How can I make an idempotent projection to an Hermitian operator in a Hilbert space?

Question

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?

Answer 1

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}

Answer 2

Let $\{|v_j\rangle\}_{j=1}^m \subset \mathbb R^3$, where $0 \le m \le 3$ and $n = 3$. Let the subspace spanned by $\{|v_j\rangle\}_{j=1}^m$ be denoted $$ V \equiv \text{span}\{|v_j\rangle\}_{j=1}^m $$ First one has to find an orthonormal basis in $V$, for the tensor ket-bra notation to work. Denote this ONB in $V$ by $\{|u_j\rangle\}_{j=1}^{m'}$, where $m' \le m$.

The projector operator (in the tensor bra-ket notation or the like) that projects the space onto $V$ is $$ P:= \sum_{j=1}^{m'} |u_j\rangle\langle u_j| $$ and is clearly hermitian.

For the case at hand if the given kets $| 0\rangle,| 1\rangle,| 2\rangle$ are the standard basis in $\mathbb R^3$, then we orthonormalise the vectors $\{|v_j\rangle\}_{j=1}^m$ (by the Gram-Schmidt process) to get the $\{|u_j\rangle\}_{j=1}^{m'}$, which are in terms of {$| 0\rangle,| 1\rangle,| 2\rangle$}. So it remains to substitute these $\{|u_j\rangle\}_{j=1}^{m'}$ in the operator and then expand the ket-bras to get the matrix elements wrt {$| 0\rangle,| 1\rangle,| 2\rangle$}.

the answer is $$ \begin{pmatrix} \frac{2}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} \\ \end{pmatrix} $$ which is an idempotent (checked by Mathematica) and hermitian (symmetric in $\mathbb R^{3\times3}$). If I have to add the details of the solution, tell me in the comment below.