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Theme 2 — Hilbert Spaces and Dirac Notation

Adjoints and projectors

Keywords: adjoint · conjugate transpose · orthogonal projector · idempotent

Exercise 1 : Adjoints and projectors

Question 1
Let HH be a complex Hilbert space (antilinear in the first argument). Recall that the adjoint A†A^\dagger of an operator AA is defined by: ⟨u,A†v⟩=⟨Au,v⟩∀ u,v∈H.\langle u, A^\dagger v \rangle = \langle Au, v \rangle \qquad \forall\, u, v \in H.

Question 2
Show that if AA is represented by a matrix (Aij)(A_{ij}) in an orthonormal basis, then (A†)ij=Aji∗(A^\dagger)_{ij} = A_{ji}^*.

Solution
Let ∣ei⟩|e_i\rangle be an orthonormal basis, with Aij=⟨ei,Aej⟩A_{ij} = \langle e_i, A e_j\rangle. By the definition of the adjoint:

(A†)ij=⟨ei,A†ej⟩=⟨Aei,ej⟩=⟨ej,Aei⟩∗=Aji∗.(A^\dagger)_{ij} = \langle e_i, A^\dagger e_j \rangle = \langle A e_i, e_j \rangle = \langle e_j, A e_i \rangle^* = A_{ji}^*.

Question 3
Let ∣ψ⟩∈H|\psi\rangle \in H be a normalised vector. We define P=∣ψ⟩⟨ψ∣P = |\psi\rangle\langle\psi|, that is, the operator P ⁣:∣v⟩↦⟨ψ,v⟩ ∣ψ⟩P \colon |v\rangle \mapsto \langle \psi, v\rangle\, |\psi\rangle. Show that PP is self-adjoint (P†=PP^\dagger = P) and idempotent (P2=PP^2 = P).

Solution
Self-adjointness. For all ∣u⟩,∣v⟩∈H|u\rangle, |v\rangle \in H:

⟨u,P†v⟩=⟨Pu,v⟩=⟨⟨ψ,u⟩ψ, v⟩=⟨ψ,u⟩∗ ⟨ψ,v⟩=⟨u,ψ⟩ ⟨ψ,v⟩=⟨u,Pv⟩,\langle u, P^\dagger v \rangle = \langle Pu, v \rangle = \langle \langle \psi, u \rangle \psi,\, v \rangle = \langle \psi, u \rangle^*\, \langle \psi, v \rangle = \langle u, \psi \rangle\, \langle \psi, v \rangle = \langle u, Pv \rangle,

and hence P†=PP^\dagger = P.

Idempotence.

P2∣v⟩=P(⟨ψ,v⟩∣ψ⟩)=⟨ψ,v⟩ P∣ψ⟩=⟨ψ,v⟩ ⟨ψ,ψ⟩ ∣ψ⟩=⟨ψ,v⟩ ∣ψ⟩=P∣v⟩,P^2|v\rangle = P\bigl(\langle\psi,v\rangle|\psi\rangle\bigr) = \langle\psi,v\rangle\, P|\psi\rangle = \langle\psi,v\rangle\,\langle\psi,\psi\rangle\,|\psi\rangle = \langle\psi,v\rangle\,|\psi\rangle = P|v\rangle,

where we used ⟨ψ,ψ⟩=1\langle\psi,\psi\rangle = 1.

Question 4
Show that 1−P\mathbf{1} - P is also an orthogonal projector, and interpret it geometrically.

Solution
Set Q=1−PQ = \mathbf{1} - P. Then Q†=1−P†=QQ^\dagger = \mathbf{1} - P^\dagger = Q and Q2=1−2P+P2=1−P=QQ^2 = \mathbf{1} - 2P + P^2 = \mathbf{1} - P = Q. Thus QQ is an orthogonal projector; its image is the orthogonal complement (Im P)⊥={∣ψ⟩}⊥(\mathrm{Im}\, P)^\perp = \{|\psi\rangle\}^\perp.