6 The quantum simple harmonic oscillator
One of the first systems you have seen, both and classical and quantum mechanics, is the simple harmonic oscillator: \hat{H} = \frac{\hat{p}{}^2}{2m} + \frac{1}{2} m \omega^2 \hat{x}{}^2. This is a great example in both cases; it is one of the few models that can be solved analytically in complete detail. Of course, the SHO is much more than just a textbook example; it is practically applicable to a very wide range of systems. SLAC particle theorist Michael Peskin likes to describe all of physics as “that subset of human experience that can be reduced to coupled harmonic oscillators”.
The ubiquity of the SHO is completely expected, because in physics we like to study systems which are close to equilibrium.
Classically, points of stable equilibrium occur at minima of the potential energy, where the force vanishes since dV/dx = 0. Quantum mechanically, the situation is more complicated, but it is still true that stable bound states of a particular system will be associated with a minimum of the potential; near the minimum we can identify a series of bound states.
In either case, we can make significant progress in studying the system by expanding the potential about its minimum: V(x) = V(x_0) + (x-x_0) V'(a) + \frac{1}{2} (x-x_0)^2 V''(x_0) + ... The first derivative vanishes since we’re at a minimum, and V(a) is just a constant shift, so we can see that for small displacements from equilibrium, the potential is well-approximated by a quadratic, V(x) \approx \frac{1}{2} (x-x_0)^2 V''(x_0). This is exactly a simple harmonic oscillator! Of course, this is a very simplified picture for one particle in one dimension. But many real quantum-mechanial systems are well-described by harmonic oscillators (usually coupled together) when near equilibrium, for example the behavior of atoms within a crystalline solid.
What is the best way to study the quantum SHO? One option is to simply try to solve the time-independent Schrödinger equation, i.e. try to find the energy eigenstate solutions. If we simply plug in the form of the potential above, we find the differential equation -\frac{\hbar^2}{2m} \frac{\partial^2 \psi_E}{\partial x^2} = (E - \frac{1}{2} m \omega^2 x^2) \psi_E(x). This isn’t nearly as simple as the classical SHO equation, unfortunately. This equation can be solved analytically using standard methods; the solutions involve the Hermite polynomials, which you may or may not have seen before. If you’re interested in the brute-force approach, I direct you to Merzbacher, Chapter 5, for the gory details. We’re going to do it the clever way, taking advantage of our Hilbert space formalism.
6.1 Ladder operators
We start by noticing that the Hamiltonian looks reasonably symmetric between \hat{x} and \hat{p}; if we can “factorize” it into the square of a single operator, then maybe we can find a simpler solution. If we were dealing with numbers instead of operators, we could write H = \frac{1}{2} m\omega^2 \left(x + \frac{ip}{m\omega}\right) \left(x - \frac{ip}{m\omega}\right). Unfortunately, we’re stuck with the operators \hat{x} and \hat{p}, which don’t commute; but since their commutation relation is relatively simple, we might be able to factorize anyway. Let’s define the new operator \hat{a} \equiv \sqrt{\frac{m\omega}{2\hbar}} \left(\hat{x} + \frac{i\hat{p}}{m\omega}\right) \\ \hat{a}{}^\dagger \equiv \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} - \frac{i\hat{p}}{m\omega}\right). Notice that this is not a Hermitian operator! Now, we try to get back to \hat{H}: \hat{a}{}^\dagger \hat{a} = \frac{m\omega}{2\hbar} \left( \hat{x}{}^2 + \frac{\hat{p}{}^2}{m^2 \omega^2} + \frac{i}{m\omega} [\hat{x}, \hat{p}] \right) \\ = \frac{1}{2\hbar} m\omega \hat{x}{}^2 + \frac{\hat{p}{}^2}{2\hbar m \omega} - \frac{1}{2} \\ = \frac{\hat{H}}{\hbar \omega} - \frac{1}{2}. The product operator \hat{a}{}^\dagger \hat{a} \equiv \hat{N} is called the number operator, for reasons which will become clear shortly. Since the number operator is exactly the Hamiltonian up to some constants, the two operators are simultaneously diagonalizable. In fact, it’s easy to see that they have the same eigenstates; if we let \hat{N} \ket{n} = n \ket{n} be the eigenkets of the number operator, then from above we have \hat{H} \ket{n} = \left( \hat{N} + \frac{1}{2} \right) \hbar \omega \ket{n} \\ = \left(n + \frac{1}{2}\right) \hbar \omega \ket{n}, so \ket{n} are also the energy eigenstates, with eigenvalues E_n = \left(n + \frac{1}{2}\right) \hbar \omega. This is a clever construction, but can we get more than just the energy levels? Since our goal was factorization, we need to study the individual operators \hat{a} and \hat{a}{}^\dagger. Their commutator is easily derived: [\hat{a}, \hat{a}{}^\dagger] = \frac{1}{2\hbar} \left(-i[\hat{x}, \hat{p}] + i[\hat{p}, \hat{x}] \right) = 1. Then [\hat{N}, \hat{a}] = [\hat{a}{}^\dagger \hat{a}, \hat{a}] = \hat{a}{}^\dagger [\hat{a}, \hat{a}] + [\hat{a}{}^\dagger, \hat{a}] \hat{a} = -\hat{a} and similarly, [\hat{N}, \hat{a}{}^\dagger] = \hat{a}{}^\dagger. The commutation relations are enough to tell us how the \hat{a} act on the eigenstates of \hat{N}: notice that [\hat{N}, \hat{a}] \ket{n} = \hat{N} \hat{a} \ket{n} - \hat{a} \hat{N} \ket{n} \\ -\hat{a} \ket{n} = \hat{N} \hat{a} \ket{n} - \hat{a} n \ket{n} \\ \hat{N} \hat{a} \ket{n} = (n-1) \hat{a} \ket{n} and similarly, \hat{N} \hat{a}{}^\dagger \ket{n} = (n+1) \hat{a}{}^\dagger \ket{n}. So the operators \hat{a} and \hat{a}{}^\dagger map eigenstates of the number operator into one another! Specifically, \hat{a} \ket{n} is proportional to the state \ket{n-1}, and \hat{a}{}^\dagger \ket{n} to \ket{n+1}. For obvious reasons, the operators \hat{a}{}^\dagger and \hat{a} are called the raising and lowering operators, or sometimes creation and annihilation operators. (You’ll also hear them called ladder operators as a pair, since they raise and lower the \ket{n} states by one unit.)
Assuming that all of the basis kets \{\ket{n}\} are orthonormal is enough to fix the normalization of the raising and lowering operators, which is left as an exercise for you: the result is, assuming the normalization is real and positive (since we want to end up with real positive energies), \hat{a} \ket{n} = \sqrt{n} \ket{n-1} \\ \hat{a}{}^\dagger \ket{n} = \sqrt{n+1} \ket{n+1}. Notice that as long as the n are labeled by integers, this sequence terminates; if we try to annihilate the state \ket{0}, we get the null ket, \hat{a} \ket{0} = 0. (This assumes n is an integer, but if it’s non-integer then you will end up with arbitrarily negative energies! If our Hamiltonian system is unbounded from below, bad things will happen, like runaway solutions that will end up with infinitely high energy if we couple to another system.)
Derive the results above for the coefficients when the ladder operators act on number eigenstate \ket{n}. In other words, starting from the established results
\hat{a} \ket{n} = c_n \ket{n-1}, \\ \hat{a}{}^\dagger \ket{n} = d_n \ket{n+1}, find c_n and d_n.
Answer:
This is most easily approached not by looking at the action on individual states, but by taking an expectation value:
\bra{n} \hat{a}{}^\dagger \hat{a} \ket{n} = \bra{n-1} c_n^\dagger c_n \ket{n-1} \\ = |c_n|^2 \left\langle n-1 | n-1 \right\rangle = |c_n|^2, where the key step was acting backwards on the bra with the \hat{a}{}^\dagger, and we use orthonormality of the \ket{n} eigenkets.
But \hat{a}{}^\dagger \hat{a} = \hat{N} is the number operator too, so this is also \bra{n} \hat{N} \ket{n} = n = |c_n|^2
which immediately gives us c_n = \sqrt{n} e^{i\phi(n)} - since we only constrained the norm squared, we haven’t shown yet whether there is an extra phase or not.
To get at d_n, we need the operators to be reversed: \bra{n} \hat{a} \hat{a}{}^\dagger \ket{n} = \bra{n+1} d_n^\dagger d_n \ket{n+1} \\ = |d_n|^2. But this isn’t the number operator anymore, to get that we need a commutator: \bra{n} \hat{a} \hat{a}{}^\dagger \ket{n} = \bra{n} (\hat{a}{}^\dagger \hat{a} + 1) \ket{n} \\ = n+1 = |d_n|^2 so d_n = \sqrt{n+1} e^{i\chi(n+1)}.
To deal with the phases, we can just look at the two operators acting on the same state as follows: \hat{a}{}^\dagger \hat{a} \ket{n} = \hat{a}{}^\dagger c_n \ket{n-1} \\ = c_n (d_{n-1}) \ket{n}. But this is also \hat{N} \ket{n} = n \ket{n}, so we have c_n d_{n-1} = n. This fixes the phases to be equal and opposite, \chi(n) = -\phi(n).
How do we get rid of the final phase? I’m honestly not entirely sure…suggestions are welcome!The state \ket{0} corresponds to the lowest possible energy of the system, E_0 = \hbar \omega/2; we call this the ground state. We can use the ladder operators to construct any other state from the ground state, making sure to normalize properly: \hat{a}{}^\dagger \ket{0} = \ket{1} \\ \hat{a}{}^\dagger \ket{1} = \sqrt{2} \ket{2} \Rightarrow \frac{(\hat{a}{}^\dagger)^2}{\sqrt{2}} \ket{0} = \ket{2} \\ ... \\ \ket{n} = \frac{(\hat{a}{}^\dagger)^n}{\sqrt{n!}} \ket{0}.
6.2 Finding the SHO wavefunctions
So we’ve found a solution for the SHO energy eigenstates that completely circumvents solving the Schrödinger equation. But what if we want the position-space wavefunction that Schrödinger would have given us?
In fact, we have all the information we need to reconstruct the position-space wavefunctions for these states. The ground state is easily derived: \bra{x} \hat{a} \ket{0} = 0 \\ \sqrt{\frac{m \omega}{2\hbar}} \bra{x} \left[ \hat{x} + \frac{i\hat{p}}{m \omega} \right] \ket{0} = 0 \\ \left[x + \frac{\hbar}{m\omega} \frac{\partial}{\partial x} \right] \left\langle x | 0 \right\rangle = 0. This is a much easier differential equation to solve: the extra factor of x suggests a solution of the form e^{x^2}, and indeed we find that \left\langle x | 0 \right\rangle = \left( \frac{m\omega}{\pi \hbar}\right)^{1/4} \exp \left[-\frac{1}{2} \frac{m\omega}{\hbar} x^2 \right]. From here we don’t have to solve any more differential equations; the raising operator is all we need to construct the rest of the states, like so: \left\langle x | 1 \right\rangle = \bra{x} \hat{a}{}^\dagger \ket{0} = \sqrt{\frac{m\omega}{2\hbar}} \left( x - \frac{\hbar}{m\omega} \frac{\partial}{\partial x} \right) \left\langle x | 0 \right\rangle \\ = \left( \frac{m\omega}{\pi \hbar}\right)^{1/4} \sqrt{\frac{m\omega}{2\hbar}} \left(2x \right) \exp \left( -\frac{1}{2} \frac{m\omega}{\hbar} x^2 \right). We can keep going to obtain the higher states, but taking the derivatives becomes more tedious; we’ll see a better way shortly.
Although we can derive the wavefunctions, the fact of the matter is that we don’t need them; the states and the ladder operators are all we need to calculate almost anything we want to about the SHO. The operators \hat{x} and \hat{p} can be rewritten as combinations of ladder operators: \hat{x} = \sqrt{\frac{\hbar}{2m\omega}} (\hat{a}{}^\dagger + \hat{a}) \\ \hat{p} = i \sqrt{\frac{\hbar m \omega}{2}} (\hat{a}{}^\dagger - \hat{a}). The matrix elements of the ladder operators themselves are particularly simple: \bra{n'} \hat{a} \ket{n} = \sqrt{n} \delta_{n',n-1} \\ \bra{n'} \hat{a}{}^\dagger \ket{n} = \sqrt{n+1} \delta_{n',n+1}. and so, for example, the matrix elements of \hat{x} are \bra{n'} \hat{x} \ket{n} = \sqrt{\frac{\hbar}{2m \omega}} \left( \sqrt{n} \delta_{n',n-1} + \sqrt{n+1} \delta_{n',n+1} \right). Similar expectation values for powers of \hat{x} and \hat{p} require use of the commutation relations, but are all straightforward to construct. Although there are an infinite number of states \ket{n}, here they’re labelled by integers instead of real numbers, so we’ve gone back to our discrete notation. In fact, we can think of \hat{x} as an infinite-dimensional matrix with entries just off the diagonal: \hat{x} = \sqrt{\frac{\hbar}{2m \omega}} \left( \begin{array}{ccccc} 0& 1 & 0 & 0 & ...\\ 1&0&\sqrt{2}&0&...\\ 0&\sqrt{2}&0&\sqrt{3}&...\\ 0&0&\sqrt{3}&0&...\\ ...&...&...&...&...\end{array}\right) and we can do the same for \hat{p} easily, or for \hat{a} and \hat{a}{}^\dagger.
The matrix form of this operator suggests a better way to calculate the position-space wavefunction, in fact. We know that the eigenket \ket{x} satisfies the eigenvalue equation \hat{x} \ket{x} = x \ket{x}; in matrix notation, we can write this as \sqrt{\frac{\hbar}{2m \omega}} \left( \begin{array}{ccccc} 0& 1 & 0 & 0 & ...\\ 1&0&\sqrt{2}&0&...\\ 0&\sqrt{2}&0&\sqrt{3}&...\\ 0&0&\sqrt{3}&0&...\\ ...&...&...&...&...\end{array}\right) \left( \begin{array}{c} c_0 \\ c_1 \\ ... \\ ... \\ ... \end{array} \right) = x \left( \begin{array}{c} c_0 \\ c_1 \\ ... \\ ... \\ ... \end{array} \right), where the components c_n = \left\langle n | x \right\rangle. The inner products obviously satisfy a set of recurrence relations, c_1 = \sqrt{\frac{2m\omega}{\hbar}} x c_0, \\ \sqrt{2} c_2 + c_0 = \sqrt{\frac{2m\omega}{\hbar}} x c_1, and so on; it’s easy to spot the pattern, \sqrt{n+1} c_{n+1} + \sqrt{n} c_{n-1} = \sqrt{\frac{2m\omega}{\hbar}} x c_n. (If this is too much hand-waving for you, note that you can derive the same recurrence relation by just using the form of \hat{x} in terms of the ladder operators, and inserting a complete set of states.)
Fortunately for us, this turns out to be a famous recurrence relation, defining a sequence of functions known as the Hermite polynomials, H_n(x). The first few are: H_0(x) = 1 \\ H_1(x) = 2x \\ H_2(x) = 4x^2 - 2 \\ H_3(x) = 8x^3 - 12x Keeping track of all of the constant factors, the solution in full becomes \left\langle n | x \right\rangle = \frac{1}{\sqrt{2^n n!}} \left( \frac{m\omega}{\pi \hbar}\right)^{1/4} \exp \left(-\frac{m\omega}{2\hbar} x^2 \right) H_n \left( \sqrt{\frac{m\omega}{\hbar}} x \right).
Now equipped with both the wavefunctions and the expressions for \hat{x} and \hat{p} in terms of ladder operators, we have a couple of different ways to evaluate expectation values of functions of x and p. One immediate thing we can notice is that the energy eigenstates satisfy \bra{n} \hat{x} \ket{n} = \bra{n} \hat{p} \ket{n} = 0. If we try to think about the classical harmonic oscillator, this is clearly not something that maps onto it easily: the only solution for the classical oscillator that has zero momentum and position is if the mass is at rest at the bottom of the potential, but in that case the energy is fixed to the minimum. If we plot the wavefunctions, we find they are symmetric around x=0 but localized and more spread out as n increases. Obviously there is no direct classical analogue, but I think the closest analogy is that you should think of the energy eigenstates of particles moving in both directions in the well at the same time.
Of course, for other states that aren’t energy eigenstates we can find non-trivial time evolution. The following tutorial will give some practice in a simple example, and we will study some other particularly interesting states below
Here, you should complete Tutorial 3 on “Time evolution of the quantum harmonic oscillator”. (Tutorials are not included with these lecture notes; if you’re in the class, you will find them on Canvas.)
6.3 Bound states and zero-point energy
If we inspect the form of the wavefunction obtained above, we see that it is localized around the potential minimum at x=0; regardless of what the Hermite polynomials do, all wavefunctions die off as \exp(-x^2) at large x. Moreover, the set of allowed energy levels is discrete, with only E_n = \hbar \omega (n+1/2) as allowed values. (Both of these properties are in contrast to the simple plane-wave solution we found before, in which any positive value of the energy is allowed continuously.)
This turns out to be a very generic feature of quantum wave mechanics: bound states, in which the wavefunction is confined to a finite region in space, have a discrete set of solutions for energy eigenfunctions (i.e., the energy is quantized.) In fact, it’s not too difficult to argue why this has to be the case. Suppose we have an arbitrary “potential well”, meaning an unknown potential V(x) which is negative for some region near the origin x=0, but satisfies \lim_{x \rightarrow \pm \infty} V(x) = 0. Then for any bound state with energy E < 0, we know that outside of the well region at large x, the solutions must take the form we found for negative-energy plane waves, \psi_E(x) \rightarrow_{x \rightarrow \pm \infty} B e^{\kappa x} + C e^{-\kappa x}. Let’s consider just positive x to begin with. Then the C term is perfectly fine; it will go to zero at large x, and we can normalize our wavefunction. But the B term diverges - is not normalizable - so the only possibility is that B must satisfy the equation B(E) = 0. This is the source of quantization; the function B(E) is itself continuous, but will generally only have a finite number of zeroes. This is similar to how quantization arises in classical physics - wave modes must satisfy some boundary condition, e.g. for electromagnetic waves in a cavity or a string tied between two posts. But in quantum mechanics, everything is a wave, so the effect occurs more often!
If you’re paying very close attention, you might wonder exactly what the difference is between the “not normalizable” exponential states above, which we reject as “unphysical”, and the infinite plane waves which we’ve already noted are also not normalizable in the usual sense. Since we’re relying heavily on normalizability for this “proof”, it’s worth being a little careful. Let’s restrict ourselves to a single “right-moving” solution for simplicity, \psi_k(x) = A e^{ikx}. We’ll consider both real k, which gives a normal plane wave which is oscillating, and imaginary k = i \kappa corresponding to negative energy, which gives real exponentials. We assume the system is free (V(x) = 0.)
There are two ways to see why an oscillating plane-wave state is acceptable (the vocabulary term is that they are “improperly normalized”), while the real exponential is “non-normalizable” and therefore not physical. The first argument is more physical, and relies on the idea of a cutoff. In the real world, we’re never actually integrating all the way out to infinity; our experimental apparatus will have some size L, and the practical limit of “infinity” in all of our position integrals is cut off by this size. This means that our normalization integrals become \int_{-L/2}^{L/2} |\psi_k(x)|^2 = 1. Let’s plug in our plane wave: \int_{-L/2}^{L/2} (A^\star e^{-ikx}) A e^{ikx} = 1 \\ \int_{-L/2}^{L/2} |A|^2 = 1 \Rightarrow |A| = \frac{1}{\sqrt{L}}. This is a perfectly fine normalization, which goes to zero as L \rightarrow \infty, reflecting the fact that in the “infinite box” limit the integral over the plane-wave itself is divergent.
On the other hand, in the real exponential case, \int_{-L/2}^{L/2} (A^\star e^{-\kappa x}) A e^{-\kappa x} = 1 \\ |A|^2 \int_{-L/2}^{L/2} e^{-2\kappa x} = 1 \\ |A|^2 \frac{\sinh(\kappa L)}{\kappa} = 1 \Rightarrow |A| = \sqrt{\frac{\kappa}{\sinh(\kappa L)}}.
Now, at first glance this seems like the same thing: in either case, |A| goes to zero as L \rightarrow \infty, compensating for the divergence. But the strength of the divergence being exponential in the latter case leads to unphysically strong dependence on the value of our cutoff L. For example, we can ask in both cases what the probability is of finding |x| < x_0 for some x_0 \ll L. For the ordinary plane wave, p(|x| < x_0) = \int_{-x_0}^{x_0} |\psi_k(x)|^2 = 2x_0 |A|^2 \\ = \frac{2x_0}{L}. For the real exponential, on the other hand, p(|x| < x_0) = \int_{-x_0}^{x_0} |A|^2 e^{-2\kappa x} \\ = \frac{\kappa}{\sinh(\kappa L)} \frac{\sinh(2\kappa x_0)}{\kappa} \\ \approx \frac{2\kappa x_0}{\sinh(\kappa L)}. If we keep x_0 / L fixed while taking L \rightarrow \infty, then the probability of finding our particle in the region near the origin stays fixed in the plane-wave case, but it vanishes exponentially fast in the real exponential case. In fact, we find in the latter case that the probability of our particle being anywhere is concentrated at the boundaries x = \pm L. Because the exponential divergence is so strong, if we try to calculate any physics at positions x \ll L, we basically find that this solution is useless - it’s basically equivalent to the trivial \psi(x) = 0, since as we try to remove the cutoff our particle just sits out at the cutoff x \approx L. For the plane wave, the probability remains evenly distributed.
Here’s the second way to discriminate “improper normalizability” from “non-normalizability”. Recall that the context we’re working in here is finding energy eigenstate solutions, which we use to construct an orthonormal basis. In the case of a finite Hilbert space, this means that our solutions satisfy \left\langle E_i | E_j \right\rangle = \delta_{ij} = \int_{-\infty}^\infty \psi_{i}^\star(x) \psi_j(x). This reduces to the normalization being 1 in the case i = j, but the orthogonality of the states is also important (the integral should be zero if i \neq j.) If we re-do our integrals in the box for two different values of k or \kappa, we find the following: \left\langle k' | k \right\rangle = \int_{-L/2}^{L/2} \psi_{k'}^\star(x) \psi_k(x) \\ = \frac{2}{k-k'} |A|^2 \sin \left( \frac{1}{2} (k-k') L \right). If k \neq k', then for this to be zero we need to satisfy a boundary condition, namely \frac{(k-k') L}{2} = n \pi \Rightarrow k-k' = \frac{2\pi n}{L}. So we see that in a finite box, we can only have a finite number of possible momentum eigenstates. But the spacing between these eigenstates goes to zero as L \rightarrow \infty, in which case we can think of recovering the continuous limit where any k \neq k' causes the above integral to vanish. (The integral seems indeterminate but as mentioned in the plane waves section above, you can show it vanishes by using contour integration.)
What about the negative energy states? We have \left\langle \kappa' | \kappa \right\rangle = \int_{-L/2}^{L/2} |A|^2 e^{-(\kappa + \kappa') x} \\ = \frac{2}{\kappa + \kappa'} |A|^2 \sinh \left( \frac{(\kappa + \kappa') L}{2} \right). For real \kappa and \kappa', there are no solutions that cause this to vanish! So our states are no longer orthogonal to one another at finite L, which means there is no sensible L \rightarrow \infty limit that lets us recover an orthonormal energy basis. In fact, this leads us to another simple way to understand the problem: in a free system, if E < 0, then the momentum p must be imaginary. This means that our Hamiltonian \hat{H} is no longer Hermitian! As we have seen, Hermitian operators are guaranteed to give us an orthonormal basis; the loss of Hermiticity at negative energy is why we no longer have a good basis from energy eigenstates. So the problems with these solutions are much more profound than just an integral being infinite!
(A final aside on the aside: you might worry about whether this means the Hamiltonian is non-Hermitian even if E < V(x) in a finite region. It isn’t; in the more general case, the real-exponential solutions simply are not momentum eigenstates, so the Hamiltonian doesn’t have to have imaginary momentum to yield them. It’s only the free Hamiltonian where \hat{H} and \hat{p} are simultaneously diagonalizable that leads to this “imaginary momentum” problem.)
(I’ve ignored negative x, but the story is similar; at negative x we have one other divergent solution that has to have zero coefficient, leading to more energy quantization conditions. For a symmetrical system like the harmonic oscillator, we won’t find any new information since positive x and negative x have to look the same.)
You might wonder, what if E > 0? Those solutions won’t be quantized; but by the way we’ve set the problem up, they also won’t be bound states, since they will connect on to oscillating plane-wave solution. We have given a very general argument that any bound solutions will have quantized energies.
6.3.1 Zero-point energy
The precise values of the bound-state energies will of course depend on the details of the potential, but there is another very general and powerful observation we can make about the minimum possible bound-state energy. Just as in classical physics, we can divide our total energy into the expectation values of kinetic and potential energy: E = \left\langle \hat{T} \right\rangle + \left\langle \hat{V} \right\rangle.
In terms of the energy eigenstate wavefunction, we can write \left\langle \hat{T} \right\rangle = \left\langle \frac{\hat{p}^2}{2m} \right\rangle = - \frac{\hbar^2}{2m} \int_{-\infty}^\infty dx\ \psi^\star(x) \frac{d^2}{dx^2} \psi(x) \\ = \frac{\hbar^2}{2m} \int_{-\infty}^\infty dx \left| \frac{d\psi}{dx} \right|^2 where we have integrated by parts to move one of the derivatives onto \psi^\star(x), and discarded the “boundary term” using the requirement that \psi(x) goes to zero as x \rightarrow \infty for a bound state. The mean potential energy is easier to find: \left\langle \hat{V} \right\rangle = \int_{-\infty}^\infty dx\ V(x) |\psi(x)|^2. The kinetic energy is the integral of a squared absolute value, so \left\langle \hat{T} \right\rangle > 0 always (the only possible exception is d\psi/dx = 0, which would imply the wavefunction is just zero everywhere.) On the other hand, \left\langle \hat{V} \right\rangle = \int_{-\infty}^\infty V(x) |\psi(x)|^2 \geq \int_{-\infty}^\infty V_{\rm min} |\psi(x)|^2 = V_{\rm min}. where V_{\rm min} is the global minimum value of V(x). So we have proved that E = \left\langle T \right\rangle + \left\langle V \right\rangle > V_{\rm min} Our result implies that for the lowest possible energy E_0, corresponding to the ground state, we have E_{ZP} = E_0 - V_{\rm min} > 0. The quantity E_{ZP} is known as the zero-point energy. For the simple harmonic oscillator, V_{\rm min} = V(0) = 0 so the zero-point energy is E_{ZP} = \hbar \omega / 2; it is exactly the extra term that doesn’t depend on n in the result E_n = \hbar \omega (n + 1/2).
There is a more physical way to think of this energy, in terms of the uncertainty principle. In classical mechanics, there is a simple and obvious solution for E = V_{\rm min}: the classical particle can simply be placed at rest at the position x_{\rm min} corresponding to the minimum of the potential. But in quantum mechanics, the Heisenberg uncertainty principles prevents us from fully specifying both the position x_{\rm min} and the momentum p=0. So zero-point energy is exactly just the fact that no matter how hard we try to put our quantum particle at the potential minimum, it will always spread out a bit and have a bit of momentum, causing its energy to increase. In other words: uncertainty between \Delta x and \Delta p means we can never simultaneously satisfy the conditions “at rest” and “at the bottom of the potential well” - so there will always be some zero-point energy.
You might ask at this point, is the zero-point energy meaningful at all? In the SHO it’s just a constant, and we can always shift what energy we call zero. But this is a global shift, and if we shift our energy definition by -\hbar \omega / 2 then we have to shift the SHO potential V(x) down as well, so the zero-point energy remains.
Of course, in general only energy differences are observable anyways - does the zero-point energy have any physical consequences, or is it just an odd feature of our theory? In a fixed simple harmonic oscillator, the answer (as far as I know) is that it has no effect - it will always cancel out. But it’s easy to imagine that if we change the potential V(x) in some smooth way, we will change the zero-point energy, and then maybe we can see some effect. The most well-known example of an experimental observation of zero-point fluctuations is the Casimir effect, in which a small force is measured in between two grounded, conducting plates with pure vacuum in between them, corresponding to a change in energy as the plate separation is varied. But understanding this effect requires knowing how to deal with fluctuations of electromagnetic fields in vacuum, which we’re a long way from!
6.4 Coherent states
One interesting feature of the SHO energy eigenstates to explore is their uncertainty in position and momentum. Due to the way the ladder operators work, as pointed out above, it’s easy to read off that \left\langle \hat{x} \right\rangle = \left\langle \hat{p} \right\rangle = 0 for any energy eigenstate \ket{n}; for example, \bra{n} \hat{x} \ket{n} = \sqrt{\frac{\hbar}{2m\omega}} \bra{n} (\hat{a}^\dagger + \hat{a}) \ket{n} = 0 since a single ladder operator will give us \left\langle n | n \pm 1 \right\rangle = 0.
What about the squared uncertainties? Let’s do the algebra out: \bra{n} \hat{x}^2 \ket{n} = \frac{\hbar}{2m\omega} \bra{n} (\hat{a}^\dagger + \hat{a})^2 \ket{n} As above, two of the four terms are automatically zero: we only need to keep the two that have one raising and one lowering operator. Using the commutation relation, we’ll write the result in normal ordering, which means all of the \hat{a}’s are to the right of the \hat{a}^\dagger’s: \Rightarrow \left\langle \hat{x}^2 \right\rangle = \frac{\hbar}{2m\omega} \bra{n} (2\hat{a}^\dagger \hat{a} + 1) \ket{n} = \frac{\hbar}{m \omega} (n + 1/2). Now for momentum: \bra{n} \hat{p}^2 \ket{n} = -\frac{\hbar m \omega}{2}\bra{n} (\hat{a}^\dagger - \hat{a})^2 \ket{n} \\ = -\frac{\hbar m \omega}{2} \bra{n} (-2\hat{a}^\dagger \hat{a} - 1) \ket{n} \\ = \hbar m \omega (n + 1/2). Putting them together, the uncertainty product is \Delta x \Delta p = \sqrt{\left\langle \hat{x}^2 \right\rangle \left\langle \hat{p}^2 \right\rangle} = \hbar (n + 1/2). Only for the ground state n = 0 does this saturate the uncertainty principle \Delta x \Delta p \geq \hbar/2 (which is actually something of a special case; in general, quantum ground states aren’t also minimum-uncertainty states.)
What if we want to try to build a minimum-uncertatinty state explicitly? What does such a state even look like? We can find some guidance by going back to the uncertainty relation. Without proving it, I will draw on a result from Merzbacher: the uncertainty relation becomes and equality, \Delta A \Delta B = \frac{1}{2} |\left\langle [\hat{A}, \hat{B}] \right\rangle|, if and only if (\Delta \hat{B}) \ket{\psi} = \frac{\left\langle [\hat{A}, \hat{B}] \right\rangle}{2(\Delta A)^2} (\Delta \hat{A}) \ket{\psi}. Note that this is a statement not just about the operators, but also about the state \ket{\psi}; if this is ever satisfied for any state, then all states are minimum-uncertainty states.
Let’s focus on position and momentum specifically. Then we know the commutator, and the relation above becomes (\hat{p} - \left\langle p \right\rangle) \ket{\psi} = \frac{i\hbar}{2(\Delta x)^2} (\hat{x} - \left\langle \hat{x} \right\rangle) \ket{\psi}. Going to position space, this is a differential equation for the wavefunction: \left( -i \hbar \frac{d}{dx} - \left\langle p \right\rangle \right) \psi(x) = \frac{i\hbar}{2(\Delta x)^2} (x - \left\langle x \right\rangle) \psi(x). The solution to this simple differential equation is very familiar: \psi(x) = \frac{1}{\pi^{1/4} \Delta x} \exp \left[ \frac{i\left\langle p \right\rangle x}{\hbar} - \frac{(x-\left\langle x \right\rangle)^2}{4(\Delta x)^2} \right]. This is exactly the wave-packet solution we wrote down before! So minimum-uncertainty states (with regard to momentum and position) are precisely wave packets - plane waves with a Gaussian envelope applied. The position uncertainty is a free parameter, but the product of uncertainties is fixed.
However, we got this by modulating a plane wave, which was a free solution. Is this also a solution to the simple harmonic oscillator? We could use the differential equation to find out, but there’s an easier way using operators. In operator form, the equation for a minimum-uncertainty state was, rearranging slightly, \left( \hat{p} - \frac{i\hbar}{2(\Delta x)^2} \hat{x} \right) \ket{\psi} = \left( \left\langle p \right\rangle - \left\langle x \right\rangle \right) \ket{\psi}.
So this is saying that wave packets are eigenstates of a linear combination of \hat{p} and \hat{x}. That should sound familiar, because the ladder operator \hat{a} is just such a combination! If we identify (\Delta x)^2 = \frac{\hbar m \omega}{2}, then the equation becomes simply \hat{a} \ket{\alpha} = \alpha \ket{\alpha}, the eigenvalue equation for the ladder operator. These states \ket{\alpha} are known as coherent states.
Now, based on what we know about the ladder operator this equation looks very funny: at first glance, it looks impossible to satisfy, since \hat{a} always changes energy eigenstates. But that’s fine as long as the expansion of \ket{\alpha} in \ket{n} is infinite: \ket{\alpha} = \sum_{n=0}^\infty c_n \ket{n}. In fact, it’s straightforward to show that we can think of the coherent state as being created by exponentiating the creation operator: the normalized formula is \ket{\alpha} = e^{-|\alpha|^2/2} e^{\alpha \hat{a}^\dagger} \ket{0}.
Derive the formula above for \ket{\alpha} in terms of the ladder operator \hat{a}^\dagger.
Answer:
Then \hat{a} can give the same state back: \hat{a} \ket{\alpha} = \sum_{n=0}^\infty c_n \sqrt{n} \ket{n-1} \\ = \sum_{n=0}^\infty c_{n+1} \sqrt{n+1} \ket{n} \\ = \alpha \ket{\alpha} which will work as long as c_{n+1} \sqrt{n+1} = \alpha c_n. It’s not too hard to just read off from this the answer c_n = \frac{\alpha^n}{\sqrt{n!}} c_0.
Next, we rewrite the states \ket{n} in terms of raising operators: from the definition, \ket{n} = \sqrt{n(n-1)(n-2)...(1)} (\hat{a}^\dagger)^n \ket{0} = \sqrt{n!} (\hat{a}^\dagger)^n \ket{0}
Now we can just plug in the exponential series to find e^{\alpha \hat{a}^\dagger} c_0 \ket{0} = \sum_{n=0}^\infty \frac{(\alpha \hat{a}^\dagger)^n}{n!} c_0 \ket{0} \\ = \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} c_0 \ket{n} = \ket{\alpha}.
To finish, we just need to fix the normalizing constant c_0 by requiring these states are normalized: \left\langle \alpha | \alpha \right\rangle = 1 \\ = |c_0|^2 \left( \sum_{m=0}^\infty \bra{m} \frac{(\alpha^\star)^m}{\sqrt{m!}} \right) \left( \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} \ket{n} \right) \\ = |c_0|^2 \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{(\alpha^\star)^m \alpha^n}{\sqrt{m!n!}} \delta_{mn} \\ = |c_0|^2 \sum_{n=0}^\infty \frac{|\alpha|^{2n}}{n!} \\ = |c_0|^2 e^{|\alpha|^2}
from which we read off c_0 = e^{-|\alpha|^2/2}, proving the result.
Notably, any complex value of \alpha is a valid eigenvalue for a coherent state. What is the difference between these states? We can see it easily by calculating the other expectation values that determine a wave packet: \left\langle \hat{x}(\alpha) \right\rangle = \bra{\alpha} \hat{x} \ket{\alpha} \\ = \sqrt{\frac{\hbar}{2m\omega}} \bra{\alpha} (\hat{a} + \hat{a}^\dagger) \ket{\alpha} \\ = \sqrt{\frac{\hbar}{2m\omega}} \bra{\alpha} (\alpha + \alpha^\star) \ket{\alpha} \\ = \sqrt{\frac{2\hbar}{m\omega}}\ {\rm Re}(\alpha) and similarly, \left\langle \hat{p}(\alpha) \right\rangle = \sqrt{2\hbar m \omega} {\rm Im}(\alpha). So the full complex range of \alpha allows us to construct a wave packet with arbitrary position and momentum, with minimum uncertainty - and with its spread in position space \Delta x fixed by the properties of the SHO.
6.4.1 Squeezed states
One of the advantages of the way that we’ve derived the coherent states is that we can immediately see another adjustment we can make. When we used the ladder operator \hat{a} to construct the coherent states, it resulted in a fixed position uncertainty \Delta x. But what if we just build a different ladder operator with any \Delta x we want? We define the following: \hat{b} \equiv \sqrt{\frac{m \omega'}{2\hbar}} \left(\hat{x} + \frac{i}{m\omega'} \hat{p} \right) Choosing the same formula as the ladder operator but with different frequency ensures the algebra is the same, i.e. [\hat{b}, \hat{b}^\dagger] = 1. Now, to understand what these operators are, we can rewrite them in terms of the ladder operators: \hat{b} = \lambda \hat{a} + \nu \hat{a}^\dagger \\ \hat{b}^\dagger = \lambda \hat{a}^\dagger + \nu \hat{a} for \nu, \lambda real (they are functions of \omega and \omega' that you can easily work out; in this notation they satisfy \lambda^2 - \nu^2 = 1.) Now the question is, what are the eigenstates of this operator? Writing \hat{b} \ket{\beta} = \beta \ket{\beta}, we can solve just as we did for the coherent states by assuming \ket{\beta} = \sum_{n=0}^\infty b_n \ket{n}. I won’t give the explicit solution for these states, but it always exists for any \omega', and the resulting \ket{\beta} are known as squeezed states. Varying the parameter \beta lets us adjust to any \left\langle x \right\rangle and \left\langle p \right\rangle, while by construction we can also adjust for any position uncertainty, since (\Delta x)^2 = \frac{\hbar}{2m\omega'} now also contains a free parameter. Squeezed states have a number of practical applications, including to interferometers and atomic clocks, but we don’t have the background to really understand how these experiments work yet so I’ll just leave this as a remark and we’ll move on for now.