Chapter 7: The Linear Harmonic Oscillator

Note

If quantum mechanics hasn’t profoundly shocked you, you haven’t understood it yet. ― Niels Bohr

Summary

Attention

In the last chapter, we highlighted the importance of finding the eigenstates of the Hamiltonian (i.e., energy) operator. In this chapter, we study in depth a specific class of Hamiltonian operators where the potential energy is a quadratic function of the position (classically):

$V(x)=\frac{1}{2} k x^{2}$

This is a very important class of problems as most energy functions are quadratic around an energy minimum. This can be seen directly by examining the Taylor series of a potential around the minimum. Here, we will limit ourselves to one-dimensional examples and this will be expanded to three dimensions in a later chapter.

Remember that the Hamiltonian operator includes both a kinetic energy term and a potential energy term, we can write:

$\hat{H}=\frac{\hat{p}_{x}^{2}}{2 m}+\frac{1}{2} m \omega^{2} \hat{x}^{2}$

We know the eigenstates of each of the operators ( $\hat{x}$ hat{p_x}`) but, because those two operators do not commute (as we saw in Chapter 6: Wave mechanics in one-dimension), the eigenstates of the two operators cannot be eigenstates of the Hamiltonian (note to students: make sure you understand the last sentence I wrote here!). This is why we develop a special method to solve the problem.

In this method, we introduce the raising and lowering operators. This is reminiscent of the operators that raise and lower the angular momentum. There is, however, an important difference: there is no maximum value for the eigen-energy (since the harmonic potential is only bounded in the minimum direction, but it is not bounded in the increasing energy values). As a result, we find the energy eigenvalues:

$E_n=\hbar \omega\left(n+\frac{1}{2}\right)$

This expression shows that (1) there is a zero-point energy (i.e., the ground state is not a zero-energy value) and (2) the energy eigenvalues are equidistant. The existence of a non-vanishing zero-point energy is related to the uncertainty relationship of the momentum and position operators: $\langle E\rangle=\frac{\left(\Delta p_{x}\right)^{2}}{2 m}+\frac{1}{2} m \omega^{2}(\Delta x)^{2}$ , which shows that the expectation value of the energy can never be zero (if it were, we would know both the position and the momentum of the particle at the same time). In the box below, we show that the ground state is a Gaussian state. Remembering that such a state is the minimum uncertainty state, we find that the expectation value of the energy is exactly the (non-zero) ground-state energy of the system.

Next, we examined the time dependence of the harmonic oscillator. Of course, we created a superposition of eigenstates (since the dynamics of a stationary states is not very illuminating or interesting!). In a superposition state, we find that the expectation value fot the position operator oscillates back and forth.

Most of the chapter dealt with the H.O. using an operator approach. We also considered the position space representation of the Schroedinger equation, to yield:

$\frac{d^{2} \psi}{d y^{2}}+\left(\varepsilon-y^{2}\right) \psi=0$

Solving this equation is an excellent example of how we solve quantum mechanical problems by building solutions that are known to be correct at various limits. Here, we seek a power series solution and realize that the series MUST stop (that is: the series is not infinite) and the solution is therefore a polynomial. In the case of the harmonic oscillator, the polynomial is knows as the Hermite polynomial and it is often defined by a recursion relationship: $\frac{a_{k+2}}{a_{k}}=\frac{2 k+1-\varepsilon}{(k+2)(k+1)}$ (see box below on how to learn about special functions).

Warning

The study of the harmonic oscillator is a good place to ask the most obvious question: Why don’t we see manifestations of Quantum Mechanics more easily in nature?. First, I like to emphasize that we may not see directly quantum mechanics, what we see can be traced back to quantum mechanics. However, we can wonder: why don’t we see the oscillations predicted for the quantum oscillator? To address this, we introduced the concept of classical turning point (CTP) and the fact the harmonic oscillator can actually go beyond the CTP. This tunneling effect can be explained using the uncertainty principle (because if the quantum mechanical oscillator were to stop at the CTP, the position and the linear momentum of the oscillator would be known perfectly, at odd with the uncertainty principle). Next, we introduced the concept of correspondence principle (or Bohr’s principle): the predictions of QM match those of classical physics when classical physics works. For instance, a macroscopic oscillator corresponds to states excited up to $N\approx 10^{26}$ : in such a regime, the number of oscillations is so large that our eyes only see the average values.

Note

Going from an operator framework to a position-space representation

As we have seen in Chapter 6: Wave mechanics in one-dimension, the simple rules to express quantum mechanical expressions in position space representation include:

$\begin{array}{l} \left\langle x\left|\hat{p}_{x}\right| \psi\right\rangle=\frac{\hbar}{i} \frac{\partial\langle x \mid \psi\rangle}{\partial x} \\ \langle x|\hat{x}| \psi\rangle=x\langle x \mid \psi\rangle \end{array}$

Of course this approach only works when it is convenient to use the position-space representation. We could consider, as an alternative, using the momentum space-representation. In the specific case of the harmonic oscillator both approaches are pretty much equally appealing (just check the mathematical form of the Hamiltonian to convince yourselves). In fact one can go from one to the other using a simple Fourier transform.

Example: effect of lowering operator on ground state

$\langle x|\hat{a}| 0\rangle=\sqrt{\frac{m \omega}{2 \hbar}}\left\langle x\left|\left(\hat{x}+\frac{i}{m \omega} \hat{p}_{x}\right)\right| 0\right\rangle=0$

This becomes:

$\frac{d\langle x \mid 0\rangle}{d x}=-\frac{m \omega x}{\hbar}\langle x \mid 0\rangle$

Solution of this equation is a Gaussian state:

$\langle x \mid 0\rangle=\left(\frac{m \omega}{\pi \hbar}\right)^{1 / 4} e^{-m \omega x^{2} / 2 \hbar}$

Warning

Special functions

Physics uses many special functions as solutions to specific problems. There is nothing to be afraid of! Special functions are usually defined as such and their properties have been worked out in tables, etc. What is important is to be able to interpret the solution. In this chapter, we see the importance of Hermite polynomials. One excellent place to learn about special functions and their properties is NIST Digital Library of Mathematical Functions:. For example, Hermite polynomial is show here. We also found that spherical Bessel functions and spherical Neumann functions play an important role in solving the nfinite potential well problems. See the page Special Mathematical Functions for a summary of all special functions encountered in this course.

Learning Material

Copy of Slides

The slides for Chapter 7 are available in pdf format here: 📂.

Screencast

Key Learning Points

Creation and annihilation operators:

$\begin{aligned} \hat{a} &=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}+\frac{i}{m \omega} \hat{p}_{x}\right) \\ \hat{a}^{\dagger} &=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}-\frac{i}{m \omega} \hat{p}_{x}\right) \end{aligned}$
Position and momentum operators:

$\begin{aligned} \hat{x} &=\sqrt{\frac{\hbar}{2 m \omega}}\left(\hat{a}+\hat{a}^{\dagger}\right) \\ \hat{p}_{x} &=-i \sqrt{\frac{m \omega \hbar}{2}}\left(\hat{a}-\hat{a}^{\dagger}\right) \end{aligned}$
Commutation relationship:

$\left[\hat{a}, \hat{a}^{\dagger}\right]=1$
Number operator:

$\hat{N}=\hat{a}^{\dagger} \hat{a}$
Hamiltonian:

$\hat{H}=\hbar \omega\left(\hat{a}^{\dagger} \hat{a}+\frac{1}{2}\right)$
Matrix elements:

$\hat{a}|n\rangle=\sqrt{n}|n-1\rangle$

$\hat{a}^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle$
A coherent state is an eigenstate of the lowering operator. $|\alpha\rangle=c_{0} \sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}}|n\rangle$ . This is a minimum uncertainty state (the uncertainty in position and momentum is not time-dependent)

Test your knowledge

If $\hat{a}\ket{n}=\sqrt{n}\ket{n-1}$ , what is $\bra{n}\hat{a}^{\dagger}$ ?
1. $\sqrt{n}\bra{n-1}$
2. $\sqrt{n}\bra{n}$
3. $\sqrt{n}\bra{n+1}$
What is $\bra{n}\hat{a}^{\dagger}\hat{a}\ket{n}$ ?
1. 0
2. n-1
3. n
4. n+1
What is the harmonic approximation?
1. It is the claim that one makes fewer mistakes in physics when listening to music.
2. It is the claim that everything is a spring in physics, even in the fall.
3. It is the claim that around a local minimum, a potential behaves approximately as a quadratic function.
4. It is the approximation that Joe Harmo, the brother of Nick, can always solve physics problems, even without opening the textbook.
What is the zero point energy?
1. It is an arbitrary constant one can eliminate by a convenient shift of the energy scale.
2. It is the energy of the ground state of the harmonic oscillator. It is a consequence of Heisenberg uncertainty.
3. It is the energy you find at the classical turning point of the harmonic oscillator.
What is the correspondence principle?
1. It is the principle that for an oscillator in a large- $n$ state, the behavior predicted by quantum mechanics matches that of classical physics.
2. It is the principle that states that quantum mechanics can never be really understood by us mere mortals.
3. It is the principle that is responsible for the zero point energy.
Calculate the expectation value of the square of the uncertainty of $\hat{p}_x$ (that is: $(\Delta p_x)^2$ ) for the eigenstate $\ket{n}$ of the harmonic oscillator. Remember that: $\hat{p}_{x}=-i \sqrt{\frac{m \omega \hbar}{2}}\left(\hat{a}-\hat{a}^{\dagger}\right)$ and $\hat{a} | n \rangle =\sqrt{n} | n-1 \rangle{\rm ; and~}\hat{a}^{\dagger} | n \rangle =\sqrt{n+1} | n+1 \rangle$ .
1. 0
2. $\left(n+\frac{1}{2}\right) m \omega \hbar$
3. $\left(n+1\right) m \omega \hbar$
4. $\frac{m \omega \hbar}{2}$
5. $\left(2n+1\right) m \omega \hbar$
If you examine the $x$ -representation of the eigenstate $\ket{n}$ of the number operator ( $\hat{N}$ ): $\bra{x}\ket{n}$ . What can you conclude?
1. The function abruptly decays to zero at the classical turning point.
2. The function is real and crosses the $x$ -axis exactly $n$ times.
3. The function is always positive but shows a number of local minima.
4. The function represents the probability density of the problem.
When prepared in a single eigenstate, the expectation value of the position of a harmonic oscillator oscillates back and forth.
1. True
2. False
3. It depends