangle_Q|1\rangle_M$. The state after the measurement will be $$\tfrac{1}{\sqrt{2}}(|0\rangle_Q|0\rangle_M+|1\rangle_Q|1\rangle_M).$$

The reduced state of the qubit will then be $$\rho_Q = \tfrac{1}{2}(|0\rangle_{QQ}\langle0|+|1\rangle_{QQ}\langle1|).$$

Now, if the qubit and the measuring apparatus may interact with you Y via the environment E in which case the state will be $$\tfrac{1}{\sqrt{2}}(|0\rangle_Q|0\rangle_M|0\rangle_E|0\rangle_Y+|1\rangle_Q|1\rangle_M|0\rangle_E|0\rangle_Y).$$ There will then be two versions of you, one of them will see 0, in which case the qubit's relative state will be $|0\rangle_Q$ and the other will see 1 in which case the relative state will be $|1\rangle_Q$. You need not know that this interaction has taken place, so the question is ill-posed since the answer could be any of $|0\rangle_Q,|1\rangle_Q,\rho_Q$.

I suspect the person who asked the question is looking for the answer $\rho_Q$, since there is a common, but mistaken, idea that the states of quantum systems depend on knowledge. Since you don't know the measurement result, you treat the system as though you have not interacted with it directly or indirectly. In reality, the state of a system depends on how it has evolved, including what interactions it has undergone.

• "The state of the qubit does not depend on what you or anyone else knows about it." -- Doesn't a quantum state describe our knowledge about a system? – Norbert Schuch Oct 19 '15 at 16:25
• No. The state of a quantum system does not represent your knowledge. In the Heisenberg picture it represents the information recorded mostly the environment about the values of decoherent observables. In the Schrodinger picture it represents a mixture of those records and how the system evolves over time. To put things another way: any physical theory worth having represents how the world actually works, not what you happen to imagine about how it works. – alanf Oct 19 '15 at 19:13
• I don't think we have a good enough insight into the measurement problem to make bold claims such as "The state of the qubit does not depend on what you or anyone else knows about it". The answer is great without that part. – Javier Oct 19 '15 at 21:23
• @alanf You should at least acknowledge that there are different opinions on what a quantum state describes, and what a physical theory describes. – Norbert Schuch Oct 20 '15 at 6:20
• I acknowledged the other opinion in the last paragraph. – alanf Oct 20 '15 at 19:17

I like alanf's answer but I'd like to expand on it somewhat. The way we represent a quantum state that is unknown to us is by using a density matrix, $\rho$. This is defined such that if there are $i$ states, and each one has probability $p_i$, the density matrix is: $$\rho = \sum_{i} | \psi_i \rangle \langle \psi_i |$$ The idea here is not necessarily that the quantum state "depends on our knowledge" but rather that the quantum state is not known to us, and we want to account for all the probabilities. It's exactly analogous to, in classical mechanics, talking about the distribution in phase space of a system. It's not that we don't (again, in classical mechanics) believe that the system has particular atomic velocities and positions--it's that we don't know them, and so to predict the results of measurements we have to average over the definite results of the indefinite possible states.

Now, in quantum mechanics, you have a little bit of ontological freedom as to what you think happens when you make a measurement. I like alanf's approach best myself--that measurement entangles systems and observers, rather than collapsing wavefunctions or anything. However, throughout QM we find that our interpretation tends not to matter. The important thing to to note about an answer in which the wavefunction separates into "the qubit was zero and the environment now reflects this" and "the qubit was one and the environment now reflects this" is that it has the same end result as a case where the qubit has "collapsed" to one state or the other because these branches can no longer reach each other, and you can deal with only one or the other fact. In this interpretation, a density matrix is an approximation to account for the fact that you are not considering an entire system--you "integrate over" or "trace out" degrees of freedom that are inaccessible to you to get an expectation value, which now contains an element of randomness/indeterminacy that's different than the usual quantum uncertainty for measurements--it truly reflects your lack of knowledge rather than different measurement outcomes in the pure state $| \psi \rangle$.

All of this is a lot of philosophy to say that when talking about an individual qubit, you want to talk about only that qubit, rather than the entire universe that may have become entangled with it. Because this inevitably involves throwing out some of the information--going from a definite state that's a superposition of different environment states to an indefinite state that refers only to the qubit--this evolution looks strange and leads to QM's measurement mysteriousness. The study of how this occurs continuously and naturally is very interesting, and the phenomenon is known as "decoherence." But it's really just us making a necessary sacrifice, just as we can't do thermodynamics by keeping track of particle trajectories. By working with density matrices, we account for all the many different possibilities we can get from measurements and the like, and this makes it very useful because it allows us to interrogate all the various branches.

For this reason, I would answer that the state "of the qubit" (rather than "of the system" or "of the universe") is $$\rho = \frac{1}{2} | 0 \rangle \langle 0 | + \frac{1}{2} | 1 \rangle \langle 1 |$$ Properly, this is a mixture of states, rather than a state itself, but in quantum information "state" is quite commonly used to refer to these mixed states.

• Thanks, sadly I can only choose one answer, however your comment was very insightful. – user138901 Oct 20 '15 at 21:11