Coherent States

Part One of Two!

A Quick Review of the Algebra

Let’s begin with the now familiar Dirac basis for the one-dimensional Harmonic oscillator. The algebra can be packaged by a single, nontrivial commutator:

\([a,a^{\dagger}] = 1.\)

The irreducible representation we’ve studied for this algebra is given by the set of complex polynomials in the variable a†. That is, the vector space upon which our algebra acts¹ is

\(H = \mathbb{C}[a^{\dagger}].\)

We’ve seen that physical states of fixed energy amount to the monomials in a†. In particular, the energy operator,

\(E = \hbar\omega\left(a^{\dagger}a + \frac{1}{2}\right).\)

You might recall that we normalized our basis of energy eigenvectors as,

\(| n \rangle = \frac{1}{\sqrt{n!}}a^{\dagger n} | \varnothing \rangle,\quad E_{n} = \hbar\omega\left(n + \frac{1}{2}\right).\)

This vector space of physical states is infinite-dimensional, but graded by the natural numbers, which means we can write down a linear decomposition into one-dimensional subspaces,

\(H = \bigoplus_{n=0}^{\infty} \,\mathsf{span}_{\mathbb{C}}\{a^{\dagger n}\}.\)

The number operator N = a†a reveals the degree of each subspace, which we can interpret as saying that the monomials are eigenvectors of the operator N whose eigenvalues are given by their degree.

The Dirac basis is akin to holomorphic coordinates on the phase space. They are related to the usual symplectic coordinates as one would taking real and imaginary parts,

\(q = \left(\frac{\hbar}{2m\omega}\right)^{1/2} \left(a + a^{\dagger}\right),\quad p = -i\left(\frac{m \hbar\omega }{2}\right)^{1/2} \left(a - a^{\dagger}\right).\)

Wave Packets and Coherence

What makes Quantum Mechanics novel is the vector space of physical states. Any linear combination of physically allowed states is physically allowed².

In the free particle model, we’ve seen that this allows us to prepare a physical state in a wave packet that spans across different momenta. This allows us to create a physically reasonable model, encoding³ our ignorance about the particle’s momentum with a Gaussian parameter σ.

\(\Psi(x,t) = N(\sigma) \int_{\mathbb{R}} dk\; \exp(ikx - iE_{k}t)\exp\left(-\frac{k^{2}}{2\sigma^{2}}\right).\)

For the case of a massive free particle, we saw that this wave packet spreads over time, so that initial uncertain in momentum - and hence position - grows. The information we know about the particle decreases. The reason for this was easy enough to understand, particles at different momenta travel faster. In the language of plane waves, we say that there is a nonlinear dispersion relationship between energy and momentum.

For a massless particle, such as a photon, we saw that wave packets do not spread. Another way to say that is light is coherent⁴.

\(\Psi(x,t) = \frac{1}{\zeta \sqrt{2\pi}}\exp\left(-\frac{1}{2\zeta^{2}}(x-ct)^{2}\right),\)

By comparison, the modes that comprise the wave packet of a massive particle are incoherent

Given that the harmonic oscillator was used a model for individual frequencies of light, you might wonder whether wave packets of harmonic oscillator states spread.

Wave Packets are Complicated!

The free particle Hilbert space was continuous. Fortunately, H is discrete. This means that a generic wave packet can be written as a sum over the basis states,

\(| c(t) \rangle = \sum_{n=0}^{\infty} c_{n} \, e^{-iE_{n}t/\hbar} \,|n\rangle.\)

Here we have included the full time dependence, anticipating the question about wave packet spreading.

Just as the Gaussian wave packet was normalized, this one should be too, so that

\(\langle c (t) | c(t) \rangle = 1.\)

In particular, this must be true at t = 0. Hence we have

\(\sum_{n=0}^{\infty} |c_{n}|^{2} = 1.\)

Warm Up: Computing Uncertainties

You’ll want to participate in these calculations to make sure you follow what’s going on.

Our basis of eigenstate is orthonormal. So computing the uncertain in position is straightforward,

\(\sigma^{2}_{x}(n) = \langle n | x^{2} | n \rangle - \langle n | x | n \rangle^{2}.\)

The second term vanishes by orthogonality,

\(\langle n | x | n \rangle = \sqrt{\frac{\hbar}{2m\omega}} \langle n | a + a^{\dagger} |n\rangle = 0.\)

While the first term has only two non vanishing terms,

\(\langle n | x^{2} | n \rangle = \frac{\hbar}{2m\omega}\langle n | aa^{\dagger} + a^{\dagger}a |n\rangle = \frac{\hbar}{2m\omega}(2n+1).\)

Hence the overall uncertainty in position is

\(\sigma_{x} = \sqrt{(2n+1)\hbar/2m\omega}.\)

The momentum computation is virtually identical, yielding,

\(\sigma_{p} = \sqrt{m(2n+1)\hbar\omega/2}.\)

Hence we have the product,

\(\sigma_{x}\sigma_{p} = (2n+1)\frac{\hbar}{2}.\)

This is clearly consistent with the Heisenberg uncertainty principle, which reads

\(\sigma_{x}\sigma_{p} \geq \frac{\hbar}{2}.\)

This lower bound on the uncertain of symplectic pairs of operators is a direct application of the Heisenberg algebra to the inner product on the Hilbert space of states H. In particular, it follows from the Cauchy-Schwarz inequality.

Hence, the uncertainties of the individual energy eigenstates of the quantum harmonic oscillator do not vary with time. They merely vary with the energy by way of the degree of the state, n.

The General Computation

We will save the full details of this computation for your own enjoyment. But, we’ll get you started.

Exercise 1: Verify that,

\(\langle c(t) | x | c(t) \rangle = \sqrt{\frac{\hbar}{2m\omega}} \sum_{n=0}^{\infty} \sqrt{n+1}\left( \overline{c}_{n+1}c_{n}e^{i\omega t} + \overline{c}_{n}c_{n+1}e^{-i\omega t}\right).\)

Exercise 2: Verify that,

\(\langle c(t) | x^{2} | c(t) \rangle = \frac{\hbar}{2m\omega} \sum_{n=0}^{\infty} \left\{ (2n+1) |c_{n}|^{2} + \sqrt{(n+1)(n+2)}\left( \overline{c}_{n+2}c_{n}e^{2i\omega t} + \overline{c}_{n}c_{n+2}e^{-2i\omega t}\right) \right\}.\)

Having computed these results, hopefully you see the difference between finding the uncertainty in a wave packet versus finding the uncertain in basis element. There are cross terms! There are infinitely many cross terms!

You’re welcome to try to find the generic uncertainty there, but as you can see from the time dependence, it’s generically not going to be coherent.

It’s very close to coherent, however. There are certainly going to be coherent parts to the uncertainty. Put differently, if we could find the right relationship between the complex coefficients in our wave packet, we could get the incoherent parts to cancel out exactly. For example we’ll need all terms with coefficients like

\((\overline{c_{n+1}}c_{n})^{2}\overline{c_{n}}c_{n+2}\)

to conspire in just the right way.

Coherence from Simple Algebra

Rather than expanding and finding an entire recurrence relationship between all those terms, let’s ask a simpler question. What if we choose coefficents⁵ so that

\(a | c(t) \rangle = \lambda | c(t) \rangle.\)

In this case we would also have the action of a† the dual space H*

\(\langle c(t) | a^{\dagger} = \langle c(t) | \overline{\lambda}.\)

In this case, we can solve our problem algebraically!

In particular,

\(\langle c(t) | x | c(t) \rangle = \sqrt{\frac{\hbar}{2m\omega}} \left(\lambda + \bar{\lambda}\right),\)

and

\(\langle c(t) | x^{2} | c(t) \rangle = \frac{\hbar}{2m\omega} \left(\bar{\lambda}^{2} + \lambda^{2} + 2|\lambda|^{2} + 1\right).\)

In this case we see that

\(\sigma^{2}_{x}(c) = \langle c(t) | x^{2} | c(t) \rangle - \langle c(t) | x | c(t) \rangle^{2} = \frac{\hbar}{2m\omega}.\)

A similar computation for momentum gives⁶,

\(\sigma^{2}_{p}(c) = \frac{m\hbar\omega}{2}.\)

Hence, this algebraically nice state - assuming it exists - also saturates the lower bound of the uncertainty relation,

\(\sigma_{x}(c)\sigma_{p}(c) = \frac{\hbar}{2}.\)

In particular, such a state is manifestly coherent. Its uncertainty does not evolve in time. Let’s attempt to build one.

Building a Coherent State

We are looking for some coefficients c_n such that

\(a | c(t) \rangle = \lambda | c(t) \rangle.\)

Let’s expand that state and see what we find.

\(a | c(t) \rangle = \sum_{n=0}^{\infty} e^{-iE_{n}t/\hbar}c_{n} a | n\rangle = e^{-\frac{1}{2}i\omega t}\sum_{n=0}^{\infty} \frac{c_{n}}{\sqrt{n!}}e^{-in\omega t} a\, a^{\dagger n}| \varnothing \rangle.\)

You might recall that

\([a,a^{\dagger n}] = n a^{\dagger n-1}.\)

Hence

\(a | c(t) \rangle = e^{-\frac{1}{2}i\omega t}\sum_{n=0}^{\infty} \frac{c_{n} n }{\sqrt{n!}}e^{-in\omega t} a^{\dagger n-1}| \varnothing \rangle.\)

Shifting the index by 1,

\(a | c(t) \rangle = e^{-\frac{3}{2}i\omega t}\sum_{n=0}^{\infty} \frac{c_{n+1}\sqrt{n+1} }{\sqrt{n!}}e^{-in\omega t} a^{\dagger n}| \varnothing \rangle.\)

So if we can set

\(c_{n+1} = \frac{\lambda e^{i\omega t}}{\sqrt{n+1}}c_{n}\)

we’d be done.

This is easily solved. Let λ_0 be a complex number.

\(c_{n} = \frac{\lambda_{0}^{n}}{\sqrt{n!}}\)

Then we have

\(| c(t) \rangle = \sum_{n=0}^{\infty} e^{-i E_{n} t/ \hbar}\frac{\lambda_{0}^{n}}{n!}a^{\dagger n} |\varnothing\rangle = e^{i\omega t/2}\sum_{n=0}^{\infty} \frac{\left( a^{\dagger}\lambda_{0}e^{-i\omega t} \right)^{n}}{n!} |\varnothing\rangle \)

The state c(t) essentially amounts to an exponential.

Now one can repeat the calculation above to find that

\(a | c(t) \rangle = \lambda_{0}e^{-i\omega t} | c(t)\rangle.\)

In other words,

\(\lambda = \lambda_{0}e^{-i\omega t}.\)

But we need to make sure this state is normalized, or at least normalizable. That is the trouble with infinite sums: the need to converge.

\(\langle c(t) | c(t) \rangle = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{\bar{\lambda}_{0}^{m}\lambda_{0}^{n}}{\sqrt{m!n!}} \langle m | n \rangle = \sum_{n=0}^{\infty}\frac{|\lambda_{0}|^{2n}}{n!}.\)

Demanding that this state is normalized amounts to the constraint

\(e^{|\lambda_{0}|^{2}} = 1.\)

Which naïvely would set λ_0 to zero. Meaning all of this work was for nothing. It’s appears as if it is the logarithmic version of ending with all the terms canceling to no avail.

But take heart! We have a way out. Suppose we simply normalize c(t) as follows,

\(| c(t) \rangle = e^{-\frac{1}{2}|\lambda_{0}|^{2}}\sum_{n=0}^{\infty} e^{-iE_{n}t/\hbar}\frac{\lambda^{n}_{0}}{n!}a^{\dagger n} |\varnothing\rangle.\)

This solves our problem immediately! And there are two benefits. First, we get a discretized version of the Gaussian wave packet we saw employed for photons. It’s always nice to see parallel structure.

Second, and better still, we’ve found an entire family of coherent state solutions that minimize uncertainty, specified by the complex number λ.

Interpreting the Coherent State

Aside from a combination of all possible energy states, what exactly does this state represent?

A basic physical question might be, what’s the overlap between the known basis functions and this coherent state? Mathematically we ask this by taking the inner product of c(t) with some fixed state n.

Hence, per the rules of quantum mechanics, the probability of measuring the state c(t) in the energy eigenstate n is:

\(P(n) = | \langle n | c(t) \rangle |^{2}.\)

Let’s compute it.

\(\langle n | c(t) \rangle = e^{-\frac{1}{2}|\lambda_{0}|^{2}}\sum_{k=0}^{\infty} \langle n | \frac{\lambda_{0}^{k}}{\sqrt{k!}}|k\rangle = e^{-\frac{1}{2}|\lambda_{0}|^{2}}\frac{\lambda_{0}^{n}}{\sqrt{n!}}.\)

Hence we have,

\(P(n) = e^{-|\lambda_{0}|^{2}}\frac{|\lambda_{0} |^{2n}}{n!}.\)

This is precisely the Poisson probability distribution with mean value |λ_0|^2. This comports well with the Planckian interpretation of the quantum harmonic oscillator as photons of a fixed frequency and therefore energy⁷. From this perspective we see that the expectation value of the degree operator N = a†a is,

\(\langle c(t) | N | c(t) \rangle = |\lambda_{0}|^{2}.\)

Put differently, the average energy observed in this state is

\(\langle c(t) | E | c(t) \rangle = \hbar\omega\left( |\lambda_{0}|^{2} + \frac{1}{2} \right).\)

Coherent Motion

Let’s revisit the expectation value of position and momentum now that we have our explicit coherent state.

\(\langle c(t) | x | c(t) \rangle = \sqrt{\frac{\hbar}{2m\omega}} \langle c(t) | \bar{\lambda}_{0}e^{i\omega t} + \lambda_{0}e^{i\omega t} | c(t)\rangle.\)

Let us resolve λ_0 into a magnitude and phase,

\(\lambda_{0} = |\lambda_{0}| e^{i\theta}.\)

Then we may write

\(\langle c(t) | x | c(t) \rangle = \sqrt{\frac{2\hbar}{m\omega}} |\lambda_{0}| \cos\left( \theta - \omega t\right).\)

Similarly,

\(\langle c(t) | p | c(t) \rangle = \sqrt{2m\hbar\omega} |\lambda_{0}| \sin\left( \theta - \omega t\right).\)

Which is exactly the kind of oscillatory behavior you would expect from the classical equation of motion!

TL;DR

The harmonic oscillator affords a particular class of solution which minimizes uncertainty and resembles the classical state in its time evolution. These solutions are parametrized by a complex number that codifies the average expected energy. It involves a nontrivial linear combination of all previously known basis states.

1. We’ve also seen that there is an equivalent, dual representation, H* given by C[a].

2. There are some subtleties to this in some physical systems with nontrivial topology or other symmetric considerations. These are so-called superselection sectors, but we don’t need to worry about these details with a system as friendly as the harmonic oscillator.

3. You can easily perform this integral to get the Gaussian spread in position, but the time variable creeps in nontrivially. See the discussion in

Uncertainty in Quantum States

4. Unlike in the massive case, here we see that everything depends on (x -ct), which resembles a classical particle’s motion at speed c.

5. Why a and not a†? The operator a is external to the Hilbert space of states for which we are taking linear combinations, H = C[a†].

6. This is why I wanted you to follow along with the computations earlier. Having done those, you hopefully see how easy these are by contrast!

7. See our previous discussion on the Planck distribution:

Grading the Electromagnetic Field