Prove that the projection operator $\mathbb P_+\equiv|+z\rangle\!\langle +z|$ is Hermitian

Question

Use Dirac notation (the properties of kets, bras and inner products) directly to establish that the projection operator $\mathbb{\hat P}_+$ is Hermitian. Use the fact that $\mathbb{\hat P}^2_+=\mathbb{\hat P}_+$ to establish that the eigenvalues of the projection operator are $1$ and $0$.

I know how to prove this using mathematical notation, i.e. for any $x,y\in V$ we must show that $\langle x, \ \mathbb{\hat P}_+y\rangle = \langle \mathbb{\hat P}_+x, \ y\rangle$ but how can I prove the way the book suggested, i.e. using Dirac notation and the properties of kets and bras?

Answer 1

Any projection operator can be written in the form $$ P = \sum_{j = 1}^r |\psi_j \rangle \langle \psi_j | $$ Where $\psi_1,\dots,\psi_n$ is an orthonormal basis of our Hilbert space. Given $\psi = c_1\psi_1 + \cdots + c_n \psi_n$, we calculate $$ \langle\psi| P = \langle \psi | \left(\sum_{j = 1}^r |\psi_j \rangle \langle \psi_j | \right) = \sum_{j = 1}^r \langle \psi \mid \psi_j \rangle \langle \psi_j | =\\ \sum_{j = 1}^r \langle \psi_j \mid \psi \rangle^* \langle \psi_j | = \sum_{j = 1}^r c_j^* \langle \psi_j | $$ This is the bra corresponding to the ket $P |\psi \rangle = \sum_{j=1}^r c_j | \psi_j \rangle$. So, $P$ is self-adjoint.

Answer 2

I assume from the comments that $P$ is a rank-1 projection, and so it is of the form $$P=|\psi\rangle\langle\psi|$$ for some vector $|\psi\rangle$ in the Hilbert space. Observe that, in a sense, $|\psi\rangle^*=\langle\psi|$ in Dirac notation, whence $$P^* = (\langle\psi|)^*(|\psi\rangle)^*,$$ and since the $*$ is involutive one has $$P^*=|\psi\rangle\langle\psi| = P.$$

The "idempotency" $P^2 = P$ comes from a direct computation $$P^2 = |\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi|=P$$ since $|\psi\rangle$ is assumed to be a vector of norm one, so that $\Vert\psi\Vert^2 = \langle\psi|\psi\rangle=1$.

Answer 3

I figured out the solution to my question. It is similar to the answers given but I will write it anyways for future reference.

If $\mathbb{\hat P}_+$ is Hermitian then $\langle \psi|\mathbb{\hat P}_+|\phi\rangle=\langle\phi|\mathbb{\hat P}_+|\psi\rangle^*.$

$$\langle\psi|\mathbb{\hat P}_+|\phi\rangle=\langle\psi|+z\rangle\langle+z|\phi\rangle=\langle+z|\psi\rangle^*\langle\phi|+z\rangle^*=\left(\langle\phi|+z\rangle\langle+z|\psi\rangle\right)^*=\left(\langle\phi|\mathbb{\hat P}_+|\psi\rangle\right)^*.$$

Therefore $\mathbb{\hat P}_+$ is Hermitian.

For an eigenstate $\mathbb{\hat P}_+|\lambda\rangle=\lambda|\lambda\rangle$ so $$\mathbb{\hat P}_+^2|\lambda\rangle=\lambda\mathbb{\hat P}_+|\lambda=\lambda^2|\lambda\rangle$$ but since $\mathbb{\hat P}_+^2=\mathbb{\hat P}_+$ we have $\lambda^2=\lambda$ therefore $\lambda=0,1.$

Answer 4

Two basic facts are that

1. if $|w\rangle$ and $|v\rangle$ are two vectors, then

$$ (|w\rangle\langle v|)^\dagger=|v\rangle\langle w| $$ (*)

2. in the general case $P=\sum\limits_{i}|i\rangle\langle i|$, where $|i\rangle$ are vectors of an othonormal basis. In your case $P=|\psi\rangle\langle \psi|$

(**)

Therefore, in order to prove that $P^\dagger=P$, we have simply that $P=|\psi\rangle\langle \psi|=(|\psi\rangle\langle \psi|)^\dagger=P^\dagger$.

To see this (*) we use the uniqueness of the Hermitian operator defined as

$$ (|v\rangle, A|w\rangle)=(A^\dagger|v\rangle, |w\rangle) $$ with $A=|v\rangle\langle w|$, by substitution is easy to see that the unique way to obatin equality is setting $A^\dagger=|w\rangle\langle v|$.

For the (**) that enable us to use the same proof scheme for the general case of $P=\sum\limits_{i}|i\rangle\langle i|$, we can use the linearity of the inner product in its second argument to obtain $P^\dagger=\sum\limits_{i}(|i\rangle\langle i|)^\dagger$:

$$ (|v\rangle, (A+B)|w\rangle)=(|v\rangle, A|w\rangle)+(|v\rangle, B|w\rangle)=(A^\dagger|v\rangle, |w\rangle)+(B^\dagger|v\rangle, |w\rangle)=(|v\rangle, (A+B)^\dagger|w\rangle) $$