# Why is $|P_0- P_1|=1$?

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

• Hi it follows from union bound donnit? Nov 6, 2022 at 13:19
• notice how one of the two will be 1 and other will be 0
– glS
Nov 6, 2022 at 15:37
• From my understanding, being sure of our results means that our state is either $|0\rangle$ or $|1\rangle$, no? Nov 6, 2022 at 17:22
• @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 $|+\rangle$
– glS
Nov 6, 2022 at 19:00
• @user206904 where does this come from? Is it a textbook? Nov 7, 2022 at 0:07

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.