7 Hermitian Operators

7.1 More Linear Algebra

We begin with a quick review of some more linear algebra. As before, we first consider a finite-dimensional complex vector space \(V\) with Hermitian inner product \(\langle\, \cdot \, , \, \cdot \, \rangle\) and an orthonormal basis \(\{e_j\}\).

A linear operator is a map \(A : V \to V\) such that \[\begin{aligned} A \cdot (a_1 v_1 + a_2 v_2) = a_1 (A \cdot v_1)+ a_2 (A \cdot v_2) \end{aligned}\] for any vectors \(v_1,v_2\in V\) and complex numbers \(a_1,a_2\in\mathbb{C}\). Any linear combination \(a_1A_1+a_2A_2\) and composition \(A_1 \cdot A_2\) of two linear operators \(A_1\), \(A_2\) is again a linear operator. The matrix elements of a linear operator in an orthonormal basis \(\{e_j\}\) are defined by \(A_{ij} = \langle e_i , A\cdot e_j\rangle\).

The adjoint \(A^\dagger\) of a linear operator \(A\) is defined by \[\langle v_1,A\cdot v_2\rangle = \langle A^\dagger \cdot v_1, v_2\rangle\] for any pair of vectors \(v_1,v_2 \in V\). Let us compute the matrix elements of \(A^\dagger\) with respect to an orthonormal basis, \[\begin{aligned} A^\dagger_{ij} & = \langle e_i ,A^\dagger e_j\rangle \\ \nonumber & = \langle A \cdot e_i ,e_j\rangle \\ & = \overline{\langle e_j , A \cdot e_i\rangle} \\ & = \overline{A_{ji}} \, . \end{aligned}\] On other words, the matrix elements of operators \(A\) and \(A^\dagger\) are related by taking the conjugate transpose of the matrix.

The adjoint operation has the following basic properties

• \((a_1A_1 + a_2A_2)^\dagger = \bar a_1 A_1^\dagger + \bar a_2 A_2^\dagger\)

• \((A_1A_2)^\dagger = A_2^\dagger A_1^\dagger\)   .

It follows from these properties that \((A^n)^\dagger = (A^\dagger)^n\) and therefore \(f(A)^\dagger = f(A^\dagger)\) for any polynomial function \(f(a)\).

A Hermitian operator is a linear operator that is equal to its adjoint, \(A = A^\dagger\). An equivalent way to say this is that a Hermitian operator obeys \[\langle v_1,A\cdot v_2\rangle = \langle A\cdot v_1, v_2\rangle \label{eq:herm-def}\] for any \(v_1,v_2\in V\). The matrix elements \(A_{ij}\) of a Hermitian operator form a Hermitian matrix, \(\overline{A_{ji}} = A_{ij}\), hence the name! We will see that Hermitian operators play an incredible important role in quantum mechanics.

As an aside: mathematicians tend to call Hermitian operators ‘symmetric operators’, and reserve the word Hermitian for matrices. There are subtle issues related to the fact that the domain of \(A\) may not be equal to the domain of \(A^\dagger\), in which case the operator is symmetric but not ‘self-adjoint’. For a lot of physics applications the theory around this is more complicated than the solution, but we may touch on some of this in a problem later. However, if this is your thing and you cannot wait, read Michael Reed and Barry Simon. Methods of Modern Mathematical Physics. Volume 1-4. Elsevier, 1975. (all four volumes of it).

7.2 Linear Differential Operators

We now return to quantum mechanics of a particle moving in one dimension \(x\in \mathbb{R}\). In the previous lecture, we learnt that continuous square-integrable wave functions \(\psi(x)\) form a complex vector space with Hermitian inner product \[\langle \psi_1,\psi_2\rangle = \int^\infty_{-\infty} \overline{\psi_1(x)} \psi_2(x) {\rm d}x \, .\] Furthermore, it often convenient to introduce a discrete orthonormal basis of wave functions \(\{ \phi_n(x)\}\). Remember that the index \(n\) may run over a set with an infinite number of elements, for example \(n \in \mathbb{Z}_{>0}\).

A linear operator now corresponds to a linear differential operator \(A\) built from derivatives with respect to \(x\). We have already encountered two important examples of linear differential operators:

• The position operator \(\hat x = x\)

• The momentum operator \(\hat p = - i \hbar \displaystyle\frac{\partial}{\partial x}\)

We define the matrix elements of \(A\) in a discrete orthonormal basis by \[A_{mn} : = \langle \phi_m , A \cdot \phi_n \rangle = \int^\infty_{-\infty} \overline{\phi_m(x)} \left( A \cdot \phi_n(x) \right) \, .\] If there is an infinite number of wave functions \(\phi_n(x)\) in the orthonormal basis, this will be an infinite-dimensional matrix.

The adjoint of a linear differential operator is defined by \[\langle \psi_1,A^\dagger\cdot \psi_2\rangle = \langle A \cdot \psi_1, \psi_2\rangle\] and has the same properties as above. It has matrix elements given by the conjugate transpose, \(A^\dagger_{mn} = \bar A_{nm}\).

A Hermitian operator again satisfies \(A^\dagger = A\) or equivalently \[\langle \psi_1,A\cdot \psi_2\rangle = \langle A \cdot \psi_1, \psi_2\rangle \, .\] for any continuous square-normalisable wave functions \(\psi_1(x)\), \(\psi_2(x)\). The matrix elements of a Hermitian operator form a Hermitian matrix \(A_{mn} = \bar A_{nm}\).

7.3 Position, Momentum and Energy

Let us prove the position and momentum operators are Hermitian. First, using the fact that \(x\) is real we have (on the real line) \[\begin{aligned} \langle \hat x \cdot \psi_1 , \psi_2 \rangle & = \int^\infty_{-\infty} \overline{x \psi_1(x)} \, \psi_2(x) \, {\rm d}x \\ \nonumber & = \int^\infty_{-\infty} \overline{\psi_1(x)} \, x \psi_2(x) \, {\rm d}x \\ & = \langle \psi_1, \hat x \cdot \psi_2\rangle \, . \end{aligned}\] Second, integrating by parts we find \[\begin{aligned} \langle \hat p \cdot \psi_1 , \psi_2 \rangle & = \int^\infty_{-\infty} \overline{-i\hbar\frac{\partial\psi_1(x)}{\partial x}} \psi_2(x) \, {\rm d}x \\ \nonumber & = \int^\infty_{-\infty} i \hbar\frac{\partial\overline{ \psi_1(x)}}{\partial x} \psi_2(x) \, {\rm d}x \\ & = \int^\infty_{-\infty} \overline{ \psi_1(x)}\left(- i \hbar\frac{\partial \psi_2(x)}{\partial x} \right) \, {\rm d}x + i \hbar \left[ \overline{\psi_1(x)} \psi_2(x) \right]^\infty_{-\infty} \\ & = \langle \psi_1, \hat p \cdot \psi_2\rangle \, . \end{aligned}\]

More general Hermitian operators can be constructed from polynomials in the position and momentum operators. An important example is the Hamiltonian operator \[\begin{aligned} \hat H & = \frac{\hat p^2}{2m} + V(x) \\ & = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \, , \end{aligned}\] corresponding to measurements of energy. This suggests that perhaps all measurable quantities are represented by Hermitian operators in quantum mechanics. We will explain why this is the case in upcoming lectures.

7.4 Example: Particle in a Box

Let us consider an infinite potential well in the region \(0\leq x \leq L\). In the last lecture, we introduced a basis of square-integrable wave functions \[\phi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right) \qquad n \in \mathbb{Z}_{>0} \, .\] The first few of these functions are displayed in figure \(\ref{f:box_ef}\); note how they satisfy the boundary conditions that the wave functions vanish at the edge of the box.

We will now compute the matrix elements of position, momentum and Hamiltonian operators in this basis.

7.5 Problems

1. Properties of the adjoint:

   The adjoint \(A^\dagger\) of a linear operator \(A\) is defined by \[\langle \psi_1 , A \psi_2 \rangle = \langle A^\dagger \psi_1 , \psi_2\rangle\] for any continuous square-integrable wavefunctions \(\psi_1(x)\) and \(\psi_2(x)\). Show that

   1. \((A_1+A_2)^\dagger = A_1^\dagger + A_2^\dagger\)

   2. \((a A)^\dagger = \bar a A^\dagger\) for any constant \(a\in \mathbb{C}\)

   3. \((A_1A_2)^\dagger = A_2^\dagger A_1^\dagger\)

   4. \((A^n)^\dagger = (A^\dagger)^n\) for any \(n \in \mathbb{N}\)

   5. If \(f(x)\) is a real analytic function of \(x\in \mathbb{R}\), \(f(A)^\dagger = f(A^\dagger)\).

   Solution ▶ These properties can be shown algebraically using the properties of the inner product in question 2.1.

   1. We have \[\begin{aligned} \langle \psi,(A_1+A_2)\psi' \rangle & = \langle \psi,A_1\psi' + A_2 \psi'\rangle \\ \nonumber & = \langle \psi,A_1\psi' \rangle + \langle \psi ,A_2\psi'\rangle \\ & = \langle A_1^\dagger \psi , \psi' \rangle + \langle A_2^\dagger \psi , \psi' \rangle \\ & = \langle A_1\psi+A_2\psi , \psi' \rangle \\ & = \langle (A_1^\dagger+A_2^\dagger)\psi , \psi' \rangle \end{aligned}\] and therefore \((A_1 +A_2)^\dagger = A_1^\dagger + A_2^\dagger\).

   2. We have \[\begin{aligned} \langle \psi,(aA)\psi' \rangle & = \langle \psi , a (A\psi')\rangle \\ \nonumber & = a \langle \psi ,A\psi' \rangle \\ & = a \langle A^\dagger \psi , \psi' \rangle \\ & = \langle \bar a(A^\dagger \psi , \psi' \rangle \\ & = \langle (\bar a A^\dagger) \psi , \psi' \rangle \end{aligned}\] and therefore \((aA)^\dagger = \bar a A^\dagger\).

   3. We have \[\begin{aligned} \langle \psi , (A_1A_2) \psi' \rangle & = \langle \psi, A_1(A_2\psi') \rangle \\ \nonumber & = \langle A_1^\dagger \psi , A_2 \psi'\rangle \\ & = \langle A_2^\dagger (A_1^\dagger) \psi , \psi'\rangle \\ & = \langle (A_2^\dagger A_1^\dagger)\psi , \psi'\rangle \end{aligned}\] and therefore \((A_1A_2)^\dagger = A_2^\dagger A_1^\dagger\).

   4. From part \((c)\), we have immediately \((A^2)^\dagger = (A^\dagger)^2\). We now proceed by induction on \(n\). Let us assume \((A^n)^\dagger = (A^\dagger)^n\). Then \[(A^{n+1})^\dagger = (A A^{n})^\dagger = (A^n)^\dagger A^\dagger = (A^\dagger)^n A^\dagger = (A^\dagger)^{n+1}\] where have again used part (c) with \(A_1 = A\) and \(A_2 = A^n\).

   5. A real analytic function \(f(x)\) has a convergent power series expansion \[f(x) = \sum_{n=0}^\infty f_n x^n\] with real coefficients \(f_n\in \mathbb{R}\). The result now follows from parts (a), (b), and (d).

2. TUTORIAL 2
   Properties of Hermitian operators Hermitian operators satisfy various properties which will be important in later chapters:

   1. Show that any linear operator \(A\) can be expressed as a sum \[A = U +V\] where \(U^\dagger = U\) is Hermitian and \(V^\dagger = -V\) is anti-Hermitian.

   2. Show that if \(A\) and \(B\) are Hermitian then \(A+B\) is Hermitian.

   3. Show that if \(A\) is Hermitian then \(aA\) is Hermitian if and only if \(a\in \mathbb{R}\). What happens if \(a \in i \mathbb{R}\)?

   4. Show that if \(A\) and \(B\) are Hermitian then the commutator \[[A,B] := AB-BA\] is anti-Hermitian.

   5. Suppose \(A\) and \(B\) are Hermitian. Under what condition is \(AB\) Hermitian?

   6. Show that if \(A\) is Hermitian then \(A^n\) is Hermitian for \(n>0\).

   Solution ▶ For this question, you will need to use properties of the adjoint from question 2.

   1. We can express any linear operator as \(A = U + V\) where \[U = \frac{1}{2}(A+A^{\dagger}) \qquad V = \frac{1}{2}(A-A^{\dagger}) \, .\] It is straightforward to see that \(U^\dagger = U\) and \(V^\dagger =- V\).

   2. Suppose \(A\) and \(B\) are Hermitian, then \((A+B)^\dagger = A^\dagger + B^\dagger = A+B\) and therefore \(A+B\) is Hermitian.

   3. Suppose \(A\) is Hermitian and \(a \in \mathbb{C}\). Then \((aA)^\dagger = \bar a A^\dagger = \bar a A\). Therefore \(aA\) is Hermitian if \(a\in \mathbb{R}\) and anti-Hermitian if \(a\in i \mathbb{R}\).

   4. The commutator is \([A,B] = AB-BA\). Computing the adjoint \[[A,B]^\dagger = (AB)^\dagger - (BA)^\dagger = B^\dagger A^\dagger - A^\dagger B^\dagger = BA - AB = - [A,B] \, .\]

   5. Suppose \(A\) and \(B\) are Hermitian then \((AB)^\dagger = B^\dagger A^\dagger = BA\). So \(AB\) is Hermitian if \(AB=BA\), or equivalently if \([A,B] = 0\).

   6. If \(A\) is Hermitian, we have \((A^n)^\dagger = (A^\dagger)^n = A^n\). This implies \(f(A)\) is also Hermitian where \(f(a)\) is a real analytic function.

3. TUTORIAL 2
   Eigenfunctions have zero uncertainty Consider a eigenfunction of a Hermitian operator \(A\) with eigenvalue \(a\). You may assume the wavefunction is normalised.

   1. Show that \(\langle A^n\rangle = a^n\) for any \(n>0\).

   2. Hence show that \(\Delta A = 0\).

   3. What is the physical interpretation of these results?

   Solution ▶

   1. Since \(A \cdot \psi = a \psi\) we also have \(A^n \cdot \psi = a^n \psi\) for any \(n>0\) and therefore \[\langle A^n\rangle = \langle \psi,A^n \cdot\psi\rangle = a^n \langle \psi,\psi\rangle = a^n\] since \(\langle \psi,\psi\rangle = 1\) for a normalised wavefunction.

   2. Using part (a), \((\Delta A)^2 = \langle A^2 \rangle - \langle A\rangle^2 = a^2 - a^2 = 0\).

   3. An eigenfunction of a Hermitian operator \(A\) has zero uncertainty for measurements of \(A\). In fact a measurement of \(A\) will yield the eigenvalue \(a\) with probability \(1\).