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I have a question

we have $ |0 \rangle, |1 \rangle, |+ \rangle$ and $|- \rangle, $ defined as usual.

Let $P_0$= probability that a state be in 0, $P_+$= probability that a state be in +, and same for $P_1$ and $P_-$

The notes mention that if we are sure of our result in the basis $\{ |0 \rangle, |1\rangle \}$ then $|P_0- P_1|=1$ and if the basis is $\{ |+ \rangle, |-\rangle \}$ then $|P_+- P_-|=1$

Can anyone explain why is that?

Also we have $ (P_0- P_1 )^2 + (P_+- P_- )^2 \leq 1$ but I don't see why neither...

Thanks

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  • $\begingroup$ Hi it follows from union bound donnit? $\endgroup$ Commented Nov 6, 2022 at 13:19
  • $\begingroup$ notice how one of the two will be 1 and other will be 0 $\endgroup$
    – glS
    Commented Nov 6, 2022 at 15:37
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    $\begingroup$ From my understanding, being sure of our results means that our state is either $|0\rangle$ or $|1\rangle$, no? $\endgroup$ Commented Nov 6, 2022 at 17:22
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    $\begingroup$ @user206904 indeed, that is only true for $P_0,P_1$ when you have either $|0\rangle$ or $|1\rangle$, and similarly for $P_\pm$ and $|\pm\rangle$. That's what I thought was the assumption by reading your post. It's obviously not true in general, just try to compute $P_0-P_1$ for $|+\rangle$ $\endgroup$
    – glS
    Commented Nov 6, 2022 at 19:00
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    $\begingroup$ @user206904 where does this come from? Is it a textbook? $\endgroup$ Commented Nov 7, 2022 at 0:07

1 Answer 1

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If you are promised that you will receive a qubit in a classical state and you also know the basis it is in, then it is certain that when you perform a measurement in that basis, the resulting probabilities will satisfy $|P_0 - P_1| = 1$. This means your information is classical, and if you pick the proper basis, you will retrieve the correct bit with probability 1.

For example, you get a qubit in state $|0\rangle$, and you are told that you should use the basis $\{|0\rangle, |1\rangle\}$. Then measuring in that basis will always get you the equality $|P_0 - P_1| = 1$. This is because you will always get the outcome $|0\rangle$, so $P_0 =1$ and $P_1 = 0$.

However, let's say that you decided to measure in $\{|+\rangle, |-\rangle\}$ basis instead. In that case, you get completely different probabilities. To see this, note that in the basis $\{|+\rangle, |-\rangle\}$ the state $|0\rangle$ is given by the following equality: $$|0\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle.$$

Since you are measuring in $\{|+\rangle, |-\rangle\}$, the probabilities for outcomes $|+\rangle$ and $|-\rangle$ are $P_+ = 1/2$ and $P_- = 1/2$ respectively. It follows, that $|P_+ - P_-| = |1/2 - 1/2| = 0$. This means your outcome is not certain because you picked a different basis.

The main point here is that the probability of a certain outcome depends on the basis you choose.

Given the discussion above, the inequality $(P_0- P_1 )^2 + (P_+- P_- )^2 \leq 1$ should be more or less clear now. When we choose between two mutually unbiased bases such as $\{|+\rangle, |-\rangle\}$ or $\{|0\rangle, |1\rangle\}$ we get that one of the squared terms is exactly 1, and the other term is exactly 0 for the state given in one of the basis states. So the inequality is satisfied. In a more general case, the two squared terms will be both non-zero but their sum will be always less than or equal to 1. This follows from the mutually unbiased relationship between $\{|+\rangle, |-\rangle\}$ and $\{|0\rangle, |1\rangle\}$.

Note: The bases $\{|+\rangle, |-\rangle\}$ and $\{|0\rangle, |1\rangle\}$ are called mutually unbiassed since if any state of one basis is measured in the other basis, the outcomes are always equally likely.

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