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Nielsen and Chuang (on page 188 exercise 4.33) says that the circuit including CNOT and Hadamard is performing a measurement in the Bell basis. But I can't see how.

enter image description here

The matrix representing the circuit functionality is:

$$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 &-1 & 0 \end{bmatrix}$$

Now, what I have found out is, that instead of measuring in the Bell state, it is actually simply converting $|00\rangle$ to $\beta_{00}$, $|01\rangle$ to $\beta_{01}$ and vice versa and so on. So it's actually transforming from one basis to another. But how is it a measurement in the Bell state?

Suppose you take the vector $(1,0,0,0)$ in the Bell state it's written $(1,0,1,0)$ then when measuring in Bell state I would expect you get either $(1,0,0,0)$ which is $|00\rangle + |11\rangle$ or $(0,0,1,0)$ which is $|00\rangle-|11\rangle$. But why isn't it doing the measurement but just converting from basis to basis and yet Wikipedia says it's a measurement?

Sanchayan Dutta
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bilanush
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2 Answers2

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  1. You're correct that unitary gates like the Hadamard and CNOT do not perform any measurement. In fact, measurement isn't a unitary operation at all! There are separate measurements gates to measure the individual qubit states. The basis which the measurement gates will use is implementation dependent. However, in your specific circuit (N&C p. 188, Ex. 4.33) they're using the $\{|0\rangle, |1\rangle\}$ basis. Notice the meter type gates in your circuit. Those are the measurement gates. That circuit is essentially the one used to distinguish between the four orthogonal Bell states. They're mapped to $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$ respectively. The mappings are (notations explained here):

$$|\Phi^+\rangle \to |00\rangle$$ $$|\Phi^-\rangle \to |10\rangle$$ $$|\Psi^+\rangle \to |01\rangle$$ $$|\Psi^-\rangle \to |11\rangle$$

  1. As for how we can distinguish the four Bell states using two measurement gates, read Craig Gidney's answer to Distinguishing between orthogonal bell states.

Typically, a Bell basis measurement is done with a CNOT and a Hadamard like so:

Bell basis measurement

It's exactly the reverse of making a Bell pair.

You can also simply do parity measurements along the X and Z axes:

Bell basis parity measurement

Further Reading:

  1. If all quantum gates must be unitary, what about measurement?

  2. What does “measurement in a certain basis” mean?

  3. How to measure relative phases of quantum states?

Sanchayan Dutta
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    @bilanush Can you give me the link to the Wikipedia page? – Sanchayan Dutta Mar 03 '19 at 11:42
  • Better look at QCQI by Chuang Nielsen page 188 excersie 4:33 . Also Wikipedia on bell state – bilanush Mar 03 '19 at 11:50
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    @bilanush I clearly see two measurement gates there (N&C page 188, exercise 4.33). What's the problem? – Sanchayan Dutta Mar 03 '19 at 11:54
  • Ok, it's not the circuit you sketcted in your answer before.. It's just CNOT followed by Hadamard and I didn't get your answer – bilanush Mar 03 '19 at 12:13
  • @bilanush 1. The specific circuit is irrelevant here. 2. You had not linked the circuit you were referring to initially, so I just gave an example. 3. We can't read your mind, so when you're talking about specific circuits, please add the image in the question itself. – Sanchayan Dutta Mar 03 '19 at 12:16
  • @bilanush Which part of the answer did you not understand? Just saying "I didn't get your answer" isn't too helpful. Please consider elaborating. A simple CNOT followed by Hadamard does not perform any measurement. The measurement gates are compulsory. – Sanchayan Dutta Mar 03 '19 at 12:17
  • I didn't understand what you said because my question was why Wikipedia and qcqi say it's performing a measurement in Bell basis? – bilanush Mar 03 '19 at 13:15
  • So what you say is, that it doesn't perform measurement but rather take (1,0,0,0) for example to (1,0,1,0) which is the same vector written in bell basis. And then meter symbol performing a measurement so it collapse to either (1,0,0,0) or to (0,0,10,)in bell basis . Is that correct? – bilanush Mar 03 '19 at 13:18
  • @bilanush It's a bit more subtle. The single-qubit measurement gates are working in the ${|0\rangle, |1\rangle}$ basis rather than the ${|0\rangle+|1\rangle, |0\rangle-|1\rangle}$ basis. The CNOT and Hadamard circuit is simply mapping the Bell states to $|00\rangle, |01\rangle, |10\rangle$ and $|11\rangle$. After that, the measurement (meter) gates measure the qubits in the standard ${|0\rangle, |1\rangle}$basis. However, as there is a clear correspondence (one-to-one mapping) between the Bell basis and Z-basis, it's equivalent to measuring in the Bell basis. – Sanchayan Dutta Mar 03 '19 at 13:22
  • So you're correct: $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$ and $(0,0,0,1)$ will correspond to the 4 different Bell basis states. – Sanchayan Dutta Mar 03 '19 at 13:28
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Just to supplement the given answer, if you want to know what the effective measurement is, you start from the measurement operators and work backwards. So, the two measurements mean that you're projecting on the states $|00\rangle,|01\rangle,|10\rangle,|11\rangle$. How do these transform under the preceding gates? Immediately before the measurement, we have a Hadamard on the first quibt. Incorporating that into the measurement basis means measuring in $|+0\rangle,|+1\rangle,|-0\rangle,|-1\rangle$. Finally, we incorporate the controlled-not, and we see that the overall thing corresponds to measurement in the Bell basis.

DaftWullie
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