I have tried to solve the following question:
Consider the Hilbert space $\mathbb R^3$. Find the projection to the 2-dimensional subspace
$$\text{span}\{|0\rangle+|1\rangle,|0\rangle+|2\rangle\}$$.
I found the following idempotent operator which projects any vectors to that subspace:
$$
\begin{bmatrix}
0 & 1 & 1\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}.
$$
However, it is not Hermitian. How can I make it Hermitian? or should I find another solution which is both Hermitian and idempotent, can anyone guide me please?
On the other hand, if I change the Hilbert space in question to $\mathbb C^3$, is the solution differ?
Best Answer
Suppose you have a subset $V$ of a Hilbert space $H$, which is spanned by $m$ orthogonal elements $v_1,\ldots,v_m$. Then, the unique hermitian projection (also called orthogonal projection) $P_V$ onto $V$ can be written as $$ P_Vx = \sum_{k=1}^m \frac{1}{\|v_k\|^2}\langle v_k,x\rangle v_k \qquad\mbox{for}\ x\in H. $$ It can be easily verified, that this yields $P_V$ to be self-adjoint (hermitian) and idempotent.
In Dirac notation, this takes the slightly easier form $$ P_V = \sum_{k=1}^m \frac{1}{\|v_k\|^2}|v_k\rangle\langle v_k|.$$
If $H=\mathbb R^n$, then $\langle v_k|=v_k^T$ is the transposed vector of $v_k$. If $H=\mathbb C^n$, then $\langle v_k|=v_k^\dagger$ is the transposed and complex conjugated vector of $v_k$.
In your example, $V$ is spanned by the vectors $u_1 = \begin{pmatrix}1\\1\\0\end{pmatrix}$ and $u_2 = \begin{pmatrix}1\\0\\1\end{pmatrix}$. (I assume you have $\{|0\rangle,|1\rangle,|2\rangle\}$ as the standard basis in $\mathbb R^3$.) As $\langle u_1,u_2\rangle = 1$, we first need to orthogonalize the two vectors. Therefore, we set $v_1=u_1$ and $$v_2 = u_2-\frac{\langle u_2,v_1\rangle}{\|v_1\|^2}v_1 = \begin{pmatrix}1\\0\\1\end{pmatrix} - \frac12 \begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}\frac 12\\-\frac 12\\1\end{pmatrix}.$$ Now, $V=\operatorname{span}\{v_1,v_2\}$ and $\langle v_1,v_2\rangle = 0$. Then we obtain \begin{align} P_V &= \frac 12 \begin{pmatrix}1\\1\\0\end{pmatrix}\begin{pmatrix}1&1&0\end{pmatrix} + \frac{2}{3} \begin{pmatrix}\frac 12\\-\frac 12\\1\end{pmatrix}\begin{pmatrix}\frac 12&-\frac 12&1\end{pmatrix} \\&= \frac 12\begin{pmatrix}1&1&0\\1&1&0\\0&0&0\end{pmatrix}+\frac23\begin{pmatrix}\frac 14&-\frac 14&\frac 12\\-\frac 14&\frac 14&-\frac12\\\frac 12&-\frac 12&1\end{pmatrix} \\&= \begin{pmatrix}\frac 23&\frac 13&\frac 13\\\frac 13&\frac 23&-\frac13\\\frac 13&-\frac13&\frac 23\end{pmatrix}. \end{align}