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eigenvalues_and_eigenvectors [2022/10/06 23:30] – i dont got one admineigenvalues_and_eigenvectors [2022/10/06 23:47] (current) – [Consequences of the Spectral Theorem] admin
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 \end{array} \right ) \end{array} \right )
 \] \]
 +
 +Finally, in section 2.iii.2, we showed that a projection operator $\hat{P}$ is Hermitian $\hat{P}^{\dagger} = \hat{P}$ and idempotent $\hat{P}^2 = \hat{P}$.  We can now use the spectral theorem to prove the converse.
 +
 +Let $\ket{\psi}$ be a nonzero eigenvector of $\hat{P}$.  Since $\hat{P}$ is Hermitian, the corresponding eigenvalue, $a$, must be real.
 +\begin{equation}
 +\label{proj1}
 +\hat{P} \ket{\psi} = a\ket{\psi}.
 +\end{equation}
 +Now apply \hat{P} to both sides again
 +\[\hat{P}^2 \ket{\psi} = a\hat{P}\ket{\psi}.\]
 +On the left hand side, we will use idempotency $\hat{P}^2 = \hat{P}$ and on the right hand side we will use the eigenvalue equation $\hat{P} \ket{\psi} = a\ket{\psi}$ again.  This gives,
 +\begin{equation}
 +\label{proj2}
 +\hat{P} \ket{\psi} = a^2 \ket{\psi}.
 +\end{equation}
 +
 +Comparing equations \eqref{proj1} and \eqref{proj2}, we have $a^2 = a$, or $a(a-1)=0$. Since $a$ is real, this means that either $a=0$ or $a=1$.
 +
 +This means that $\hat{P}$ has (at most) two eigenspaces.  Let $\hat{P}_0$ be the projector onto the $a=0$ eigenspace and $\hat{P}_1$ the projector onto the $a=1$ eigenspace.  By the spectral decomposition:
 +\[\hat{P} = 0\hat{P}_0 + 1\hat{P}_1 = \hat{P}_1.\]
 +Since $\hat{P}_1$ is a projector, this implies that $\hat{P}$ is a projector.  In fact, it is the projector onto its eigenspace corresponding to the eigenvalue $1$.
 +
 +{{:question-mark.png?nolink&50|}}
 +====== In Class Activities ======
 +
 +  - Show that, if $\hat{A}^{-1}$ exists and $a$ is an eigenvalue of $\hat{A}$, then $\frac{1}{a}$ is an eigenvalue of $\hat{A}^{-1}$.
 +  - Show that, if $\hat{A}$ is unitary, i.e. $\hat{A}^{\dagger}\hat{A} = \hat{I}$, and $a$ and $b$ are distinct eigenvalues of $a$ with eigenvectors $\ket{a}, \ket{b}$ then $\braket{b}{a} = 0$.
 +
 +HINT: Compute $\sand{b}{\hat{A}^{\dagger}\hat{A}}{a}$ in two different ways.  Once using $\hat{A}^{\dagger}\hat{A} = \hat{I}$ and once using
 +\[\hat{A}\ket{a} = a\ket{a}, \qquad\qquad \hat{A}\ket{b} = b\ket{b}.\]
 +