One of my discoveries as a physicist was that, despite all attempts at clarity, we still have different meanings for the same words and use different words to refer the the same thing. When Alice says measurement, Bob hears a `quantum to classical channel', but Alice, a hard-core Everettian, does not even believe such channels exist. When Charlie says non-local, he means Bell non-local, but string theorist Dan starts lecturing him about non-local Lagrangian terms and violations of causality. And when I say non-local measurements, you hear ???? ?????. Let me give you a hint, I do not mean 'Bell non-local quantum to classical channels', to be honest, I am not even sure what that means.

So what do I mean when I say measurement? A measurement is a quantum operation that takes a quantum state as its input and spits out a quantum state **and** a classical result as an output (no, I am not an Everettian). For simplicity I will concentrate of a special case of this operation, a projective measurement of an observable A.

The classical result of a projective measurement is an eigenvalue of \(A\), but what is the outgoing state?

## The Lüders Measurement

Even the term projective measurement can lead to confusion, and indeed in the early days of quantum mechanics it did. When von Neumann wrote down the mathematical formalism for quantum measurements he missed an important detail about degenerate observables (i.e., Hermitian operators with a degenerate eigenvalue spectrum). In the usual projective measurement the state of the system after the measurement is uniquely determined by the classical result (an eigenvalue of the observable). Consequently, if we don't look at the classical result the quantum channel is a standard dephasing channel. In the case of a degenerate observable, the same eigenvalue corresponds to two or more orthogonal eigenstates. Seemingly the state of the system should correspond to one of those eigenstates, and the channel is a standard dephasing channel. But a degenerate spectrum means that the set of orthogonal eigenvectors is not unique, instead each eigenvalue has a corresponding subspace of corresponding eigenvectors. What Lüders suggested is that the dephasing channel does nothing within these subspaces.

### Example

Consider the two qubit observable \(A=|00\rangle\langle 00 |\). It has eigenvalues \(1,0,0,0\). A 1 result in this measurement corresponds to "The system is in the state \(|{00}\rangle\)." Following a measurement with outcome \(1\), the outgoing state will be \(|00\rangle\). Similarly, a 0 result corresponds to "The system is not in the state \(|{00}\rangle\)". But here is where the Lüders rule kicks in. Given a generic input state \(\alpha|{00}\rangle+\beta|{01}\rangle+\gamma|{10}\rangle+\delta|{11}\rangle\) and a Lüders measurement of \(A\) with outcome 0, the outgoing state will be \(\frac{1}{\sqrt{|\beta|^2+|\gamma|^2+|\delta|^2}}\left[\beta|{01}\rangle+\gamma|{10}\rangle+\delta|{11}\rangle\right]\).

Non-local measurements

The relation to non-locality may already be apparent from the example, but let me start with some definitions. A system can be called non-local if it has parts in different locations, e.g., one part on Earth and the other on the moon. A measurement is non-local if it reveals something about a non-local system as a whole. In principle these definitions apply to classical and quantum systems. Classically a non-local measurement is trivial, there is no conceptual reason why we can't just measure at each location. For a quantum system the situation is different. Let us use the example above, but now consider the situation where the two qubits are in separate locations. Local measurements of \(\sigma_z\) will produce the desired measurement statistics (after coarse graining) but reveal too much information and dephase the state completely, while a Lüders measurement should not. What is quite neat about this example is that the Lüders measurement of \(|{00}\rangle\) cannot be implemented without entanglement (or quantum communication) resources and two-way classical communication. To prove that entanglement is necessary, it is enough to give an example where entanglement is created during the measurement. To show that communication is necessary, it is enough to show that the measurement (even if the outcome is unknown) can be used to transmit information. The detailed proof is left as an exercise to the reader. The lazy reader can find it here.

## Comments

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Alex

Tue, 2016-02-16 14:37

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In your measurement example I suggest substituting alpha to delta in the scalar outside the []. This substituion gives a normalized state.

Aharon Brodutch

Tue, 2016-02-16 17:17

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Thanks for spotting the error. A corrected version will appear soon.

Colin

Tue, 2016-02-16 20:57

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I once gave a quiz question, what is the out come of a projective measurement |0><0| on state |1>. Most of the students answered the measurement destroys the state, of course this is wrong, the correct answer is final state remains unchanged i.e., |1>, the projection fails. Somehow it was difficult to drill this fact into the students, your post definitely helps.

Aharon Brodutch

Wed, 2016-02-17 14:13

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Thanks for the feedback. That's a great exam question, I wonder if it was a graduate or undergraduate course.

Colin Benjamin

Wed, 2016-02-17 22:33

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In our institute we have a 5 year undergraduate Master's program. In the final two years students take elective courses on various advanced topics. The elective courses are also open to First year Ph D students. The course I am talking about was an elective course on Quantum Information and it had both students in the final 2 years of their masters program as well as first year Ph D students.

Aharon Brodutch

Thu, 2016-02-18 12:20

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What's kind of interesting about your exam question is that it is actually completely classical. If you asked, I prepare a ball in Box |1>. I open box |0>, What happens to the ball? The answer would be obvious.

Colin Benjamin

Thu, 2016-02-18 22:13

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Exactly. It's another way to explain the very obvious.