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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? Continue reading Quantum Mechanics Fails Its Relativity Exams