11 Energy Revisited
In the last two lectures, we summarised the postulates of quantum mechanics and revisited what we learnt earlier in the course about measurements of position and momentum. In this lecture, we will do the same for measurements of energy and prove a couple of very important theorems about the spectrum of the Hamiltonian operator.
11.1 Reminder about Energy Measurements
Given the wave function \(\psi(x)\) of a particle moving in some potential \(V(x)\), the basic question we want to answer is: how can we determine the possible outcomes of a measurement of energy and their probabilities?
The starting point is to construct an orthonormal basis of eigenfunctions of the Hamiltonian operator \(\hat H\). For now, let’s assume the spectrum of the Hamiltonian has the following properties:
Discrete: There is a discrete set of eigenvalues \(\{E_j\}\).
Non-degenerate: There is a unique eigenfunction \(\phi_j(x)\) solving \[\hat H \cdot \phi_j(x) = E_j \phi_j(x)\] for each eigenvalue \(E_j\).
In this case, we can choose the eigenfunctions to be orthonormal, \[\langle \psi_i , \psi_j\rangle = \delta_{ij}\, .\] We then expand the wave function \[\psi(x) = \sum_j c_j \phi_j(x)\] and, provided the wave function is normalised, the probability to measure energy \(E_j\) is \(P_j = |c_j|^2.\) As a consistency check, it is straightforward to check that \(\sum_j P_j = 1\).
Finally, we can compute expectation value of energy measurements by summing the possible outcomes of an energy measurement weighted by their probabilities, \(\langle H \rangle = \sum_j P_j E_j\).
11.2 Examples with Bound States
The Hamiltonian could in principle have a continuous spectrum or both discrete and continuous eigenvalues. It may also be degenerate with multiple linearly independent eigenfunctions with the same eigenvalue. It is therefore important to qualify when the assumptions made in the previous section are valid.
For concreteness, let us consider a particle of mass \(m\) moving on a line with potential \(V(x)\). The Hamiltonian operator is then \[\begin{aligned} \hat H & = \frac{\hat p^2}{2m} + V(x) \\ & = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \end{aligned}\] and Hamiltonian eigenfunctions are solutions to the differential equation \[-\frac{\hbar^2}{2m} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) \psi(x) = E\psi(x)\] that satisfy appropriate boundary conditions as \(x \to \pm \infty\) and at any discontinuities in the potential \(V(x)\).
The Hamiltonian operator typically has a discrete spectrum for energies \(E\) where the corresponding classical solutions \(x(t)\) would be bounded in space. For this reason, the corresponding eigenfunctions are sometimes called “bound states". This is perhaps best illustrated with examples.
Infinite square well. Consider the potential \[V(x) = \begin{cases} 0 & \; 0<x<L \\ \infty & \; \text{other} \end{cases} \, .\] The classical motion is certainly bounded for any energy \(E>0\): the particle bounces back and forth from the walls of the box. The spectrum of the Hamiltonian is indeed discrete and non-degenerate with eigenvalues \[E_n = \displaystyle\frac{\hbar^2}{2m}\left( \frac{n\pi}{L} \right)^2 \qquad n\in \mathbb{Z}_{>0} \, .\]
Simple harmonic oscillator. Consider the quadratic potential \[V(x) = \frac{1}{2}m\omega^2 x^2\, .\]
The classical motion is bounded: the particle oscillates with angular frequency \(\omega\) for any finite energy \(E>0\). Later in the course, we will show that the spectrum is again discrete and non-degenerate with eigenvalues \[E_n = \hbar \omega\left(n+\frac{1}{2}\right) \qquad n \in \mathbb{Z}_{\geq 0} \, .\]
Hydrogen Atom: The effective potential of an electron in a hydrogen atom is \[V(x) = \frac{J^2}{2mx^2} - \frac{e^2}{x}\, .\]
The classical motion is bounded for \(E<0\) and unbounded for \(E\geq 0\). Correspondingly, there is a discrete set of bound states with energy \(E<0\), given by \[E_n = -\frac{me^4}{2\hbar^2n^2} \qquad n \in \mathbb{Z}_{>0}\, .\] There is also a continuous spectrum of “scattering states" with \(E>0\) that will also be studied later in the course.
These examples illustrate some important universal features of a particle moving on a real line:
The Hamiltonian eigenvalues obey \(E > V_0\) where \(V_0\) is the minimum of the potential \(V(x)\).
The spectrum of the Hamiltonian is non-degenerate.
We prove these properties in the following sections.
11.3 Minimum Energy
Suppose the potential is bounded below, meaning \(V(x) \geq V_0\) for all \(x \in \mathbb{R}\). The point \(x_0\) where \(V(x_0) = V_0\) is typically a local minimum.
In classical mechanics there is a continuum of bound states with energy \(E\geq V_0\) with the minimum energy configuration \(E = V_0\) corresponding to a stationary particle at \(x = x_{0}\). In contrast, in quantum mechanics we must have \(E > V_0\).
Theorem: If the wave function is
normalised, \(\langle H\rangle >
V_{0}\).
Proof: The energy expectation value is
\[\begin{aligned}
\langle H\rangle & = \langle \psi,\hat H\psi\rangle \\
& = \frac{1}{2m} \langle \psi, \hat p^2\psi \rangle + \langle
\psi,V(x)\psi\rangle \\
& = \frac{1}{2m} \langle \hat p \psi, \hat p\psi \rangle + \langle
\psi,V(x)\psi\rangle \, ,
\end{aligned}\] since the momentum operator \(p\) is Hermitian. The first term is
necessarily non-negative, \[\langle \hat
p\psi , \hat p\psi\rangle = \int^\infty_{-\infty} | \hat p \psi(x)|^2
\, {\rm d}x = \int^\infty_{-\infty} \left| -i \psi'(x) \right|^2 \,
{\rm d}x \geq 0\] and therefore \[\langle H\rangle \geq \langle
\psi,V(x)\psi\rangle = \int^\infty_{-\infty} V(x) |\psi(x)|^2 \, {\rm
d}x \geq V_0 \int^\infty_{-\infty} |\psi(x)|^2 \, {\rm d}x = V_0 \,
.\] We therefore conclude that \(\langle H\rangle \geq V_0\).
The next step is to rule out equality. A necessary condition for
equality is \(\langle \hat p\psi , \hat
p\psi\rangle=0\). From positive definiteness of the inner
product, this would require that \(\hat p\psi
=0\) and therefore \[-i\hbar\frac{\partial\psi(x)}{\partial x} = 0
\quad \Rightarrow \quad \psi(x) = c\,,\] where \(c \in \mathbb{C}\) is constant. However,
square-normalisability requires \(\psi(x) \to
0\) for \(x \to \pm \infty\) and
hence \(c=0\), so such a wave function
would vanish identically. We therefore conclude that \(\langle H\rangle > V_0\).
Corollary: If \(\psi(x)\) is a normalized eigenfunction of
\(H\) with eigenvalue \(E\), then \(E
>V_0\).
Proof: This is a simple consequence of the
above theorem: \(\langle H\rangle = \langle
\psi,H\psi\rangle = E\langle \psi,\psi\rangle = E > 0\).
This means that states with the minimum energy in classical mechanics, \(E = V_0\), corresponding to a stationary particle, cannot exist in quantum mechanics. This is compatible with our intuition from Heisenberg’s uncertainty principle. It is a very important result.
The smallest eigenvalue \(E>V_0\) is sometimes known as the ‘zero-point energy’ in quantum mechanics. This is a characteristic feature of quantum mechanics that leads ultimately to some of the greatest mysteries in theoretical physics.
11.4 Non-degeneracy
Theorem: The spectrum of the
Hamiltonian is non-degenerate.
Proof: Consider a pair of
square-normalizable eigenfunctions with the same eigenvalue, \[\begin{aligned}
\hat H \cdot \psi_1(x) & = E \psi_1(x)\\
\hat H \cdot \psi_2(x) & = E\psi_2(x) \, .
\end{aligned}\] We will prove that \(\psi_1(x) \propto \psi_2(x)\), which
implies there is a unique normalised eigenfunction for each eigenvalue
\(E\).
Expanding out, we have \[\begin{aligned} -\frac{\hbar}{2m}\partial_x^2 \psi_1(x) + V(x) \psi_1(x) & = E\psi_1(x) \\ -\frac{\hbar}{2m}\partial_x^2 \psi_2(x) + V(x) \psi_2(x) & = E\psi_2(x) \, . \end{aligned}\] Subtracting the top equation multiplied by \(\psi_2(x)\) from the bottom equation multiplied by \(\psi_1(x)\), the contributions propotional to \(V(x)\) and \(E\) vanish and we find \[\begin{aligned} 0 & = \psi_1 \partial_x^2\psi_2 - \psi_2 \partial_x^2\psi_1 \\[1ex] & = \partial_x( \psi_1 \partial_x\psi_2 - \psi_2 \, \partial_x\psi_1 ) \, . \end{aligned}\] Therefore, \[\label{e:c-eq} \psi_1 \partial_x\psi_2 - \psi_2 \partial_x\psi_1 = c \in \mathbb{C} \, .\]
11.5 The Evader
We have encountered a potential counterexample to the above theorem: a free particle moving on a circle of circumference \(L\) with potential \(V(x) = 0\). There is an orthonormal basis of Hamiltonian eigenfunctions \[\phi_{\pm n}(x) = \frac{1}{\sqrt{L}} e^{\pm 2\pi i n x /L} \qquad n \in \mathbb{Z}\] with eigenvalues \[E_n = \frac{\hbar^2}{2m}\left( \frac{2\pi n}{L} \right)^2 \, .\] The spectrum is degenerate because \(E_n = E_{-n}\). The above proof fails at the stage where we required the wave function to vanish as \(x \to \pm \infty\). Here, instead we have imposed periodic boundary conditions \(\psi(x+L) = \psi(x)\), which is not strong enough to determine the constant \(c = 0\) in \(\eqref{e:c-eq}\).
11.6 Problems
- TUTORIAL 3Delta function potential So far we have seen only situations in which the energy spectrum is either continuous (when e.g. \(V(x)=0\) everywhere) or discrete (e.g. the infinite square well). That is not the generic situation, however. Most potentials lead to spectra which have both a discrete and a continuous part.
Consider particle on a real line with potential given by \[V(x) = -\alpha\,\delta(x)\,,\] where \(\alpha>0\).
Solve for the most general \(\psi(x)\) in the region \(x<0\) which is an eigenfunction of the Hamiltonian with negative eigenvalue \(E\). This should have one unknown parameter (in addition to the constant \(E\)).
Do the same for the region \(x>0\). This again should have one unknown parameter (again, in addition to \(E\)).
Impose that the wave function is continuous. Integrate the Hamiltonian eigenfunction equation over an interval \(-\epsilon < x < \epsilon\) with \(\epsilon\rightarrow 0\). Use this to relate \(E\) to \(\alpha\).
Unit-normalise the wave function to completely fix \(\psi(x)\).
There is thus a single (bound) state with \(E<0\).
We will look at the continuum part of the spectrum (eigenfunctions with \(E>0\)) later (problem in chapter 17).
Solution ▶
The eigenvalue equation is \[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi(x) + V(x)\psi(x) = E(x)\psi(x)\,.\] For \(x<0\) the potential \(V(x)=-\alpha\delta(x)\) vanishes, and the general solution reads \[x<0:\qquad \psi(x) = A e^{-\kappa x} + B e^{\kappa x}\,,\quad \kappa = \sqrt{-\frac{2mE}{\hbar^2}}\,.\] For a normalisable solution \(A=0\).
For \(x>0\) the story is similar, now we get \[\psi(x) = \tilde{B} e^{-\kappa x}\,.\]
Continuity implies \(B=\tilde{B}\). Integrating the eigenfunction equation over an infinitesimal region around \(x=0\) yields \[\int_{-\epsilon}^{\epsilon} \big(-\frac{\hbar^2}{2m}\psi''(x) -\alpha \delta(x)\psi(x) - E \psi(x)\big) {\rm d}x\,.\] The last term vanishes in the limit \(\epsilon\rightarrow 0\), while the second term gives \(-\alpha\psi(0) = -\alpha B\). The first term is a total derivative, equal to \[-\frac{\hbar^2}{2m} \psi'(x)\Big|_{x=-\epsilon}^{\epsilon} = -\frac{\hbar^2}{2m} 2 B \kappa\,.\] Together we thus find \[\kappa = \frac{m\alpha}{\hbar^2}\,,\quad\text{or}\quad E = -\frac{m\alpha^2}{2\hbar^2}\,.\]
Integrating the norm-squared of the wave function over all of space gives \[\int|\psi^2| = 2 |B|^2 \int_0^\infty e^{-2\kappa x} {\rm d}x = \frac{|B|^2}{\kappa} = 1\,,\quad\rightarrow\quad |B| = \sqrt{\kappa}\,.\]
