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--- eigenvalues_and_eigenvectors [2022/10/06 18:58] – [Eigenvalues of Functions] admin
+++ eigenvalues_and_eigenvectors [2022/10/06 23:47] (current) – [Consequences of the Spectral Theorem] admin
@@ Line 95: / Line 95: @@
   * Anti-Hermitian operators $\hat{A}^{\dagger} = -\hat{A}$: \[[\hat{A},\hat{A}^{\dagger}] = [\hat{A},-\hat{A}] = - [\hat{A},\hat{A}]= 0\]
   * Unitary Operators $\hat{A}^{\dagger}\hat{A} = \hat{A}\hat{A}^{\dagger} = \hat{I}$: \[[\hat{A},\hat{A}^{\dagger}] = \hat{A}\hat{A}^{\dagger} - \hat{A}^{\dagger}\hat{A} = \hat{I} - \hat{I} = 0.\]
+====== The Spectral Theorem ======
+The set of eigenvalues of an operator is called its //**spectrum**//.  Note, there //is// a connection between this mathematics terminology and the physicist's notion of the spectrum of an atom, i.e. the set of frequencies of electromagnetic radiation that it can emit and absorb.  The eigenvalues of the Hamiltonian operator in quantum mechanics are the possible energies of the system, so, for the Hamiltonian operator, the mathematician's //spectrum// is this set of possible energies.  But radiation is emitted and absorbed when the system makes a transition between its possible energy levels, so the set of possible energies of an absorbed or emitted photon are the magnitudes of the //differences// between the eigenvalues of the Hamiltonain.  Therefore, the physicists //spectrum// is the set of magnitudes of differences between the eigenvalues.  The mathematician's term //spectrum// is clearly inspired by the physicist's, although they are not quite the same thing.
+The spectral theorem for an operator with a discrete set of eigenvalues $a_1,a_2,\cdots$ is as follows.
+**Theorem**
+Let $a_1,a_2,\cdots$ be the (distinct) eigenvalues of an operator $\hat{A}$.  Then, $\hat{A}$ is a normal operator if an only if all of the following hold:
+  - Its eigenspaces are orthogonal: $\hat{P}_{a_j}\hat{P}_{a_k} = \delta_{jk}\hat{P}_{a_j}$.
+  - Its eigenspaces are //**complete**//: $\sum_j \hat{P}_{a_j} = \hat{I}$.
+  - The //**spectral decomposition**// holds: $\hat{A} = \sum_j a_j \hat{P}_{a_j}$.
+Note that it is possible that one of the eigenvalues $a_j = 0$, in which case the term $a_j P_{a_j} = 0$ may be omitted from the spectral decomposition.  However, the completness relation $\sum_j \hat{P}_{a_j} = \hat{I}$ only holds if we include the projector onto the eigenspace with eigenvalue zero.
+Note, if the operator $\hat{A}$ has a continuous spectrum on an interval $a_{\mathrm{min}} < a < a_{\mathrm{max}}$ then we would have to write orthogonality as $\hat{P}_a\hat{P}_{a'} = \delta(a - a') \hat{P}_a$, completeness as
+\[\int_{a_{\mathrm{min}}}^{a_{\mathrm{max}}} \hat{P}_{a} \, \mathrm{d}a = \hat{I},\]
+and the spectral decomposition as
+\[\hat{A} = \int_{a_{\mathrm{min}}}^{a_{\mathrm{max}}} a \hat{P}_{a} \, \mathrm{d}a.\]
+Even more generally, $\hat{A}$ might have a discrete set $a_1,a_2$ of eigenvalues within some interval as well as a continuous set $a_{\mathrm{min}} < a < a_{\mathrm{max}}$ in a disjoint interval, and then we would have to write the spectral decomposition as
+\[\hat{A} = \sum_j a_j \hat{P}_{a_j} + \int_{a_{\mathrm{min}}}^{a_{\mathrm{max}}} a \hat{P}_{a} \, \mathrm{d}a.\]
+The proof of the most general version of the spectral theorem is well beyond the scope of this course.  It can be found in many advanced mathematics textbooks such as
+  * [[https://doi.org/10.1007/BFb0101631|Henry Helson, "The Spectral Theorem", pp. 23--41 (Springer, 1986)]]
+In these notes, we will prove the finite-dimensional case.  That proof is fairly straightforward, but involved, so it has [[the_spectral_theorem|its own page]].  On this page, we will prove a small part of the theorem.
+**Theorem**
+The eigenvectors corresponding to distinct eigenvalues of a Hermitian operator $\hat{A}$ are orthogonal.
+**Proof**
+Let $a \neq b$ be distinct eigenvalues of $\hat{A}$ with eigenvectors $\ket{a}$ and $\ket{b}$.  Then we can calculate $\sand{b}{\hat{A}}{a}$ in two different ways.  First, acting with $\hat{A}$ on the right,
+\[\sand{b}{\hat{A}}{a} = \bra{b} \left ( \hat{A} \ket{a}\right ) = a \braket{b}{a}.\]
+Second, using hermiticity and acting with $\hat{A}$ to the left,
+\[\sand{b}{\hat{A}}{a} = \sand{b}{\hat{A}^{\dagger}}{a} = \left ( \bra{b}\hat{A}^{\dagger} \right ) \ket{a} = b^* \braket{b}{a} = b\braket{b}{a},\]
+where the last step follows because the eigenvalues of a Hermitian operator are real.
+Equating the two expressions gives $a \braket{b}{a} = b\braket{b}{a}$, which we can rearrange to
+\[(a-b) \braket{b}{a} = 0.\]
+In order to satisfy this equation, it must be the case that either $(a-b) = 0$ or $\braket{b}{a} = 0$, but we have assumed that $a$ and $b$ are distinct eigenvalues, so $(a-b) \neq 0$ and hence $\braket{b}{a} = 0$.
+====== Consequences of the Spectral Theorem ======
+One of the most important consequences of the spectral theorem is that any vector can be written in an eigenbasis of a normal operator, and, in that basis, the action of the operator is easy to compute.
+Let $\ket{a_j,1},\ket{a_j,2},\cdots$ be an orthonormal basis for the eigenspace corresponding to eigenvalue $a_j$.  Orthogonality and completeness imply that $\ket{a_1,1},\ket{a_1,2},\cdots,\ket{a_2,1},\ket{a_2,2},\cdots$ is a complete orthonormal basis for the whole Hilbert space:
+\[\braket{a_j,k}{a_n,m} = \delta_{j,n}\delta_{km},\qquad\qquad\qquad \sum_{j,k} \proj{a_j,k} = \hat{I}\]
+This implies that any vector $\ket{\psi}$ can be written as
+\[\ket{\psi} = \sum_{j,k} b_{jk} \ket{a_j,k},\]
+where $b_{jk} = \braket{a_j,k}{\psi}$.
+In this basis, the action of $\hat{A}$ can be computed as follows:
+\[
+  \hat{A}\ket{\psi} = \sum_{j,k} b_{jk} \hat{A}\ket{a_j,k} = \sum_{j,k} b_{jk} a_j \ket{a_j,k}.
+\]
+Another way of saying the same thing is to note that the matrix representation of $\hat{A}$ in the $\ket{a_j,k}$ basis is diagonal.  The matrix elements are:
+\[\sand{a_j,k}{\hat{A}}{a_n,m} = a_n \braket{a_j,k}{a_n,m} = a_n \delta_{jn}\delta_{km},\]
+and if we choose the ordering of basis vectors $\ket{a_1,1},\ket{a_1,2},\cdots,\ket{a_2,1},\ket{a_2,2},\cdots$, this means that the matrix is
+\[
+\left ( \begin{array}{cccccc}
+a_1 & 0 & \cdots & 0 & 0 & \cdots \\
+& a_1 & \cdots & 0 & 0 & \cdots \\
+\vdots & \vdots & \ddots & \vdots & \vdots & \ddots \\
+& 0 & \cdots & a_2 & 0 & \cdots \\
+& 0 & \cdots & 0 & a_2 & \cdots \\
+\vdots & \vdots & \ddots & \vdots & \vdots & \ddots
+\end{array} \right )
+\]
+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
+\[\hat{A}\ket{a} = a\ket{a}, \qquad\qquad \hat{A}\ket{b} = b\ket{b}.\]