15 Simple Harmonic Oscillator
“The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.” – Sydney Coleman.
The simple harmonic oscillator potential is \[V(x) = \frac{1}{2}m\omega^2x^2\] where \(\omega\) is known as the “angular frequency”.
In the present chapter we will see how many quantum mechanical properties of a particle in a harmonic oscillator potential can be found exactly using very simple methods, without recourse to approximations.
15.1 Importance
The harmonic oscillator potential is an extremely important example in mathematical physics and quantum mechanics in particular:
It can be solved exactly!
It is universal: any physical system involving small fluctuations around equilibrium is described by a collection of simple harmonic oscillators. Concretely, we can expand a potential around a local minimum \(x = x_0\) as \[V(x) = V(x_0) + \frac{1}{2}V''(x_0)(x-x_0)^2 + \cdots \, .\] Provided the distance \(|x-x_0|\) remains small this is well approximated by a simple harmonic oscillator with \[m\omega^2 = \frac{1}{2} V''(x_0) \, .\]
It is prototypical: the techniques we will introduce to in solving the simple harmonic oscillator can be applied to a wide range of other problems.
15.2 Classical Solution
Hamilton’s equation for the simple harmonic oscillator are \[\begin{aligned} \dot x & = \frac{\partial H}{\partial p} = \frac{p}{m}\,, \\[1ex] \dot p & = -\frac{\partial H}{\partial x} = - m \omega^2 x\,, \end{aligned}\] with general solution \[\begin{aligned} x(t) & = \sqrt{\frac{2E}{m\omega^2}} \sin(\omega t+\delta ) \\[1ex] p(t) & = \sqrt{2mE} \cos(\omega t +\delta ) \, . \end{aligned}\] The conserved energy \(E \geq 0\) and phase \(\delta\) are determined by the initial conditions \((x(0), p(0))\). The motion oscillates between the points \(x=\pm \sqrt{\frac{2E}{m\omega^2}}\) where the kinetic energy vanishes and \(V(x) = E\).
15.3 Energy Spectrum - First Attempt
In quantum mechanics, the basic question is to determine the spectrum of eigenvalues and eigenfunctions of the Hamiltonian operator \[\hat H = \frac{\hat p^2}{2m}+V(x) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \, .\] There can gain some information on the spectrum from general theorems we proved in lecture \(\ref{sec:energy}\) about square-normalisable Hamiltonian eigenfunctions.
The Hamiltonian eigenvalues obey \(E > V_{\mathrm{min}}\) where \(V_{\mathrm{min}}\) is the minimum value of the potential. We immediately conclude that \(E>0\).
The spectrum is non-degenerate: there is one linearly independent eigenfunction for each energy \(E>0\).
Furthermore, our experience with confining potentials leads us to expect that the spectrum will be discrete.
To make further progress, we could try to construct the Hamiltonian eigenfunctions and eigenvalues using the following method.
Look for solutions to the differential equation \[-\frac{\hbar^2}{2m} \frac{\partial^2\phi}{\partial x^2} + \frac{1}{2}m\omega^2x^2\phi(x) = E\phi(x) \, .\]
Determine \(E>0\) such that there exist square-normalizable solutions \(\phi(x)\).
This is perfectly reasonable but unilluminating.
15.4 Ladder Operators
Instead, we follow a more powerful approach using “ladder operators”. We emphasise that this method is completely algebraic: we can find the spectrum of the Hamiltonian operator without solving any differential equations!
Recall that physical observables in quantum mechanics are represented by Hermitian operators obeying \(A = \hat{a}^\dagger\). For example, position \(\hat x\) and momentum \(\hat p\). Let us now introduce the ladder operators, \[\begin{aligned} \hat{a}& = \frac{1}{\sqrt{2\hbar m \omega}} \left( m \omega \hat x + i \hat p\right)\,,\\[1ex] \hat{a}^\dagger & = \frac{1}{\sqrt{2\hbar m \omega}} \left( m \omega \hat x - i \hat p\right) \, . \end{aligned}\] They are not Hermitian operators, rather \(\hat{a}^\dagger\) is the adjoint of \(a\).
Why did we introduce these strange combinations? The reason is it allows the Hamiltonian operator to be expressed in a particularly useful form. To see this, we first invert these relations to express position and momentum in terms of the ladder operators, \[\begin{aligned} \hat x & = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^\dagger)\,, \\[1ex] \hat p & = -i \sqrt{\frac{\hbar m \omega}{2}}(\hat{a}-\hat{a}^\dagger) \, . \end{aligned}\] We then write the Hamiltonian operator \[\begin{aligned} \hat H & = \frac{\hat p^2}{2m} + \frac{1}{2}m\omega^2 \hat x^2 \\ & = - \frac{1}{2m}\frac{\hbar m \omega}{2}(\hat{a}-\hat{a}^\dagger)^2 + \frac{1}{2}m\omega^2 \frac{\hbar}{2m\omega} (\hat{a}+\hat{a}^\dagger)^2 \\ & = \frac{\hbar\omega}{4} \left( (\hat{a}+\hat{a}^\dagger)^2 - (\hat{a}-\hat{a}^\dagger)^2 \right) \\ & = \frac{\hbar \omega}{2}\left( \hat{a}\hat{a}^\dagger + \hat{a}^\dagger \hat{a}\right) \, . \end{aligned}\] This form of the Hamiltonian operator will allow us to determine tge spectrum of eigenvalues \(E\). First, however, it is important to understand the relation between the combinations \(\hat{a}\hat{a}^\dagger\) and \(\hat{a}^\dagger \hat{a}\) appearing in the Hamiltonian. In other words, we need to compute the commutator \([\hat{a},\hat{a}^\dagger]\).
15.5 Commutators
Recall that the commutator is defined by \[{}[ A, B] : = AB-BA \, .\] Using the canonical commutator between position and momentum, \([x,p] = i\hbar\), we can compute the commutation relations between the ladder operators, \[\begin{aligned} \big[ \hat{a},\hat{a}^\dagger \big] & = \frac{1}{2\hbar m \omega} [ m\omega x+ip, m\omega x-ip] \\[1ex] & = \frac{1}{2\hbar}\left( -i [x,p]+i [p,x] \right) \\[1ex] & = 1 \, , \end{aligned}\] in addition to \([\hat{a},\hat{a}] = [\hat{a}^\dagger , \hat{a}^\dagger]=0\). This is known as a the Heisenberg algebra. We can use it to express the Hamiltonian operator in two more equivalent ways \[\hat{H} = \hbar\omega\left(\hat{a}^\dagger \hat{a}+\frac{1}{2} \right) = \hbar\omega\left( \hat{a}\hat{a}^\dagger -\frac{1}{2}\right) \, .\]
Finally, let us compute the commutators between the Hamiltonian and ladder operators using \([a,\hat{a}^\dagger] = 1\). We find \[\begin{aligned} \big[\hat{H},a\big] & = \hbar \omega [ \hat{a}^\dagger \hat{a} , \hat{a}] \\[1ex] & = \hbar \omega ( \hat{a}^\dagger [\hat{a},\hat{a}] + [\hat{a}^\dagger ,\hat{a}]\hat{a}) \\[1ex] & = - \hbar \omega \hat{a}\,, \end{aligned}\] and similarly \[\begin{aligned} \big[\hat{H},\hat{a}^\dagger\big] & = \hbar \omega [ \hat{a}^\dagger \hat{a} , \hat{a}^\dagger] \\[1ex] & = \hbar \omega ( \hat{a}^\dagger [\hat{a},\hat{a}^\dagger] + [\hat{a}^\dagger ,\hat{a}^\dagger]\hat{a}) \\[1ex] & = \hbar \omega \hat{a}^\dagger\,. \end{aligned}\] These equations are extremely important to determine the spectrum.
15.6 Energy Spectrum Revisited
Now, suppose \(\phi(x)\) is a Hamiltonian eigenfunction with energy \(E\). Then we can construct more eigenfunctions by acting with the ladder operators. For example, \[\begin{aligned} \hat{H} \hat{a}^\dagger \phi & = [\hat{H},\hat{a}^\dagger]\phi + \hat{a}^\dagger \hat{H}\phi \\[1ex] & = \hbar \omega \hat{a}^\dagger \phi + \hat{a}^\dagger E\phi \\[1ex] & = (E+\hbar\omega) \hat{a}^\dagger \phi\,, \end{aligned}\] and similarly, \[\begin{aligned} \hat{H} \hat{a} \phi & = [\hat{H},\hat{a}]\phi + \hat{a} \hat{H}\phi \\[1ex] & = -\hbar \omega \hat{a} \phi + \hat{a} E\phi \\[1ex] & = (E-\hbar\omega) \hat{a}\phi \, . \end{aligned}\] By induction, we find
\((\hat{a}^\dagger)^n\phi(x)\) is a Hamiltonian eigenfunction with energy \(E+n\hbar \omega\).
\(\hat{a}^n\phi(x)\) is a Hamiltonian eigenfunction with energy \(E-n\hbar \omega\),
For this reason, \(\hat{a}^\dagger\) and \(\hat{a}\) are sometimes called ‘creation’ and ‘annihilation’ operators respectively: they create and annihilate energy in units of \(\hbar \omega\).
However, recall that normalizable eigenfunctions must have \(E>0\). This means we cannot act indefinitely with \(\hat{a}\) because the eigenvalue \(E\) would eventually become negative. There must therefore exist an eigenfunction \(\phi_0(x)\) with the property that it is annihilated by \(\hat{a}\), \[\hat{a} \, \phi_0 = 0 \, .\] We have proven that square-normalizable Hamiltonian eigenfunctions are non-degenerate, so this wave function is unique up to normalisation.
Since the Hamiltonian can be expressed \[\hat{H} = \hbar\omega(\hat{a}^\dagger \hat{a}+\tfrac{1}{2}) \, ,\] we immediately find that \(\phi_0(x)\) has eigenvalue \(\tfrac{1}{2}\hbar\omega\). It is known as the “ground state” of the simple harmonic oscillator. We can then construct eigenfunctions \((\hat{a}^\dagger)^n \phi_0(x)\) with eigenvalues \(\hbar\omega(n+\tfrac{1}{2})\) where \(n>0\) by acting with the creation operator \(\hat{a}^\dagger\). They are known as “excited states”.
In summary, the spectrum of the Hamiltonian operator for the simple harmonic oscillator is \[E_n = \hbar\omega(n+\tfrac{1}{2}) \qquad n \in \mathbb{Z}_{\geq 0} \, .\] This is the key result of this lecture. We will construct the corresponding Hamiltonian eigenfunctions and study their properties more systematically next.
15.7 Normalisation
Let us first determine the normalisation of the wave functions \(\phi_n(x)\) systematically. We are going to proceed by induction on \(n\). We therefore first assume that the ground state wave function is correctly normalized, \(\langle \phi_0,\phi_0\rangle = 1\). We then introduce constants \(C_n\) such that \[\hat{a}^\dagger \phi_{n-1}(x) = C_{n} \phi_{n}(x)\] and therefore \[\begin{aligned} \phi_n(x) & = \frac{1}{C_n} \hat{a}^\dagger \phi_{n-1}(x) \\ & \qquad \qquad \vdots \\ & = \frac{1}{C_n\cdots C_1} (\hat{a}^\dagger)^n \phi_0(x) \, . \end{aligned}\] We want to determine \(C_n\) such that \(\phi_n(x)\) is normalised for all \(n>0\).
Let us assume that the \((n-1)\)-th wave function is correctly normalized, \(\langle \phi_{n-1},\phi_{n-1}\rangle = 1\), and compute the normalisation of the \(n\)-th wave function. Using the inner product notation, \[\begin{aligned} \langle \phi_{n} , \phi_{n}\rangle & = \langle \frac{1}{C_{n}}\hat{a}^\dagger \phi_{n-1} , \frac{1}{C_{n}}\hat{a}^\dagger \phi_{n-1}\rangle \\ & = \frac{1}{|C_{n}|^2} \langle \phi_{n-1} , \hat{a} \hat{a}^\dagger \phi_{n-1}\rangle \\ & = \frac{1}{|C_{n}|^2} \langle \phi_{n-1}, \left(\frac{\hat H}{\hbar\omega}+\frac{1}{2}\right) \phi_{n-1} \rangle \\ & = \frac{1}{|C_{n}|^2} n \langle \phi_{n-1},\phi_{n-1}\rangle \\ & = \frac{1}{|C_{n}|^2} n \, . \end{aligned}\] We can therefore choose \(C_{n} = \sqrt{n}\) giving the normalisation \[\phi_n(x) = \frac{1}{\sqrt{n!}} (\hat{a}^\dagger)^n \phi_0(x) \, .\]
The action of the ladder operators on the wave functions \(\phi_n\) is \[\begin{aligned} \hat{a}^\dagger \phi_n(x) & = \sqrt{n+1} \, \phi_{n+1}(x)\,, \\[1ex] \hat{a} \, \phi_n (x) & = \sqrt{n} \, \phi_{n-1}(x) \, . \label{eq:raise-lower} \end{aligned}\] The first equation is our definition of \(C_n\). The second equation follows from the first by acting with \(\hat{a}\), expressing the result in terms of the Hamiltonian operator \(\hat H = \hbar\omega(a\hat{a}^\dagger-\frac12)\), using the fact that \(\phi_n(x)\) is a Hamiltonian eigenfunction with energy \(E_n = \hbar \omega(n+\tfrac12)\) and finally shifting \(n \to n-1\).
Notice that \[\hat{a}^\dagger \hat{a} \, \phi_n(x) = n \phi_n(x) \, .\] For this reason, \(\hat{a}^\dagger a\) is sometimes called the “number operator”. In particular, we can recover the correct Hamiltonian eigenvalues \[\begin{aligned} \hat H \phi_n(x) & = \hbar\omega(\hat{a}^\dagger \hat{a}+\frac 12) \phi_n(x) \\[1ex] & = \hbar\omega(n+\frac12)\phi_x(x) \\[1ex] & = E_n\phi_n(x)\,, \end{aligned}\] as expected.
15.8 Orthogonality
In the lecture on Hermitian operators, we proved that square-normalisable eigenfunctions of a Hermitian operator with different eigenvalues are orthogonal. It is worth verifying this fact here using the ladder operators.
Without loss of generality, we will show that \(\langle \phi_{n+j},\phi_n\rangle = 0\) for all \(n\geq 0\) and \(j>0\). Since the result must vanish, we do not keep track of the normalisation. Therefore, \[\begin{aligned} \langle \phi_{n+j} , \phi_n \rangle & \propto \langle (\hat{a}^\dagger)^{n+j} \phi_0 , \phi_n \rangle \\[1ex] & = \langle \phi_0 , \hat{a}^{n+j} \phi_n \rangle \\[1ex] & \propto \langle \phi_0 , \hat{a}^j \phi_0 \rangle \\[1ex] & = 0\, , \end{aligned}\] since \(\hat{a} \, \phi_0(x) = 0\). In summary, we indeed find that \[\langle \phi_n,\phi_m\rangle = \delta_{nm} \label{eq:orthonormal}\] for any \(n,m\in \mathbb{Z}_{\geq0}\).
15.9 Expectation Values
We can now compute the expectation value of physical observables \(A(x,p)\) in any Hamiltonian eigenfunction. First, we express of position and momentum in terms of the ladder operators, \[\begin{aligned} \hat x & = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^\dagger)\,, \\[1ex] \hat p & = -i \sqrt{\frac{\hbar m \omega}{2}}(\hat{a}-\hat{a}^\dagger) \, . \end{aligned}\] Using this result, we can express \(A(x,p)\) in terms of the ladder operators \(\hat{a}\), \(\hat{a}^\dagger\). Finally, \(\langle A(x,p)\rangle\) can be computed using the known action of \(\hat{a}\), \(\hat{a}^\dagger\) (or the Hamiltonian \(\hat H\) if it is more convenient) on \(\phi_n(x)\) and the orthonormality of \(\phi_n(x)\).
For example, let us compute the expectation value of position in the \(n\)-th stationary wave function as follows \[\begin{aligned} \langle x \rangle & = \sqrt{\frac{\hbar}{2m\omega}} \langle \phi_n,(\hat{a}+\hat{a}^\dagger) \phi_n \rangle \\ \nonumber & = \sqrt{\frac{\hbar}{2m\omega}} \langle \phi_n, \sqrt{n} \, \phi_{n-1} +\sqrt{n+1} \, \phi_{n+1} \rangle \\ & = \sqrt{\frac{\hbar}{2m\omega}} ( \sqrt{n} \, \delta_{n,n-1} +\sqrt{n+1} \, \delta_{n,n+1}) \\ & = 0 \, . \end{aligned}\] The expectation value of position squared is computed similarly, \[\begin{aligned} \langle x^2 \rangle & = \frac{\hbar}{2m\omega} \langle \phi_n,(\hat{a}+\hat{a}^\dagger)^2 \phi_n \rangle \\ \nonumber & = \frac{\hbar}{2m\omega} \langle \phi_n,(\hat{a}^2+a^{\dagger 2} + \hat{a}\hat{a}^\dagger+\hat{a}^\dagger \hat{a}) \phi_n \rangle \\ & = \frac{\hbar}{2m\omega} \langle \phi_n,(\hat{a}^2+a^{\dagger 2}+ 2\hat{a}^\dagger \hat{a}+1) \phi_n \rangle \\ & = \frac{\hbar}{2m\omega} ( \sqrt{n(n-1)}\delta_{n,n-2}+\sqrt{(n+1)(n+2)}\delta_{n,n+2}+ (2n+1)\delta_{n,n} ) \\ & = \frac{\hbar}{2m\omega} (2n+1) \\ & = \frac{\hbar}{m\omega}\left(n+\frac1 2\right) \, . \end{aligned}\] By similar computations, the expectation values of momentum and momentum squared are \[\begin{aligned} \langle p\rangle =0\,, \qquad \langle p^2 \rangle = \hbar m \omega\left(n+\frac1 2\right) \, . \end{aligned}\] Note the following points:
We have \(\Delta x \Delta p = \hbar(n+\frac{1}{2})\), compatible with Heisenberg’s uncertainty principle. Furthermore, Heisenberg’s uncertainty principle is saturated for \(n=0\). Indeed, we show below that \(\phi_0(x)\) is a Gaussian.
As a consistency check, we can reproduce the expectation value of the Hamiltonian, \[\begin{aligned} \langle H\rangle & = \frac{1}{2m}\langle p^2\rangle + \frac{1}{2}m\omega^2 \langle x^2\rangle \\ & = \frac{1}{2m} \hbar m \omega \left(n+\frac1 2\right) +\frac{1}{2}m\omega^2 \frac{\hbar}{m\omega} \left(n+\frac1 2\right)\\ & = \hbar \omega\left(n+\frac12\right) \\ & = E_n \, . \end{aligned}\]
15.10 Wave functions
Everything in the simple harmonic oscillator can be computed using the properties of \(\hat{a}\), \(\hat{a}^\dagger\), \(\hat{H}\) and how they act on the wave functions \(\phi_n(x)\). That said, we could not finish without briefly exploring what these wave functions look like!
First, using the momentum operator \(\hat p = -i\hbar\partial_x\), the ladder operators become differential operators acting on wave functions \[\begin{aligned} a & = \frac{1}{\sqrt{2\hbar m \omega}} \left( m \omega x + \hbar \frac{\partial}{\partial x} \right)\,,\\[1ex] \hat{a}^\dagger & = \frac{1}{\sqrt{2\hbar m \omega}} \left( m \omega x -\hbar \frac{\partial}{\partial x}\right) \, . \end{aligned}\] The defining equation \(\hat{a}\, \psi_0(x) = 0\) for the ground state wave function becomes a differential equation, \[\left( m \omega x + \hbar \frac{\partial}{\partial x} \right) \phi_0(x) = 0 \, ,\] whose normalised solution is the Gaussian \[\phi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x^2/2\hbar} \, .\] This explains the observation that the ground state wave function saturates Heisenberg’s uncertainty principle.
The normalized exited state wave functions are then given by \[\phi_n(x) = \frac{1}{\sqrt{n!}} \frac{1}{(2\hbar m \omega)^{n/2}} \left( m \omega x -\hbar \frac{\partial}{\partial x}\right)^n\phi_0(x) \, .\] They take the form of a polynomial in \(x\), known as a “Hermite polynomial”, multiplying a Gaussian wave function. Some comments:
The wave functions are even or odd: \(\phi(-x) = (-1)^n\phi(x)\). This explains why \(\langle x\rangle = 0\).
The wave functions are real. This explains why \(\langle p\rangle = 0\).
15.11 Problems
- Virial theorem:
(Do this problem only after you have read the last chapter, on Ehrenfest’s theorem).
Consider the Hamiltonian \(H = T +V\) where \(T = \frac{p^2}{2m}\) is the kinetic energy and \(V(x)\) is the potential energy.
Use Ehrenfest’s theorem to show that \(\partial_t\langle x p\rangle = 2 \langle T \rangle - \langle x\partial_xV\rangle\).
Now consider a stationary solution of Schrödinger’s equation in the simple harmonic oscillator with potential \[V(x) = \frac{1}{2}m\omega^2 x^2 \, .\] Show that \[\langle T\rangle = \langle V\rangle \, .\]
Solution ▶
Ehrenfest’s theorem is \[\partial_t \langle A \rangle = \frac{i}{\hbar}\langle [H,A]\rangle\] for any Hermitian operator \(A\). We apply this theorem to \(A = xp\). We therefore compute the commutator on the right, \[\begin{aligned} \left[H,xp\right] & = x[H,p] + [H,x]p \\ \nonumber & = x[V(x),p] + [\frac{p^2}{2m},x]p \\ & = i\hbar \left( x \partial_xV(x) - \frac{p^2}{m} \right) \, , \end{aligned}\] where we used the fundamental commutator \([x,p] = i\hbar\). We therefore find \[\partial_t \langle xp\rangle =2\langle T\rangle - \langle x \partial_xV\rangle\] where \(T = p^2/2m\) is the kinetic energy.
First, for the simple harmonic oscillator potential, \(x\partial_x V(x) = 2V(x)\). Second, for a stationary solution of Schrödinger’s equation, the expectation value of any observable is independent of time. Combining with part (a), we find \(\langle T\rangle = \langle V\rangle\).
- Ground state of the harmonic oscillator:
Consider the initial wavefunction \[\psi(x,0) = C \exp\left( -\frac{m \omega x^2}{2\hbar} \right) \, .\] In this question, there is no need to determine \(C\)!
Show that \(\psi(x,0)\) is an eigenfunction of a Hamiltonian of the form \[\hat H = \frac{\hat p^2}{2m}+ \frac{1}{2} k \hat x^2\] and determine the constant \(k\).
What is the corresponding energy eigenvalue?
Hence write down the wavefunction \(\psi(x,t)\) for \(t>0\).
Solution ▶
Compute the action of the Hamiltonian operator on the initial wavefunction, \[\begin{aligned} \nonumber \hat H \cdot \psi(x,0) & = \left( -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + \frac{1}{2}kx^2 \right) \psi(x,0) \\ & = -C \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} e^{ -\frac{m \omega x^2}{2\hbar}} + \frac{1}{2}kx^2\psi(x,0) \\ %& = C \frac{\hbar\omega}{2}\frac{\partial}{\partial x} \left(xe^{ -\frac{m \omega x^2}{2\hbar}} \right) + \frac{1}{2}kx^2\psi(x,0)\\ & = C \frac{\hbar\omega}{2} \left(e^{ -\frac{m \omega x^2}{2\hbar}} - \frac{m\omega x^2}{\hbar} e^{ -\frac{m \omega x^2}{2\hbar}} \right) + \frac{1}{2}kx^2\psi(x,0) \\ & = \frac{\hbar \omega}{2} \psi(x,0) +\left(k- \frac{1}{2}m\omega^2\right)x^2 \psi(x,0) \, . \end{aligned}\] For the initial wavefunction to be a Hamiltonian eigenfunction, this must equal \(E \, \psi(x,0)\) where \(E\) is a real constant that is independent of \(x\). We therefore require \(k = m \omega^2\) to cancel the term proportional to \(x^2\psi(x,0)\).
The energy is therefore \(E = \hbar\omega / 2\).
The wavefunction at later times is \[\begin{aligned} \psi(x,t) & = C \exp\left( -\frac{m \omega x^2}{2\hbar} \right) \exp\left( - \frac{i E t}{\hbar} \right) \\ \nonumber & = C \exp\left( -\frac{m \omega x^2}{2\hbar} \right) \exp\left( - \frac{i \omega t}{2}\right) \, . \end{aligned}\] The constant \(\omega\) has units of inverse time and specifies the frequency of the harmonic oscillator.
- PROBLEMS CLASS 4The coherent state:
Jupyter notebook (on Google Colab) to explore the time-dependence of coherent states of the simple harmonic oscillator.Consider a normalized wave function obeying \[\hat{a} \psi(x,t) = \alpha_0e^{-i\omega t} \psi(x,t) \, .\] where \[\hat{a} = \frac{1}{\sqrt{2m\hbar\omega}}(m\omega \hat x + i \hat p)\] is the annihilation operator and \(\alpha_0 \in \mathbb{R}\) is a real constant.
Show that \(\langle H\rangle = \hbar\omega (\alpha_0^2+\tfrac{1}{2})\).
Show that \[\langle x \rangle = \sqrt{\frac{2\hbar}{m\omega}} \, \alpha_0 \cos(\omega t) \qquad \langle p \rangle = - \sqrt{2m\hbar\omega} \alpha_0 \sin (\omega t) \, .\]
Show that the above result is a solution of Hamilton’s equations.
How does the classical energy of the solution of Hamilton’s equations compare to \(\langle H\rangle\)?
Hint (a): use the form of the hamiltonian \(H = \hbar\omega(a^\dagger a+\tfrac{1}{2})\).
Hint (b): first express \(x\), \(p\) in terms of \(a\), \(a^\dagger\).
General hint: remember the definition of the adjoint \(\langle \psi_1,a^\dagger \psi_2\rangle = \langle a \psi_1,\psi_2\rangle\).
Solution ▶
The Hamiltonian can be expressed \(H = \hbar\omega(a^\dagger a+\frac12)\). The expectation value of the hamiltonian is therefore \[\langle H \rangle = \hbar \omega \langle a^\dagger a+\frac12 \rangle \, .\] We evaluate the two terms in this expression as follows:
First, \(\langle 1\rangle := \langle \psi,\psi\rangle = 1\) for a normalized wave function.
Second, using the definition of the adjoint, \[\langle a^\dagger a \rangle := \langle \psi ,a^\dagger a \psi \rangle = \langle a\psi , a\psi\rangle = \langle \alpha_0e^{-i\omega t}\psi , \alpha_0e^{-i\omega t}\psi\rangle = \alpha_0^2 \langle \psi,\psi\rangle = \alpha_0^2\] where in the final step we again used that the wave function is normalized.
We conclude that \(\langle H\rangle = \hbar\omega(\alpha_0^2+\frac 12)\).
Let us first compute the expectation values of the ladder operators \[\begin{aligned} \langle a\rangle & = \langle \psi,a\psi\rangle =\langle \psi, \alpha_0e^{-i\omega t} \psi\rangle = \alpha_0e^{-i\omega t} \nonumber \\ \langle a^\dagger \rangle & = \langle \psi,a^\dagger\psi\rangle =\langle a\psi,\psi\rangle =\langle \alpha_0e^{-i\omega t}\psi, \psi\rangle = \alpha_0e^{+i\omega t} \, . \end{aligned}\] We now express position and momentum in terms of the ladder operators \[\begin{aligned} x & = \sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger) \\ \nonumber p & = -i \sqrt{\frac{\hbar m \omega}{2}}(a-a^\dagger) \, . \end{aligned}\] The expectation values of position and momentum are then \[\begin{aligned} \langle x \rangle & = \sqrt{\frac{\hbar}{2m\omega}}\alpha_0 (e^{-i\omega t} + e^{+i\omega t}) = \sqrt{\frac{2\hbar\alpha_0^2}{m\omega}} \cos(\omega t) \\ \nonumber \langle p \rangle & = -i \sqrt{\frac{\hbar m \omega}{2}} (e^{-i\omega t} - e^{+i\omega t}) = - \sqrt{2 m \hbar \omega \alpha_0^2} \sin(\omega t) \, . \end{aligned}\]
This is a solution of Hamilton’s equations with classical energy \(E = \hbar \omega \alpha_0^2 \geq 0\).
If \(E \geq 0\) denotes the classical energy of the corresponding solution of Hamilton’s equations then \(\langle H\rangle = E+ \frac{1}{2}\hbar\omega\). This is consistent with the fact that \(\langle H\rangle >0\) for normalizable wave functions in quantum mechanics. Note that in the limit \(\alpha_0 \to0\) we recover the ground state of the simple harmonic oscillator with ‘zero-point’ energy \(\frac{1}{2}\hbar\omega\).
- TUTORIAL 4Properties of Hamiltonian eigenfunctions The ladder operators are defined by \[\begin{aligned} a & = \frac{1}{\sqrt{2m\hbar\omega}}(m\omega \hat x + i \hat p) \\ \nonumber a^\dagger & = \frac{1}{\sqrt{2m\hbar\omega}}(m\omega \hat x - i \hat p) \, . \end{aligned}\]
Using the canonical commutator \([\hat x, \hat p ] = i \hbar\), show that \([a,a^\dagger]=1\).
What property does the ground state \(\phi_0(x)\) obey?
Write down an expression for the excited wave functions \(\phi_n(x)\) in terms of creation operators acting on \(\phi_0(x)\).
Compute the expectation values \(\langle x\rangle\), \(\langle x^2\rangle\) and the uncertainty \(\Delta x\) for \(\phi_n(x)\).
Compute the expectation values \(\langle p\rangle\), \(\langle p^2\rangle\) and the uncertainty \(\Delta p\) for \(\phi_n(x)\).
Check consistency with Heisenberg’s uncertainty principle.
Check that \(\langle T \rangle = \langle V\rangle\), where \(T\) is the kinetic energy.
Solution ▶
Using the canonical commutation relation \([x,p] = i\hbar\), \[\big[ a ,a^\dagger \big] = \frac{1}{2\hbar m \omega} [ m\omega x+ip, m\omega x-ip] = \frac{1}{2\hbar}\left( -i [x,p]+i [p,x] \right) = 1 \, .\]
The ground state wave function obeys \(a \cdot \phi_0(x) = 0\).
The Hamiltonian eigenfunctions are \[\phi_n(x) := \frac{1}{\sqrt{n!}} (a^\dagger)^n\phi_0(x) \,.\] They obey the following important properties
Orthonormality: \(\langle \phi_n,\phi\rangle_m =\delta_{nm}\)
Annihilation: \(a \phi_n = \sqrt{n} \phi_{n-1}\)
Creation: \(a^\dagger \phi_n = \sqrt{n+1} \phi_{n+1}\)
which are used freely below.
We compute the expectation value of \(x\) using ladder operators as follows, \[\begin{aligned} \langle x \rangle & = \sqrt{\frac{\hbar}{2m\omega}} \langle \phi_n,(a+a^\dagger) \phi_n \rangle \\ \nonumber & = \sqrt{\frac{\hbar}{2m\omega}} \langle \phi_n, \sqrt{n} \phi_{n-1} +\sqrt{n+1}\phi_{n+1} \rangle \\ & = \sqrt{\frac{\hbar}{2m\omega}} ( \sqrt{n} \delta_{n,n-1} +\sqrt{n+1} \delta_{n,n+1}) \\ & = 0 \, . \end{aligned}\] The expectation value of \(x^2\) is computed similarly, \[\begin{aligned} \langle x^2 \rangle & = \frac{\hbar}{2m\omega} \langle \phi_n,(a+a^\dagger)^2 \phi_n \rangle \\ \nonumber & = \frac{\hbar}{2m\omega} \langle \phi_n,(a^2+a^{\dagger 2} + aa^\dagger+a^\dagger a) \phi_n \rangle \\ & = \frac{\hbar}{2m\omega} \langle \phi_n,(a^2+a^{\dagger 2}+ 2a^\dagger a+1) \phi_n \rangle \\ & = \frac{\hbar}{2m\omega} ( \sqrt{n(n-1)}\delta_{n,n-2}+\sqrt{(n+1)(n+1)}\delta_{n,n+2}+ (2n+1)\delta_{n,n} ) \\ & = \frac{\hbar}{2m\omega} (2n+1) \\ & = \frac{\hbar}{m\omega}(n+\frac1 2) \, . \end{aligned}\] The uncertainty is therefore \[\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \sqrt{\frac{\hbar}{m\omega}(n+\frac1 2)} \, .\] Note that the hamiltonian eigenfunctions are even or odd functions of \(x\), \(\phi_n(-x) = (-1)^n\phi_n(x)\). We therefore have \(|\phi_n(-x)|^2 = |\phi_n(x)|^2\) and should have expected \(\langle x \rangle = 0\).
We compute the expectation values of \(p\) and \(p^2\) in the same way, \[\begin{aligned} \langle p \rangle & =-i \sqrt{\frac{\hbar m \omega}{2}}\langle \phi_n,(a-a^\dagger) \phi_n \rangle \\ \nonumber & = -i \sqrt{\frac{\hbar m \omega}{2}} \langle \phi_n, \sqrt{n} \phi_{n-1} - \sqrt{n+1}\phi_{n+1} \rangle \\ & = -i \sqrt{\frac{\hbar m \omega}{2}} ( \sqrt{n} \delta_{n,n-1} -\sqrt{n+1} \delta_{n,n+1}) \\ & = 0 \end{aligned}\] \[\begin{aligned} \langle p^2 \rangle & = -\frac{\hbar m \omega}{2} \langle \phi_n,(a-a^\dagger)^2 \phi_n \rangle \\ \nonumber & = \frac{\hbar m \omega}{2} \langle \phi_n,(-a^2-a^{\dagger 2} + aa^\dagger+a^\dagger a) \phi_n \rangle \\ & = \frac{\hbar m \omega}{2} \langle \phi_n,(-a^2-a^{\dagger 2}+ 2a^\dagger a+1) \phi_n \rangle \\ & = \frac{\hbar m \omega}{2} ( -\sqrt{n(n-1)}\delta_{n,n-2}-\sqrt{(n+1)(n+1)}\delta_{n,n+2}+ (2n+1)\delta_{n,n} ) \\ & = \frac{\hbar m \omega}{2} (2n+1) \\ & = \hbar m \omega(n+\frac1 2) \, . \end{aligned}\] The uncertainty is therefore \[\Delta p = \sqrt{\langle p^2\rangle - \langle p\rangle^2} = \sqrt{\hbar m \omega(n+\frac1 2)} \, .\] Note that the hamiltonian eigenfunctions \(\phi_n(x)\) are all real functions and therefore we should have expected \(\langle p\rangle = 0\).
The product of position and momentum uncertainties is \[\Delta x \Delta p = \hbar (n+\frac12) \geq \frac{\hbar}{2}\] for \(n\geq 0\). Note that the ground state wave function \(\phi_0(x)\) saturates the Heisenberg uncertainty principle - it is a Gaussian wave function.
The expectation values of kinetic and potential energy are \[\begin{aligned} \langle T \rangle = \frac{1}{2m} \langle p^2\rangle = \frac{\hbar\omega}{2}(n+\frac12) \\ \nonumber \langle V \rangle = \frac{1}{2}m\omega^2 \langle x^2\rangle = \frac{\hbar\omega}{2}(n+\frac12) \end{aligned}\] and therefore \(\langle T\rangle = \langle V\rangle\) in agreement with the quantum mechanical Virial theorem.
- TUTORIAL 4Instantaneous shift of frequency For \(t<0\), a particle is in the ground state of a simple harmonic oscillator of frequency \(\omega\) with stationary wave function \[\psi(x,t) = e^{-i t \omega/2} \phi_0(x) \, .\]
At \(t=0\) the frequency suddenly doubles to \(\omega' = 2 \omega\) leaving the wave function momentarily unchanged.
Explain why a measurement at \(t<0\) yields energy \(\tfrac{1}{2}\hbar \omega\) with probability \(1\).
Explain why the probability of measuring energy \(\tfrac{1}{2} \hbar \omega\) at \(t>0\) is zero.
What is the probability of measuring energy \(\hbar \omega\) just after \(t=0\)?
Solution ▶ The hamiltonian eigenvalues before are \(E_n =\hbar\omega(n+\frac12)\) and after are \(E' = \hbar\omega'(n+\frac12) = \hbar\omega (2n+1)\).
For \(t<0\), we have a stationary wave function for the ground state with energy \(E_0 = \frac{1}{2}\hbar\omega\).
For \(t>0\), the possible outcomes of energy measurements are \(E'_n = \hbar\omega(2n+1)\) with \(n\geq 0\). The probability to measure \(\frac{1}{2}\hbar\omega\) at \(t>0\) is therefore zero.
Note that \(E_0' = \frac{1}{2}\hbar\omega' = \hbar\omega\) is the energy of the ground state wave function for \(t>0\). Since the wave function is momentarily unchanged, the probability of measuring this energy just after \(t=0\) is \[P = | \langle \phi_0 , \phi_0'\rangle|^2\] where \[\begin{aligned} \nonumber \phi_0(x) & = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x^2/2\hbar} \\ \phi_0'(x) & = \left(\frac{m\omega'}{\pi\hbar}\right)^{1/4} e^{-m\omega' x^2/2\hbar} \, . \end{aligned}\] are the ground state eigenfunctions functions before and after. We find \[\begin{aligned} \nonumber \langle \phi_0 , \phi_0'\rangle & = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(\frac{m\omega'}{\pi\hbar}\right)^{1/4}\int^\infty_{-\infty} dx e^{-m\omega x^2/2\hbar} e^{-m\omega' x^2/2\hbar} \\ & = \left(\frac{m}{\pi\hbar}\right)^{1/2}(\omega\omega')^{1/4} \int^\infty_{-\infty} dx e^{-m(\omega+\omega') x^2/2\hbar} \\ & = \left(\frac{m}{\pi\hbar}\right)^{1/2}(\omega\omega')^{1/4} \left(\frac{2\hbar}{m(\omega+\omega')} \right)^{1/2} \int^\infty_{-\infty} dy e^{-y^2} \\ & = \left(\frac{m}{\pi\hbar}\right)^{1/2}(\omega\omega')^{1/4} \left(\frac{2\hbar}{m(\omega+\omega')} \right)^{1/2} \sqrt{\pi} \\ & = \left[ \frac{4\omega\omega'}{(\omega+\omega')^2} \right]^{1/4} \, . \end{aligned}\] The probability is therefore \[P = \left[ \frac{4\omega\omega'}{(\omega+\omega')^2} \right]^{1/2} = \frac{2\sqrt{\omega\omega'}}{\omega+\omega'} = \frac{2\sqrt{2}}{3} \sim 0.943 \, .\]
