Are projective measurement bases always orthonormal?

up vote 7 down vote accepted

Yes.

Remember that you require several properties of a projective measurement including $P_i^2=P_i$ for each projector, and $$ \sum_iP_i=\mathbb{I}. $$ The first of these show you that the $P_i$ have eigenvalues 0 and 1. Now take a $|\phi\rangle$ that is an eigenvector of eigenvalue 1 of a particular projector $P_i$. Use this in the identity relation: $$ \left(\sum_jP_j\right)|\phi\rangle=\mathbb{I}|\phi\rangle $$ Clearly, this simplifies to $$ |\phi\rangle+\sum_{j\neq i}P_j|\phi\rangle=|\phi\rangle. $$ Hence, $$ \sum_{j\neq i}P_j|\phi\rangle=0. $$ The $P_j$ are all non-negative, so the only way that this can be 0 is if $P_j|\phi\rangle=0$ for all $j\neq i$. (To expand upon this, assume there's a $P_k$ such that $P_k|\phi\rangle=|\psi\rangle\neq 0$. This means that $$ \sum_{j\neq i,k}\langle\psi|P_j|\phi\rangle=-\langle\psi|P_k|\phi\rangle, $$ so some terms must be negative, which is impossible if the eigenvalues are all 0 and 1.)

Here is another way to see this.

A projection $P$ is an operator such that $P^2=P$.

This directly implies that we can attach to each projector $P$ a set of orthonormal states that represent it, by choosing any orthonormal base for its range. More precisely, if $P_i$ has trace $\operatorname{tr}(P_i)=n$, then we can represent $P_i$ as a set of orthonormal states $\{\lvert\psi_{i,j}\rangle\}_{j=1}^n$. Note in particular that if $\operatorname{tr}(P_i)=1$ then this choice is unique, meaning that there is always a bijection between trace-1 projections and states.

The projector $P_i$ and the corresponding states are connected through $$P_i=\sum_{j=1}^n \lvert\psi_{ij}\rangle\!\langle \psi_{ij}\rvert.$$ In the simpler case of $\operatorname{tr}(P_i)=1$ this reads $P_i=\lvert\psi_i\rangle\!\langle\psi_i\rvert$.

Now, if you are asking for a projective measurement basis, then you require a set of operators which describes every possible outcome of your state. This condition is expressed mathematically by requiring $$\sum_i P_i=I,$$ which in terms of the associated ket states reads $$\sum_{ij}\lvert\psi_{ij}\rangle\!\langle\psi_{ij}\rvert=I,$$ which is the completeness relation for the vectors $\{\lvert\psi_{ij}\rangle\}_{ij}$. This immediately implies that this is also an orthonormal set (to see it, take for example the sandwich of this expression with any $\lvert\psi_{ij}\rangle$).

Orthogonality of $P_i$ is equivalent to orthogonality of the corresponding $\lvert\psi_{ij}\rangle$, thus the conclusion.

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