Differences

This shows you the differences between two versions of the page.

--- eigenvalues_and_eigenvectors [2022/10/06 18:13] – [Eigenspaces and Degeneracy] admin
+++ eigenvalues_and_eigenvectors [2022/10/06 23:47] (current) – [Consequences of the Spectral Theorem] admin
@@ Line 7: / Line 7: @@
 The importance of eigenvalues and eigenvectors is that //**normal**// operators, which we shall define shortly, can be completely characterized in terms of them.  The most important types of operators for quantum mechanics are normal, including Hermitian and unitary operators.
+====== Eigenvalues of Hermitian, anti-Hermitian and Unitary Operators ======
+Hermitian, anti-Hermitian and unitary operators are the three classes of operators that are used most frequently in quantum mechanics.  We can say the following about their eigenvalues.
+  * **The eigenvalues of a Hermitian operator are real:**
+To prove this, Let $\ket{a}$ be an eigenvector of $\hat{A}$ with eigenvalue $a$ and compute $\sand{a}{\hat{A}}{a}$ in two different ways. First, by acting with $\hat{A}$ to the right,
+\[\sand{a}{\hat{A}}{a} = \bra{a} \left ( \hat{A} \ket{a}\right ) = a \braket{a}{a} = a.\]
+Second, using hermiticity and acting with $\hat{A}$ to the left,
+\[\sand{a}{\hat{A}}{a} = \sand{a}{\hat{A}^{\dagger}}{a} = \left ( \bra{a} \hat{A}^{\dagger} \right ) \ket{a} = a^* \braket{a}{a} = a^*.\]
+Equating the two expressions gives $a=a^*$, and hence $a$ is real.
+  * **The eigenvalues of an anti-Hermitian operator are imaginary:**
+Again we let $\ket{a}$ be an eigenvector of $\hat{A}$ with eigenvalue $a$ and compute $\sand{a}{\hat{A}}{a}$ in two different ways. First, by acting with $\hat{A}$ to the right,
+\[\sand{a}{\hat{A}}{a} = \bra{a} \left ( \hat{A} \ket{a}\right ) = a \braket{a}{a} = a.\]
+Second, using anti-hermiticity and acting with $\hat{A}$ to the left,
+\[\sand{a}{\hat{A}}{a} = -\sand{a}{\hat{A}^{\dagger}}{a} = -\left ( \bra{a} \hat{A}^{\dagger} \right ) \ket{a} = -a^* \braket{a}{a} = -a^*.\]
+Equating the two expressions gives $a=-a^*$, and hence $a$ is pure imaginary.
+  * **The eigenvalues of a unitary operator have modulus $1$:**
+Again we let $\ket{a}$ be an eigenvector of $\hat{A}$ with eigenvalue $a$, but this time we compute $\sand{a}{\hat{A}^{\dagger}\hat{A}}{a}$ in two different ways.  Firstly, by unitarity, $\hat{A}^{\dagger}\hat{A} = \hat{I}$, so
+\[\sand{a}{\hat{A}^{\dagger}\hat{A}}{a} = \sand{a}{\hat{I}}{a} = \braket{a}{a} = 1.\]
+Second, we use the eigenvalue equation $\hat{A}\ket{a} = a\ket{a}$, and act with $\hat{A}$ to the right and $\hat{A}^{\dagger}$ to the left.
+\[\sand{a}{\hat{A}^{\dagger}\hat{A}}{a} = \left ( \bra{a} \hat{A}^{\dagger}\right ) \left ( \hat{A} \ket{a}\right ) = a^* a\braket{a}{a} = a^*a = |a|^2.\]
+Equating the two expressions gives $|a|^2 = 1$.  As a consequence, $a = e^{i\theta}$ for some real-valued angle $\theta$.
 ====== Eigenvalues of Functions ======
@@ Line 47: / Line 75: @@
 For a degenerate eigenvalue, we need more than one vector to form a basis for the eigenspace, so we are going to need another label in addition to $a$.  We can construct an orthonormal basis for the eigenspace and label the vectors $\ket{a,1}, \ket{a_2}, \cdots$ where we should read $\ket{a,j}$ as "the $j^{\text{th}}$ vector in an orthonormal basis for the eigenspace corresponding to eigenvalue $a$.
+====== Eigenspace Projectors ======
 Since the eigenspace corresponding to an eigenvalue $a$ is a subspace, we can form the projection operator $\hat{P}_a$ onto that subspace.  This is defined as follows.  For all eigenvectors $\ket{\psi}$ with eigenvalue $a$, $\hat{P}_a\ket{\psi} = \ket{\psi}$ and for any vector $\ket{\phi}$ that is orthogonal to all of these vectors $\hat{P}_a\ket{\phi} = 0$.
@@ Line 55: / Line 85: @@
 \[\hat{P}_a = \sum_j \proj{a,j}.\]
 These projection operators act like the identity operator on the eigenspace and the zero operator on the orthogonal complement.
+====== Normal Operators ======
+An operator $\hat{A}$ is //**normal**// if $[\hat{A},\hat{A}^{\dagger}] = 0$.  The relevance of normal operators is that they are completely characterized by their eigenvalues and the projectors onto the corresponding eigenspaces.  This makes them simple to deal with.  We will discuss this in the next section.
+For now, we note that all the important types of operators used in quantum mechanics are normal.  For example:
+  * Hermitian operators $\hat{A}^{\dagger} = \hat{A}$: \[[\hat{A},\hat{A}^{\dagger}] = [\hat{A},\hat{A}] = 0\]
+  * 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
+\[\hat{A}\ket{a} = a\ket{a}, \qquad\qquad \hat{A}\ket{b} = b\ket{b}.\]