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Theme 2 — Hilbert Spaces and Dirac Notation

General properties of orthogonal projectors

Keywords: orthogonal projector · eigenvalues · sum of projectors · product of projectors · commutation · orthogonal subspaces

Exercise 1 : General properties of orthogonal projectors

In a finite-dimensional Hilbert space H\H, an orthogonal projector is any operator P^\hat P satisfying P^2=P^\hat P^2=\hat P and P^†=P^\hat P^\dagger=\hat P.

Question 1
Show that the only possible eigenvalues of an orthogonal projector are 00 and 11.

Solution
If P^∣v⟩=λ∣v⟩\hat P\ket v=\lambda\ket v with ∣v⟩≠0\ket v\neq0, then P^2∣v⟩=λ2∣v⟩\hat P^2\ket v=\lambda^2\ket v. Since P^2=P^\hat P^2=\hat P, we also have P^2∣v⟩=λ∣v⟩\hat P^2\ket v=\lambda\ket v. Thus λ2=λ\lambda^2=\lambda, that is, λ=0\lambda=0 or λ=1\lambda=1. The eigenspace associated with 11 is the image of P^\hat P, and that associated with 00 is its kernel.

Question 2
Show that every vector ∣v⟩\ket v can be written uniquely as ∣v⟩=P^∣v⟩+(1−P^)∣v⟩\ket v=\hat P\ket v+(\mathbf{1}-\hat P)\ket v, where the two terms are orthogonal. Deduce that ∥P^v∥≤∥v∥\norm{\hat Pv}\leq\norm v and that ⟨v∣P^∣v⟩=∥P^v∥2\bra v\hat P\ket v=\norm{\hat Pv}^2.

Solution
The equality is immediate. The two terms are orthogonal because

⟨P^v|(1−P^)v⟩=⟨v∣P^†(1−P^)∣v⟩=⟨v∣(P^−P^2)∣v⟩=0,\braket{\hat Pv}{(\mathbf{1}-\hat P)v}=\bra v\hat P^\dagger(\mathbf{1}-\hat P)\ket v=\bra v(\hat P-\hat P^2)\ket v=0,

where we have used P^†=P^\hat P^\dagger=\hat P and then P^2=P^\hat P^2=\hat P. The Pythagorean theorem gives ∥v∥2=∥P^v∥2+∥(1−P^)v∥2≥∥P^v∥2\norm v^2=\norm{\hat Pv}^2+\norm{(\mathbf{1}-\hat P)v}^2\geq\norm{\hat Pv}^2. Finally, ∥P^v∥2=⟨v∣P^†P^∣v⟩=⟨v∣P^2∣v⟩=⟨v∣P^∣v⟩\norm{\hat Pv}^2=\bra v\hat P^\dagger\hat P\ket v=\bra v\hat P^2\ket v=\bra v\hat P\ket v. For a normalised vector, ⟨v∣P^∣v⟩\bra v\hat P\ket v is therefore a number between 00 and 11: this is what allows it to be interpreted as a probability in the measurement postulates.

Question 3
Let P^1\hat P_1 and P^2\hat P_2 be two orthogonal projectors. Show that P^1+P^2\hat P_1+\hat P_2 is an orthogonal projector if and only if P^1P^2=0\hat P_1\hat P_2=0. Interpret this condition.

Hint
For the forward implication, multiply the resulting relation on the left, then on the right, by P^1\hat P_1.

Solution
The sum is self-adjoint in all cases. Its square is (P^1+P^2)2=P^1+P^2+P^1P^2+P^2P^1(\hat P_1+\hat P_2)^2=\hat P_1+\hat P_2+\hat P_1\hat P_2+\hat P_2\hat P_1. The sum is therefore a projector if and only if

P^1P^2+P^2P^1=0.\hat P_1\hat P_2+\hat P_2\hat P_1=0 .

If P^1P^2=0\hat P_1\hat P_2=0, then P^2P^1=(P^1P^2)†=0\hat P_2\hat P_1=(\hat P_1\hat P_2)^\dagger=0, and the condition is satisfied. Conversely, suppose that the condition is satisfied. Multiplying it on the left by P^1\hat P_1 gives P^1P^2+P^1P^2P^1=0\hat P_1\hat P_2+\hat P_1\hat P_2\hat P_1=0; multiplying it on the right by P^1\hat P_1 gives P^1P^2P^1+P^2P^1=0\hat P_1\hat P_2\hat P_1+\hat P_2\hat P_1=0. Subtraction yields P^1P^2=P^2P^1\hat P_1\hat P_2=\hat P_2\hat P_1, and the condition then gives 2P^1P^2=02\hat P_1\hat P_2=0.

The condition P^1P^2=0\hat P_1\hat P_2=0 means that the image of P^2\hat P_2 is contained in the kernel of P^1\hat P_1, that is, the two images are orthogonal. The sum is then the projector onto the orthogonal direct sum of the two images. This is the case, for example, for two projectors onto eigenspaces of an observable associated with different eigenvalues.

Question 4
Show that if P^1\hat P_1 and P^2\hat P_2 commute, their product P^1P^2\hat P_1\hat P_2 is an orthogonal projector. What is its image? Give an example in R2\R^2 showing that the product of two non-commuting projectors is not an orthogonal projector.

Solution
If P^1P^2=P^2P^1\hat P_1\hat P_2=\hat P_2\hat P_1, then (P^1P^2)†=P^2P^1=P^1P^2(\hat P_1\hat P_2)^\dagger=\hat P_2\hat P_1=\hat P_1\hat P_2, and (P^1P^2)2=P^1P^2P^1P^2=P^12P^22=P^1P^2(\hat P_1\hat P_2)^2=\hat P_1\hat P_2\hat P_1\hat P_2=\hat P_1^2\hat P_2^2=\hat P_1\hat P_2. It is an orthogonal projector. Its image is the intersection of the two images: a vector in the image satisfies ∣w⟩=P^1P^2∣w⟩=P^2P^1∣w⟩\ket w=\hat P_1\hat P_2\ket w=\hat P_2\hat P_1\ket w, so it belongs to both images; conversely, if ∣w⟩\ket w belongs to both images, P^1P^2∣w⟩=P^1∣w⟩=∣w⟩\hat P_1\hat P_2\ket w=\hat P_1\ket w=\ket w.

In R2\R^2, take the projectors onto the xx-axis and onto the first angle bisector:

P1=(1000),P2=12(1111),P1P2=12(1100).\begin{aligned} P_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad P_2=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix},\qquad P_1P_2=\frac12\begin{pmatrix}1&1\\0&0\end{pmatrix}. \end{aligned}

This product is not symmetric, hence not self-adjoint; we also verify that (P1P2)2=12P1P2≠P1P2(P_1P_2)^2=\frac12P_1P_2\neq P_1P_2. In quantum mechanics, performing two successive measurements associated with non-commuting projectors is not equivalent to performing a single projective measurement: this is the origin of the incompatibility of observables.