Chapter 4: Energy and the Hamiltonian

In this Chapter we will consider now the final important physical property of a particle we will study in this unit — the energy. Not only is this one of the most important physical properties in nature, due to its conservation law, it is central to the time evolution in quantum mechanics, as we will begin investigating in the next Chapter.

We will reintroduce the Hamiltonian operator (the total-energy operator), in the context of mechanics, its associated eigenvalue equation, and discuss the important notion of bound and non-bound states, and their distinct description within quantum mechanics.

4.1The Hamiltonian¶

What different types of energy can a particle have? In the context of the mechanics of a single particle, there are two primary contributions to the energy: the kinetic energy and the potential energy. The total energy of a particle is just the sum of these two. We call the total energy the Hamiltonian (after Hamilton), which will be

H = K + V,

(4.1)

where $K$ is the kinetic energy, and $V$ is the potential energy.

While up until now you may have found it more customary to write the kinetic energy as $K = \frac{1}{2}Mv^2$ , where $M$ is the mass of the particle and $v$ is the velocity, since $p = Mv$ , we can alternatively write the kinetic energy as

K = \frac{p^2}{2M},

(4.2)

which turns out to be the ‘correct’ way to express the kinetic energy in QM.

The potential energy is associated to the forces acting on the particle, through $F = -\frac{\partial V}{\partial x}$ . This highlights that the potential energy of a particle is usually a function of its position $V(x)$ . For example, this might be the gravitational potential energy of a particle (associated to the force of gravity), or the energy stored in a spring for a particle undergoing harmonic motion (associated to the restorative force $F(x) = - kx$ , with $k$ the spring constant).

In quantum mechanics, we associate an operator with the total energy, the Hamiltonian operator which is given by

\begin{align*} \hat{H} &= \hat{K} + \hat{V},\\ &= \frac{\hat{P}^2}{2M} + \hat{V}. \end{align*}

(4.3)

As previously, it will be very useful to introduce the associated operator $\op{H}$ that acts on wavefunctions $\psi(x)$ . It is given by

\op{H} = \frac{\op{P}^2}{2M} + \op{V}.

(4.4)

To understand the kinetic energy operator, we see can see that

\begin{align*} \frac{\op{P}^2}{2M}\psi(x) &= \frac{1}{2M}\op{P}\left(\op{P}\psi(x)\right),\\ &= \frac{1}{2M}\left(-i\hbar \frac{\partial}{\partial x}\right)\left(-i\hbar \frac{\partial}{\partial x}\psi(x)\right),\\ &= -\frac{\hbar^2}{2M}\frac{\partial^2}{\partial x^2}\psi(x), \end{align*}

(4.5)

where in the second line we used the explicit form of the momentum operator $\op{P}$ from (2.27). This shows that

\op{K} = \frac{\op{P}^2}{2M} = -\frac{\hbar^2}{2M}\frac{\partial^2}{\partial x^2},

(4.6)

is the kinetic energy operator (acting on wavefunctions). For the potential energy, the situation is simpler. We have

\op{V}\psi(x) = V(x)\psi(x),

(4.7)

that is, the potential energy operator multiplies the wavefunction at each position $x$ by the value of the potential energy at that position, namely $V(x)$ , just as we saw in (1.19) that the position operator multiplies the wavefunction by the position everywhere. Thus, we can write

\op{V} = V(x).

(4.8)

Putting everything together, we can now write down the Hamiltonian operator acting on wavefunctions:

\vph\op{H} = -\frac{\hbar^2}{2M}\frac{\partial^2}{\partial x^2} + V(x).

(4.9)

As you will show in an exercise, the kinetic energy and potential energy do not commute, $[\hat{K},\hat{V}] \neq 0$ in general. This leads to a profound aspect of quantum mechanics:

The kinetic energy and potential energy of a quantum particle are incompatible properties.

That is, a particle cannot have both a well defined kinetic energy and a well defined potential energy at the same time! One intuitive way to see why this had to be the case is because the kinetic energy depends only upon the momentum of the particle, while the potential energy depends only upon the position, and we already know that these two properties are incompatible. Much of the intricacies of how quantum particles behave derives, ultimately, from the fact that these two contributions to energy are incompatible with each other.

4.2Energy eigenstates¶

A quantum state where the particle has definite energy corresponds to an eigenstate of the Hamiltonian $\hat{H}$ . That is, a state $\ket{E}$ of energy $E$ that satisfies the eigenvalue equation

\hat{H}\ket{E} = E\ket{E}.

(4.10)

As with momentum eigenstates, we can write these eigenstates in terms of their associated wavefunctions,

\vph \ket{E} = \infint u_E(x) \ket{x} dx,

(4.11)

where $u_E(x)$ are wavefunctions of the particle corresponding to states of definite energy. We call these the energy eigenfunctions. These wavefunctions will satisfy the associated eigenvalue equation

\op{H}u_E(x) = E u_E(x).

(4.12)

For reasons that will become clear in the following chapter, this equation is often referred to as the Time-independent Schrödinger Equation or TISE for short. This will in fact be our central equation for studying the energy of a particle in quantum mechanics, essentially because of the fact that the potential energy is a function of position.

4.3Bound and non-bound states¶

We end this chapter with an important discussion about qualitatively distinct situations that can arise, and how this affects the allowed energies a particle can have, and the properties of their eigenstates.

4.3.1Bound states¶

In situations where a particle does not have enough energy to leave a finite region of space we say that it is in a bound state. This is a very generic situation, which occurs across all of physics, ranging from planets in bounded orbits around stars, all the way down to electrons bound to a nucleus. What unites all such situations is that the object will be stuck in the corresponding potential well (whether that be the gravitational potential well of the star, or the electrostatic potential well of the nucleus). This is most readily depicted graphically:

Picture of a two potential wells — Figure 4.1:Two examples of potential wells. On the left, the well is symmetric. On the right, the well is assymetric. If the energy of the particle is $E<E_B$ , then the particle will be ‘trapped’ inside the well, which we call a bound state. In this case, in QM the particle will have discrete energy levels. If $E > E_B$ the particle is not bound to the well, but will escape to infinity (on the left in either direction; on the right, only towards the right). In QM, in this regime there is no discretisation of energies; energy is continuous but eigenstates becomes unphysical.

For any potential well $V(x)$ there will be a value $E_B$ , such that if the energy of the particle is $E < E_B$ , then it will not be able to escape from the well, and is hence trapped inside it / bound to it.

In QM, when a particle is in a bound state, we find that it can only have certain special and discrete energies $E_k$ . These are the famous energy levels (e.g. of atoms). More specifically and mathematically, we will see that in this case there are only physical solutions to the TISE (4.12) for specific values $E_k$ . The corresponding eigenstates $\ket{E_k}$ are normalised and orthogonal to each other,

\inner{E_k}{E_\ell} = \begin{cases} 1 &\text{ if } k = \ell \\ 0 &\text{ if } k \neq \ell \end{cases} = \delta_{k,\ell}.

(4.13)

4.3.2Non-bound states¶

The alernative situation is where a particle is not confined to one region of space, either because it has too much energy to be bound in the well ( $E > E_B$ ), or because there are no forces, i.e. $V(x) = 0$ and there is no well to be trapped in. Classically, this corresponds to non-closed orbits, such as asteroids passing a star, or an electron that isn’t bound to a nucleus. In these cases, in principle, the object will eventually escape to infinity (travel as far as possible in some direction).

In QM in this case we find that there is no discretisation (no energy levels), but rather the particle can have any energy $E > E_B$ (just like it can have any momentum or position in general), and energy remains continuous as it is classically.

However, in this case, just like for position and momentum states, the energy eigenstates are unnormalisable and therefore unphysical. We take them to satisfy the analogous orthonormality condition as for position and momentum, namely

\inner{E'}{E} = \delta(E-E'),

(4.14)

where $\delta(E-E')$ is the Dirac delta function. Just as in those cases, even though the states $\ket{E}$ themselves are unphysical, they are a very useful basis of states, since superpositions can be normalised states.

We end by noting that in general, a quantum particle could be in a superposition of being bound and being non-bound. That is, it could have a superposition of energies, including energy levels below $E_B$ and energies above $E_B$ . This causes no problem. It will nevertheless be easier to study the two regimes separately at first, before combining them after, if required by the physical situation under consideration.

4.4Exercises¶

Exercise 4.1

In this exercise we will show that in quantum mechanics, the kinetic energy and the potential energy are incompatible observables. In this question, we will consider an arbitrary wavefunction $\psi(x)$ .

Calculate $\op{V}\op{K}\psi(x).$
Calculate $\op{K}\op{V}\psi(x).$
Using parts 1. and 2. show that

\vph[\op{K},\op{V}]\psi = -\frac{\hbar^2}{2M}\left(\frac{\partial^2 V}{\partial x^2}\psi(x) + 2 \frac{\partial V}{\partial X} \frac{\partial}{\partial x}\psi(x) \right)

(4.15)

We will denote by $\hat{V}'$ and $\hat{V}''$ the operators satisfying

\begin{align*} \hat{V}'\ket{\psi} &= \infint \frac{\partial V}{\partial X} \psi(x) \ket{x} dx,\\ \hat{V}''\ket{\psi} &= \infint \frac{\partial^2 V}{\partial X^2} \psi(x) \ket{x} dx. \end{align*}

(4.16)

Hence, show that

\vph[\hat{K},\hat{V}] = -\frac{\hbar^2}{2M}\hat{V}'' - \frac{i\hbar}{M}\hat{V}'\hat{P}.

(4.17)

The operators $\hat{V}'$ and $\hat{V''}$ can vanish (can satisfy $\hat{V}'\ket{\psi} = 0$ ). When does this happen? This is the only case in which $[\hat{K},\hat{V}] = 0$ . Does this make sense physically?

Answers

$-\frac{\hbar^2}{2M}V(x)\frac{\partial^2 \psi}{\partial x^2}$
$-\frac{\hbar^2}{2M}\left(\frac{\partial^2 V}{\partial x^2}\psi(x) + 2 \frac{\partial V}{\partial x}\frac{\partial \psi}{\partial x} + V(x)\frac{\partial^2 \psi}{\partial x^2}\right)$
n/a (show that)
n/a (show that)
$\hat{V}' = 0$ if $V(x) =$ constant (in which case $\hat{V}'' = 0$ also). This does make sense, as if the potential is constant, the force vanishes, so the particle is free.

Chapters

Ch. 3: Incomp. Obs. & HUP

Chapters

Ch. 5: Evolution