Oblique projector and orthogonal projector

We calculate $Q^2=\begin{pmatrix}1\cdot1+1\cdot0&1\cdot1+1\cdot0\\0&0\end{pmatrix}=Q$: the operator is idempotent, so it is a projector. For $\ket v=(x,y)$, we have $Q\ket v=(x+y,0)$. The image is therefore the line spanned by $(1,0)$, and the kernel is the line $x+y=0$, spanned by $(1,-1)$. These two lines are not orthogonal, since $\bigl\langle(1,0),(1,-1)\bigr\rangle=1\neq0$. We say that $Q$ is an oblique projector: it projects onto the $x$-axis parallel to the direction $(1,-1)$, rather than perpendicularly. Finally, $Q^\dagger=\begin{pmatrix}1&0\\1&0\end{pmatrix}\neq Q$: it is not self-adjoint.

We have $Q\ket v=\frac{1}{\sqrt2}(2,0)=(\sqrt2,0)$, whose norm is $\sqrt2>1=\norm v$. Unlike an orthogonal projector, which can only decrease the norm of a vector, an oblique projector can increase it.

Suppose that $\hat P^\dagger=\hat P$. If $\ket a=\hat P\ket u$ is in the image and $\ket b$ is in the kernel, then $\braket ab=\braket{\hat Pu}{b}=\braket{u}{\hat P^\dagger b}=\braket{u}{\hat Pb}=0$: the image and the kernel are orthogonal.

Conversely, suppose that the image $F$ is orthogonal to the kernel $G$. Since $\hat P$ is idempotent, every vector can be written as $\ket v=\hat P\ket v+(\mathbf{1}-\hat P)\ket v$, with $\hat P\ket v\in F$ and $(\mathbf{1}-\hat P)\ket v\in G$, since $\hat P(\mathbf{1}-\hat P)=0$. For two vectors $\ket v$ and $\ket w$, the orthogonality of $F$ and $G$ gives

$$\braket{w}{\hat Pv}=\braket{\hat Pw}{\hat Pv}=\braket{\hat Pw}{v},$$

by expanding $\ket w$ in the first equality and $\ket v$ in the second. Thus $\braket{w}{\hat Pv}=\braket{\hat Pw}{v}$ for all $\ket v,\ket w$, which means that $\hat P^\dagger=\hat P$.

We have $Q\ket v=\frac{1}{\sqrt2}(1+i,0)$, hence

$$\bra vQ\ket v=\frac12\bigl(1\cdot(1+i)+(-i)\cdot0\bigr)=\frac{1+i}{2}, \qquad \norm{Qv}^2=\frac{|1+i|^2}{2}=1 .$$

The number $\bra vQ\ket v$ is complex, and differs from $\norm{Qv}^2$. It clearly cannot represent a probability. For an orthogonal projector, by contrast, we showed in the preceding exercise that $\bra v\hat P\ket v=\norm{\hat Pv}^2$ is a real number between $0$ and $1$ for a normalised vector. Moreover, the orthogonal projectors onto the eigenspaces of an observable have orthogonal images and their sum is the identity, which ensures that the probabilities of the different outcomes are positive and sum to one. This is why the Born rule and the collapse postulate use orthogonal projectors.