Quantum Mechanics Fails Its Relativity Exams

When people talk about quantum mechanics, they’re usually referring to applying the principles of quantum mechanics to particles which are moving with non-relativistic energies. That’s fine, and quantum mechanics (or really, non-relativistic quantum particle mechanics) is extremely useful for huge ranges of problems, but there are also whole fields of physics for which we must take special relativity into account. Further, we know that particles are created and destroyed all the time—particle accelerators produce showers of daughter particles, which themselves decay into still other particles, and so on. So there is a definite need for some quantum framework which can account for Einstein’s special relativity, as well as for particle creation and destruction. This encompassing quantum framework is known as Quantum Field Theory.

In developing a relativistic quantum theory, it turns out that we can’t just quantize relativistic particles the way we can quantize non-relativistic particles, as I’ll detail below. Rather, we will need to construct a quantum field in order to take quantum mechanics into the relativistic regime. Here, we’ll take it for granted that a quantum field $\phi(x)$ can be constructed which satisfies certain fundamental requirements.

In this notation, $x$ is a 4-vector $x = \langle{t}; \vec{x}\rangle$  representing a point in spacetime, $\vec{x} = \langle{x}_{1}, {x}_{2}, {x}_{3}\rangle$ is a spatial vector in $\mathbf{R}^3$, and I’ll use boldface for operators (like the Hamiltonian, $\textbf{H}$). In the end our quantum field ($\phi$) will be operator-valued at each spacetime point—I just won’t use boldface for the fields themselves, even though they are operators. So $\phi$ associates each spacetime point $x$ with an operator called $\phi(x)$.

So, why doesn’t familiar quantum mechanics play well with relativity?

The easy way out and negative energy

In non-relativistic quantum mechanics, the Schrödinger equation (SEq),

$i\hbar\dfrac{\partial \Psi(\vec{x}, t)}{\partial t} = \textbf{H}\Psi(\vec{x}, t)$,

describes the evolution of a particle’s state in time. Unfortunately for physicists, the SEq is not “relativistically invariant.” This means that if a wave function $\Psi(\vec{x}, t)$ satisfies the SEq in one reference frame, it will not generally satisfy the equation after we perform a Lorentz boost or a rotation to a different reference frame.

There is a wave equation, however, which is relativistically invariant and does naturally arise as the description for a certain type of quantum field (although here we’ll treat it as an equation for a classical field $\Phi(x)$). This is the Klein-Gordon equation,

$\left(\dfrac{\partial^{2}}{\partial t^{2}} - \nabla^{2} + m^{2}\right)\Phi(x) = 0$,

which looks just like a classical wave equation with an added mass term. We can solve the Klein-Gordon equation, starting with an educated guess for $\Phi(x)$:

$\Phi(x) = \Phi(\vec{x},t) \propto {e}^{-{i}(Et-\vec{p}\cdot\vec{x})}.$

Using the Klein-Gordon equation,

$\left(\dfrac{\partial^{2}}{\partial t^{2}} - \nabla^{2} + m^{2}\right)\Phi(x) = (-E^{2} + \lvert\vec{p}^{2}\rvert + m^{2})\Phi(t,\vec{x}) = 0$

$\Longrightarrow (-E^{2} + \lvert\vec{p}^{2}\rvert + m^{2}) = 0.$

Solving for our allowed energies, we see that $E = \pm\sqrt{\lvert\vec{p}^{2}\rvert + m^{2}}$, and our solution takes the form

$\Phi(x) = N_{1}{e}^{-{i}(\sqrt{\lvert\vec{p}^{2}\rvert + m^{2}}-\vec{p}\cdot\vec{x})}+N_{2}{e}^{-{i}(-\sqrt{\lvert\vec{p}^{2}\rvert + m^{2}}-\vec{p}\cdot\vec{x})}$ ($N_{1}$ and $N_{2}$ are just normalization constants).

There’s something fishy going on here: the term $E = -\sqrt{\lvert\vec{p}^{2}\rvert + m^{2}}$ corresponds to solutions with negative energy. That means treating the Klein-Gordon equation as the equation of motion for a single relativistic quantum particle cannot be the correct interpretation. If $\Phi(x)$ is viewed as a field, the negative energy solutions can be interpreted as antiparticle solutions–take a look at Section 8.1 of Sakurai’s “Modern Quantum Mechanics” for more on this interpretation. In any case, a single-particle interpretation of $\Phi(x)$ is incorrect.

Problems come in pairs

We know from Einstein’s mass-energy equivalence E = mc2 that mass can be created if a suitable cost in energy is paid. If we scatter two pions, for example, we may expect the following process:

$\pi^{+} + \pi^{-} \longrightarrow \pi^{+} + \pi^{-}$

That’s not too exciting on the surface. But if the incoming pions have sufficiently high energy, we will have no choice but to consider the processes

$\pi^{+} + \pi^{-} \longrightarrow 2\pi^{+} + 2\pi^{-}$ and $\pi^{+} + \pi^{-} \longrightarrow 3\pi^{+} + 3\pi^{-}$, and so on,

because the energy of the incoming pions may be high enough to produce numerous daughter pions.

Generally, we have to worry about pair-production of particles whenever a particle is confined to a very small space–comparable to its Compton wavelength $\lambda = \frac{\hbar}{mc}$, where m is the particle’s mass. From the uncertainty relation $\Delta{x}\Delta{p} \geq \frac{\hbar}{2}$, if the particle is confined so that the uncertainty in its position is $\Delta{x} \leq \frac{\hbar}{mc}$, then $\Delta{p} \geq \frac{\hbar}{2}\frac{1}{\Delta{x}} = \frac{mc}{2}$. From E = mc2, the threshold on momentum for creating a particle is $p \geq mc$. So if $\Delta{p}$ approaches $mc$ as $\Delta{x}$ is squeezed, pair-production is a relevant problem.

Incidentally, this problem of pair-production might make it seem silly to study the hydrogen atom in the scrutinizing detail that non-relativistic quantum mechanics affords it. After all, the electron is confined to an awfully small space. But it turns out that the characteristic size of hydrogen’s orbitals is around the Bohr radius, $a_{0} = 0.0529nm$, which can be related to the Compton wavelength $\lambda_{e}$ of an electron by the fine-structure constant $\alpha$: $\alpha_{0} = \frac{\lambda_{e}}{\alpha} \approx 137\lambda_{e}$. So $\Delta{x} \sim \alpha_{0}$ is sufficiently large that pair-production doesn’t come into play, and we can use non-relativistic quantum mechanics to study the hydrogen atom.

Anyway, whatever quantum field theory we use, it must be constructed to account for creation and destruction of particles.

Faster than light?

We run into another problem in single-particle quantum mechanics when we calculate the amplitude $\textit{M}$ (which represents how probable it is for a process to occur) for a particle to travel from a point $\vec{x}_{0}$ to $\vec{x}_{1}$ in a time t. Because the speed of light c is our universal speed limit, we would expect $\textit{M}$ to be zero for points which are separated by a distance farther than light could travel in a time t. In special relativity, the boundary of points which are accessible without exceeding c defines a light cone, shown in Figure 1.

Figure 1. For a particle or person at the origin, the speed limit c means that only points within the light cone are accessible. It is impossible to travel or send information to points outside the light cone. For you here on Earth, a spot on Jupiter one second in the future is outside your light cone. You cannot influence that spacetime point—no matter what.

So if $\lvert \vec{x}_{0}-\vec{x}_{1}\rvert \geq ct$, the particle cannot possibly travel from $\vec{x}_{0}$ to $\vec{x}_{1}$ in a time t.

A word about spacetime intervals: distances in spacetime are called intervals, as they incorporate both space and time. The spacetime interval $s^{2}$ between two events is independent of reference frame, so it’s very useful. The interval is defined as $s^{2} = \Delta{x}^{2} - c^{2}\Delta{t}^{2}$ (in the sign convention I’m using), where $\Delta{x}$ and $\Delta{t}$ are respectively the spatial and temporal distances between two events. If two events have a spacetime interval $s^{2} > 0$ between them, they are said to be spacelike-separated. For spacelike-separated events we have $\Delta{x}^{2} > c^{2}\Delta{t}^{2}$, meaning the events are separated by enough space that one cannot have a causal influence on the other.

Returning to the problem of particles traveling outside their light cone: if we actually use the free-particle Hamiltonian $\textbf{H}$ to calculate the amplitude for a particle to travel from $\vec{x}_{0}$ to $\vec{x}_{1}$, we start with

$\textit{M} = \langle{\vec{x}_{1}}\lvert{e}^{-i\textbf{H}{t}/\hbar}\rvert\vec{x}_{0}\rangle = \langle{\vec{x}_{1}}\lvert{e}^{-i(\vec{\textbf{p}}^{2}/2m)t/\hbar}\rvert\vec{x}_{0}\rangle$.

Inserting an identity operator $\displaystyle\int \rvert{\vec{p}}\rangle\langle{\vec{p}}\lvert\,d^{3}\vec{p}$ gives

$\textit{M} = \displaystyle\int \langle{\vec{x}_{1}}\lvert{e}^{-i(\vec{\textbf{p}}^{2}/2m)t/\hbar}\rvert\vec{p}\rangle \langle\vec{p}\rvert\vec{x}_{0}\rangle\,d^{3}\vec{p}$,

and the exponential term acts on $\rvert{\vec{p}}\rangle$ to give

$\textit{M} = \displaystyle\int {e}^{-i(\vec{p}^{2}/2m)t}\langle{\vec{x}_{1}}\rvert\vec{p}\rangle \langle\vec{p}\rvert\vec{x}_{0}\rangle\,d^{3}\vec{p}$.

The inner-product term $\langle{\vec{x}_{1}}\rvert\vec{p}\rangle$ is just the wave function for a momentum eigenstate, $\langle{\vec{x}_{1}}\rvert\vec{p}\rangle = \dfrac{1}{\sqrt{2\pi\hbar}}{e}^{{i}(\vec{x}_{1}\cdot\vec{p})/\hbar}$. After replacing $\langle\vec{p}\rvert\vec{x}_{0}\rangle = \langle\vec{x}_{0}\rvert\vec{p}\rangle^{*}$ similarly, we have our result:

$\textit{M} = \left(\dfrac{1}{2\pi\hbar}\right)^{3}\displaystyle\int {e}^{-{i}(\vec{p}^{2}/2m){t}}{e}^{{i}\vec{p}\cdot(\vec{x}_{1}-\vec{x}_{0})/\hbar}\,d^{3}\vec{p} = \left(\dfrac{m}{2\pi{i}\hbar{t}}\right)^{\frac{3}{2}}{e}^{{i}m\lvert\vec{x}_{1}-\vec{x}_{0}\rvert^{2}/2\hbar{t}}$,

which is nonzero. That means that in the familiar framework of quantum mechanics, it’s possible for a particle to travel between any two points in any amount of time. Even if you replace the Hamiltonian operator $\textbf{H}$ in the calculation above with its relativistic form $\textbf{H} = \sqrt{\vec{p}^{2}c^{2}+m^{2}c^{4}}$, you still find that $\textit{M}$ is nonzero outside the light cone! That’s no good; faster-than-light travel has to be prohibited by QFT.

As an aside, notice that the $\textit{M}$ we calculated above corresponds to the propagation of a free particle between any two points in space. Strictly speaking, $\langle{\vec{x}_{1}}\rvert$ represents the final state of interest and ${e}^{-i\textbf{H}{t}/\hbar}\rvert\vec{x}_{0}\rangle$ represents the actual final state of the system which starts out localized at $\vec{x}_{0}$, so taking their inner product $\textit{M}$ tells us the overlap of the desired final state with the actual final state. In other words, squaring $\textit{M}$ gives the probability (density) of the propagation from $\vec{x}_{0}$ to $\vec{x}_{1}$. Our result for $\textit{M}$ is a 3D Gaussian centered around the initial point, which flattens out as time moves on. That makes some sense; a free particle initially localized at $\vec{x}_{0}$ will have a totally undetermined momentum (from the uncertainty principle), and so its motion will not have a preferred direction.

Long-distance commutation

As a final point in the problem of applying special relativity to quantum mechanics, we’ll take a look at causality from the point of view of operators.

First, remember that spacelike-separated points in spacetime cannot have a causal influence on each other.

In non-relativistic quantum mechanics, every Hermitian operator is associated with an observable quantity (say, the component of angular momentum in the z-direction). If two such operators A and B commute, then any observer can measure the quantities associated with A and B simultaneously with well-defined results. This is not so for non-commuting observables such as position and momentum.

Now think about two observers in spacelike-separated regions R1 and R2. The experimenter in R1 is trying to measure the observable A1, and the exprimenter in R2 is trying to measure an the observable A2. If A1 and A2 commute, the experimenters can succeed.

But what if A1 and A2 do not commute? Well, then the outcomes of the measurements will depend on the order in which the experimenters make their respective measurements—as is necessarily the case with non-commuting observables—meaning that a measurement taking place in R1 superluminally affects what outcomes may be seen in R2.

A technical note: this is a stranger problem than standard quantum entanglement. Real measurements on entangled states involve commuting operators (for example, S1z of particle 1 and S2z of particle 2, where particles 1 and 2 are in an entangled state). These operators are said to operate in different “subspaces”, and are defined to commute, so both S1z and S2z can be simultaneously known. That is, measuring S2z after measuring S1z doesn’t change the system’s state out of an eigenstate of S1z. But if the operators A1 and A2 don’t commute, measuring A1 in R1 and subsequently measuring A2 in R2 will change the system out of a state for which A1 is definite. If an observer in R1 then measures A1 again (after A2 has been measured), she may see a different result than she saw previously, and know superluminally that a measurement was made in R2 which disturbed her system. There is no analog for her obtaining this information in standard entangled measurements. Measuring non-commuting A1 and A2 in spacelike-separated violates causality.

So QFT has a new requirement:

$[A_{1}, A_{2}] = 0$ for any observables A1 in R1 and A2 in R2, where R1 and R2 are spacelike-separated regions. Spacelike-separated operators must commute.

This requirement squares nicely with a framework of fields rather than particles. In a field theory, we can treat the fields themselves as operators which depend on spacetime points like $x$ and $y$. Any observable quantities we want to measure will be built up of the fields. Then, if we want operators corresponding to spacelike-separated observables to commute, we just have to enforce that the fields themselves commute:

$[\phi(x),\phi(y)] = 0$ for any spacelike-separated points $x$ and $y$.

This is one of about four fundamental requirements for constructing a quantum field. The others deal with their behavior under spacetime translations and Lorentz transformations, as well as the specifics of how they handle creation and annihilation of particles. But that’s a story for another post! Quantum mechanics has totally failed its special relativity exams.