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Can anyone explain how the CNOT matrix below is a valid presentation for the four-qubit states that follow after?

enter image description here 

|0 0> -> |0 0> 
|0 1> -> |0 1>
|1 0> -> |1 1>
|1 1> -> |1 0>

Source: Wikipedia

glS
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DrHamed
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  • how what? Can you clarify what you do not find clear in the wiki page? – glS Oct 27 '18 at 18:41
  • Is your confusion about the choice of ordering the basis for which of 00,01,10 and 11 go with rows/columns 1,2,3,4 of the matrix? So you know which rows/columns to put 1s vs 0s. – AHusain Oct 27 '18 at 19:09

1 Answers1

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The one concept that I think would really help you is knowing how to turn those 4 states, $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, into vectors, so that you can do the matrix multiplication.

Let me show you.

$$ \begin{align} |00\rangle = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix},|01\rangle = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, |10\rangle = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, |11\rangle = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \end{align} $$

Now if you do the matrix multplication: $\rm{CNOT} \times |00\rangle$
You will see that you will get exactly what you said, which is $|00\rangle$, and the same is true for the rest of them!

This is using the convention that $|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $|1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, and $|ab\rangle = |a\rangle \otimes |b\rangle$ where $\otimes$ is the left Kronecker product.

user1271772 No more free time
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  • Thank you -- can you give me a bit more details on how you turned |11> into a vector of [0 0 0 1] (imagine is as a column vector please!) – DrHamed Oct 27 '18 at 19:42
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    I gave the formula |ab> = a $\otimes$ b. So please try |11> = |1> $\otimes$ |1> ! – user1271772 No more free time Oct 27 '18 at 19:58
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    @DrHamed: Please look at the formula after the words "We can write out the matrix form", in this PDF: https://www.cs.cmu.edu/~odonnell/quantum15/lecture02.pdf . I believe that concludes my answer to this question. – user1271772 No more free time Oct 27 '18 at 20:12
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    Great, I have the Michael Nielsen book, it completely skipped this step, which I found it frustrating. Thanks again for this detailed answer! – DrHamed Oct 27 '18 at 20:13
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    Typo in 00 state, too small to make as suggested edit. – AHusain Oct 27 '18 at 20:53
  • @DrHamed I had the same point of confusion with Nielsen and Chuang. This answer clarified things for me, but, looking back at the text, Nielsen and Chuang demonstrate the Kronecker product in equations 2.50, 2.51, and 2.52. Their notation for |vw> is given a page earlier at the start of the section on tensor products. – 1ijk Dec 09 '18 at 00:19