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In one of the answers to this question on measuring one qubit it is explained that given a general two-qubit state $$ |\psi\rangle = \begin{bmatrix} \alpha_{00} \\ \alpha_{01} \\ \alpha_{10} \\ \alpha_{11} \end{bmatrix} = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle $$ one measures the most-significant (leftmost) qubit in the computational basis as follows:

  • the probability that the measured qubit collapses to $|0\rangle$ is $$ P\left[|0\rangle\right] = |\alpha_{00}|^2 + |\alpha_{01}|^2 $$

  • the normalized state after the measurement is $$ |\psi\rangle = \frac{\alpha_{00}|00\rangle + \alpha_{01}|01\rangle}{\sqrt{|\alpha_{00}|^2 + |\alpha_{01}|^2}}. $$

My question is: How does one normalize in the case the probability $|\alpha_{00}|^2 + |\alpha_{01}|^2$ is zero? (This happens, of course, if $\alpha_{00} = \alpha_{01} = 0$.)

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    $\begingroup$ Well how I interpret it is that if we are saying $|\alpha_{00}|^2 + |\alpha_{01}|^2 = 0$ , there won't be a state $|\psi\rangle = \frac{\alpha_{00}|00\rangle + \alpha_{01}|01\rangle }{\sqrt{|\alpha_{00}|^2 + |\alpha_{01}|^2 }}$(mentioned above) that our system can collapse to. If that is the case, is the question about normalizing still valid? $\endgroup$ Jul 22, 2021 at 10:02
  • $\begingroup$ What originally bothered me (and is still bothering) that one find formulas like this and nobody explains the not unlikely case where the denominator is zero. $\endgroup$ Jul 22, 2021 at 15:55
  • $\begingroup$ It's easier if you consider it a two-step procedure: find the state, then normalize. If the state is $0$ (not $|0\rangle$, but $0$), because it is orthogonal to the measurement device, then it is not a state, so you don't have to normalize it. It's like asking "what is the resultant state when $1=2$" - it doesn't matter what you say as an answer, no matter how absurd, because you that step never has any relevance $\endgroup$ Jul 22, 2021 at 18:02
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    $\begingroup$ I think I like this comment by Quantum Mechanic to be the answer. $\endgroup$ Jul 23, 2021 at 6:53

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The projection operator can be writen as $\hat{P}=|0\rangle\langle 0|$, the probability of collapse as $\langle \psi|\hat{P}|\psi\rangle$, and the measurement update rule as $$|\psi\rangle\to \frac{\hat{P}|\psi\rangle}{\sqrt{\langle \psi|\hat{P}|\psi\rangle}}.$$ Both the numerator and the denominator vanish, so one could consider taking limits to see what happens.

Suppose we take $|\psi\rangle=|0\rangle+\epsilon|\phi\rangle$ for some small $\epsilon$. We find $\hat{P}|\psi\rangle=\epsilon|\phi\rangle$ and $\langle\psi|\hat{P}|\psi\rangle=\epsilon^2 \langle \phi|\hat{P}|\phi\rangle$ such that the measurement update rule becomes $$|\psi\rangle\to\frac{\epsilon|\phi\rangle}{\sqrt{\epsilon^2\langle \phi|\hat{P}|\phi\rangle}}=\frac{|\phi\rangle}{\sqrt{\langle \phi|\hat{P}|\phi\rangle}}.$$ This is a properly normalized state, no matter how small we take $\epsilon$ to be! So the case of $\epsilon=0$ must be dealt with prescriptively, as opposed to continuing with the earlier mathematical manipulations, by saying that the update rule is only defined with normalization for results with nonzero success probability. An impossible measurement result need not have a physically well-defined state.

In sum: It's easier if you consider the whole process to be a two-step procedure: find the state, then normalize. If the state is 0 (not $|0⟩$, but 0), because it is orthogonal to the measurement device, then it is not a state, so you don't have to normalize it. It's like asking "what is the resultant state when $1=2$" - it doesn't matter what you say as an answer, no matter how absurd, because that state never has any relevance.

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  • $\begingroup$ So if you were to subsequently measure this 0 state afterwards with the projection operator $|1\rangle \langle1|$, which would give probability 1 if you had measured it before, what would you then get? Also, physically, what happens to this qubit if it collapses to this 0 state? Thanks! $\endgroup$
    – H_Boofer
    May 26, 2023 at 17:08
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    $\begingroup$ @H_Boofer if you get a result 0 there is no state left - it's as though you had a filter and you rejected the state, so nothing gets through on the other side. You will never measure any property of something that didn't get through a filter. Imagine you sort a bunch of particles into going left if they are $|0\rangle$ and right if they are $|1\rangle$. If none of the states you see are $|0\rangle$, then you go to the left and try to measure something about the states there, you won't have any states to measure! $\endgroup$ May 26, 2023 at 18:07
  • $\begingroup$ Thanks for your response. The answer does leave me confused however on how things like quantum key distribution work. It is a bit of a tangential question but in something like BB84, if an infiltrator in the channel measures in the correct basis for a qubit but the orthogonal state, what you said implies that the qubit will have no state. Then how would the receiver then be able to get the qubit and perform a measurement on it? $\endgroup$
    – H_Boofer
    May 26, 2023 at 18:34
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    $\begingroup$ @H_Boofer you should probably ask a new question on this site. But there is a difference between a measurement and a filter. If you measure, you get some measurement result, which cannot be $|0\rangle$ if your state is $|1\rangle$; if you filter, you sometimes let a state through and sometimes do not $\endgroup$ May 26, 2023 at 19:13

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