0
$\begingroup$

For context, this is from Page 27 of Nielsen, M. A., & Chuang, I. L. (2010). Quantum Computation and Quantum Information: 10th Anniversary Edition. Cambridge University Press:

She then sends the first qubit through a Hadamard gate, obtaining: Blockquote
This state may be re-written in the following way, simply by regrouping terms: Blockquote
This expression naturally breaks down into four terms. The first term has Alice’s qubits in the state |00〉, and Bob’s qubit in the state α|0〉 + β|1〉 – which is the original state |ψ〉. If Alice performs a measurement and obtains the result 00 then Bob’s system will be in the state |ψ〉. Similarly, from the previous expression we can read off Bob’s post- measurement state, given the result of Alice’s measurement: enter image description here
Depending on Alice’s measurement outcome, Bob’s qubit will end up in one of these four possible states.

Regrouping the terms in expression 1.31 in Quantum Computing and Quantum Information, Nielsen and Chuang
I understand how the calculations are being done here, but I still don't understand as how did the information regarding the transferred qubit reach Bob. What physical significance does the distributive law have? Through this regrouping, how did we manage to influence Bob's qubit?

What exactly is taking place? To use a classical analogy, let us say that A has 2 containers, one with x balls and the other with either 0 or 1 ball, A knows that if A's box has 0 balls then B's box must also have 0 balls (same for 1).
Now, since we have a total of either (x+0+0) balls or (x+1+1) balls, we also have (0+0+x) balls or (1+0+[x+1]) balls. Voila, B now has a different number of balls in his box! If we open our first box and see 0 balls, then B's box has x balls! So what's happening here physically?

$\endgroup$
1
  • $\begingroup$ Hi and welcome to Quantum Computing SE. Please do not post scans of text. Rather use citation. This helps search machines to find your post. $\endgroup$ Commented May 31, 2023 at 6:35

1 Answer 1

1
$\begingroup$

Consider the following circuit:

enter image description here

enter image description here


Forcing measurement now on qubits 1 & 2 will collapse the yet unmeasured qubit 3 into exactly 1 of 4 possible states:

$$|00\rangle \rightarrow a|0\rangle + b|1\rangle$$ $$|01\rangle \rightarrow b|0\rangle + a|1\rangle$$ $$|10\rangle \rightarrow a|0\rangle - b|1\rangle$$ $$|11\rangle \rightarrow -b|0\rangle + a|1\rangle$$

Thus, measuring qubits 1 & 2 has transferred the proportionally identical coefficients $a,b$ onto the yet unmeasured qubit 3.

$\endgroup$
7
  • $\begingroup$ Thank you for the calculations, but this leaves me with the same doubt: What exactly is taking place? To use a classical analogy, let us say that A has 2 containers, one with x balls and the other with either 0 or 1 ball, A knows that if A's box has 0 balls then B's box must also have 0 balls (same for 1). Now, since we have a total of either (x+0+0) balls or (x+1+1) balls, we also have (0+0+x) balls or (1+0+[x+1]) balls. Voila, B now has a different number of balls in his box! If we open our first box and see 0 balls, then B's box has x balls! So what's happening here physically? $\endgroup$ Commented Jun 1, 2023 at 6:20
  • $\begingroup$ @AlanWhitteaker i have no idea what is happening internally, just as with the double slit experiment and many other quantum experiments. But externally we could test if this procedure can actually replicate the coefficients $\alpha, \beta$ or not. It was done some years ago I believe, and the procedure indeed worked. $\endgroup$
    – James
    Commented Jun 1, 2023 at 7:30
  • $\begingroup$ In all the other cases, atleast we have some theoretical framework in which this fits into (for example, wave nature in double slit leading to explanation via interference). Is there any such explanation here? $\endgroup$ Commented Jun 1, 2023 at 10:03
  • $\begingroup$ @AlanWhitteaker The algebra rules above leads to this result, so the question is I guess whether we trust and put complete faith in this quantum algebra? If we trust the manipulation rules to correctly describe reality, then the result presented itself. There are much better experts here than me who can probably explain why the above algebra rules correctly describe this 3 qubit quantum system... $\endgroup$
    – James
    Commented Jun 1, 2023 at 11:08
  • $\begingroup$ @AlanWhitteaker On the other hand, even the simplest entanglement case, ie. open one box to see 0, then open the other box and we are guaranteed to see 1, is not explainable classically, don't you think? $\endgroup$
    – James
    Commented Jun 1, 2023 at 11:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.