QM II: Energy Revisited

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
\begin{equation} \hat H \cdot \phi_j(x) = E_j \phi_j(x) \end{equation}
for each eigenvalue $E_j$.

In this case, we can choose the eigenfunctions to be orthonormal,

\begin{equation} \langle \psi_i , \psi_j\rangle = \delta_{ij}\, . \end{equation}

We then expand the wave function

\begin{equation} \psi(x) = \sum_j c_j \phi_j(x) \end{equation}

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

Examples of bound states suggest two important properties of the energy spectrum which hold in a variety of situations (but not all).

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{align} \hat H & = \frac{\hat p^2}{2m} + V(x) \\ & = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \end{align}

and Hamiltonian eigenfunctions are solutions to the differential equation

\begin{equation} -\frac{\hbar^2}{2m} \frac{\partial^2\psi(x)}{\partial x^2} + V(x) \psi(x) = E\psi(x) \end{equation}

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
\begin{equation} V(x) = \begin{cases} 0 & \; 0\lt{}x\lt{}L \\ \infty & \; \text{other} \end{cases} \, . \end{equation}
The classical motion is certainly bounded for any energy $E\gt{}0$: 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
\begin{equation} E_n = \displaystyle\frac{\hbar^2}{2m}\left( \frac{n\pi}{L} \right)^2 \qquad n\in \mathbb{Z}_{\gt{}0} \, . \end{equation}
Simple harmonic oscillator. Consider the quadratic potential
\begin{equation} V(x) = \frac{1}{2}m\omega^2 x^2\, . \end{equation}

Figure 11.1: Simple harmonic oscillator potential.

The classical motion is bounded: the particle oscillates with angular frequency $\omega$ for any finite energy $E\gt{}0$. Later in the course, we will show that the spectrum is again discrete and non-degenerate with eigenvalues
\begin{equation} E_n = \hbar \omega\left(n+\frac{1}{2}\right) \qquad n \in \mathbb{Z}_{\geq 0} \, . \end{equation}
Hydrogen Atom: The effective potential of an electron in a hydrogen atom is
\begin{equation} V(x) = \frac{J^2}{2mx^2} - \frac{e^2}{x}\, . \end{equation}

Figure 11.2: Effective potential of an electron in a hydrogen atom.

The classical motion is bounded for $E\lt{}0$ and unbounded for $E\geq 0$. Correspondingly, there is a discrete set of bound states with energy $E\lt{}0$, given by
\begin{equation} E_n = -\frac{me^4}{2\hbar^2n^2} \qquad n \in \mathbb{Z}_{\gt{}0}\, . \end{equation}
There is also a continuous spectrum of ``scattering states" with $E\gt{}0$ 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 $E \gt{} V_0$ 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

In quantum mechanics, the lowest lying energy eigenvalue is always strictly larger than the minimal value of the potential. This is called the zero-point 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.

Figure 11.3: A potential with a local minimum at $x=x_0$.

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 $E \gt{} V_0$.

Theorem: If the wave function is normalised, $\langle H\rangle \gt{} V_{0}$.
Proof: The energy expectation value is

\begin{align} \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{align}

since the momentum operator $p$ is Hermitian. The first term is necessarily non-negative,

\begin{equation} \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 \end{equation}

and therefore

\begin{equation} \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 \, . \end{equation}

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

\begin{equation} -i\hbar\frac{\partial\psi(x)}{\partial x} = 0 \quad \Rightarrow \quad \psi(x) = c\,, \end{equation}

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 $\langle H\rangle \gt{} V_0$.

Corollary: If $\psi(x)$ is a normalized eigenfunction of $H$ with eigenvalue $E$, then $E \gt{}V_0$.
Proof: This is a simple consequence of the above theorem: $\langle H\rangle = \langle \psi,H\psi\rangle = E\langle \psi,\psi\rangle = E \gt{} 0$.

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 $E\gt{}V_0$ 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

For a particle on the real line, the energy spectrum is non-degenerate: there exists only one eigenfunction for each energy eigenvalue. The proof generalises to some other situations, but not all.

Theorem: The spectrum of the Hamiltonian is non-degenerate.
Proof: Consider a pair of square-normalizable eigenfunctions with the same eigenvalue,

\begin{align} \hat H \cdot \psi_1(x) & = E \psi_1(x)\\ \hat H \cdot \psi_2(x) & = E\psi_2(x) \, . \end{align}

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{align} -\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{align}

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{align} 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{align}

Therefore,

\begin{equation} \label{e:c-eq} \psi_1 \partial_x\psi_2 - \psi_2 \partial_x\psi_1 = c \in \mathbb{C} \, . \end{equation}

Boundary conditions are crucial in proving non-degeneracy of the energy spectrum.

But since the wave functions are normalizable, $\psi_1(x)$, $\psi_2(x)$ necessarily vanish at $x \to \pm\infty$. Evaluating the equation at infinity, we therefore determine that $c = 0$. We can then solve the equation,

\begin{align} \, & \psi_1 \partial_x\psi_2 - \psi_2 \partial_x\psi_1 = 0 \\[1ex] \Rightarrow \; & \frac{\partial_x\psi_1}{\psi_1} - \frac{\partial_x\psi_2}{\psi_2} = 0 \\ \Rightarrow \; & \partial_x (\log \psi_1 - \log \psi_2 ) = 0 \\ \Rightarrow \; & \log \frac{\psi_1}{\psi_2} = A \\ \Rightarrow \; & \psi_1 = e^A \psi_2 \,, \end{align}

for some constant $A \in \mathbb{C}$. This shows that $\psi_1(x)$ and $\psi_2(x)$ are the same eigenfunction.

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

\begin{equation} \phi_{\pm n}(x) = \frac{1}{\sqrt{L}} e^{\pm 2\pi i n x /L} \qquad n \in \mathbb{Z} \end{equation}

with eigenvalues

\begin{equation} E_n = \frac{\hbar^2}{2m}\left( \frac{2\pi n}{L} \right)^2 \, . \end{equation}

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

3
Delta 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
\begin{equation} V(x) = -\alpha\,\delta(x)\,, \end{equation}
where $\alpha\gt{}0$.
1. Solve for the most general $\psi(x)$ in the region $x\lt{}0$ which is an eigenfunction of the Hamiltonian with negative eigenvalue $E$. This should have one unknown parameter (in addition to the constant $E$).
2. Do the same for the region $x\gt{}0$. This again should have one unknown parameter (again, in addition to $E$).
3. Impose that the wave function is continuous. Integrate the Hamiltonian eigenfunction equation over an interval $-\epsilon \lt{} x \lt{} \epsilon$ with $\epsilon\rightarrow 0$. Use this to relate $E$ to $\alpha$.
4. Unit-normalise the wave function to completely fix $\psi(x)$.
There is thus a single (bound) state with $E\lt{}0$. We will look at the continuum part of the spectrum (eigenfunctions with $E\gt{}0$) later (problem in chapter 17).
Solution:
▶
1. The eigenvalue equation is
  \begin{equation} -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi(x) + V(x)\psi(x) = E(x)\psi(x)\,. \end{equation}
  For $x\lt{}0$ the potential $V(x)=-\alpha\delta(x)$ vanishes, and the general solution reads
  \begin{equation} x\lt{}0:\qquad \psi(x) = A e^{-\kappa x} + B e^{\kappa x}\,,\quad \kappa = \sqrt{-\frac{2mE}{\hbar^2}}\,. \end{equation}
  For a normalisable solution $A=0$.
2. For $x\gt{}0$ the story is similar, now we get
  \begin{equation} \psi(x) = \tilde{B} e^{-\kappa x}\,. \end{equation}
3. Continuity implies $B=\tilde{B}$. Integrating the eigenfunction equation over an infinitesimal region around $x=0$ yields
  \begin{equation} \int_{-\epsilon}^{\epsilon} \big(-\frac{\hbar^2}{2m}\psi''(x) -\alpha \delta(x)\psi(x) - E \psi(x)\big) {\rm d}x\,. \end{equation}
  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
  \begin{equation} -\frac{\hbar^2}{2m} \psi'(x)\Big|_{x=-\epsilon}^{\epsilon} = -\frac{\hbar^2}{2m} 2 B \kappa\,. \end{equation}
  Together we thus find
  \begin{equation} \kappa = \frac{m\alpha}{\hbar^2}\,,\quad\text{or}\quad E = -\frac{m\alpha^2}{2\hbar^2}\,. \end{equation}
4. Integrating the norm-squared of the wave function over all of space gives
  \begin{equation} \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}\,. \end{equation}

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1.	Why Quantum Mechanics?
2.	The Double Slit Experiment
3.	Wave function and Probabilities
4.	Momentum and Planck's constant
5.	Schrödinger's Equation
6.	The Hilbert Space
7.	Hermitian Operators
8.	The Spectrum of a Hermitian Operator
9.	Postulates of Quantum Mechanics
10.	Commutators and Uncertainty Principle
11.	Energy Revisited
11.1.	Reminder about Energy Measurements
11.2.	Examples with Bound States
11.3.	Minimum Energy
11.4.	Non-degeneracy
11.5.	The Evader
11.6.	Problems
12.	Stationary states
13.	Case Study: The Free Particle
14.	Two-particle systems
15.	Simple Harmonic Oscillator
16.	The Continuity Equation
17.	Scattering Problems
18.	Tunnelling
19.	Momentum-space Wave function
20.	Ehrenfest's Theorem
21.	Bibliography