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I am a physics undergraduate. I am working in the world of textbook (non-relativistic) Quantum Mechanics. Say we have a wave function $\Psi(x,t)$. Must $\Psi(x,t)$ be square-integrable or normalizable?

To my understanding, normalizable implies square-integrable. However, square-integrable does not imply normalizable. For example, $f(x,t) = 0$ is square-integrable, but it is not normalizable.

This leads me to think that a wave function $\Psi(x,t)$ must be normalizable, a stricter requirement than just square-integrable. In particular, $f(x,t) = 0$ is not a wave function and thus cannot represent a physical state.

And, as an application of this tentative result, could one explain why particle annihilation is not built into (non-relativistic) textbook Quantum Mechanics by appealing to the fact that the evolution of a wave function $\Psi(x,t)$ into the wavefunction $f(x,t) = 0$ (after some time has elapsed) is not possible since $0$ itself is not a wave function?

Qmechanic
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Silly Goose
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2 Answers2

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That is a really nice remark. States are actually not quite vectors on the Hilbert space, but rather rays of vectors. Hence, indeed, they should be normalizable rather than square-integrable. In practice, this means that the allowed wavefunctions are the non-vanishing square-integrable functions (notice all of these are indeed normalizable, where I take "function" to mean actually an equivalence class of functions that are equal almost everywhere).

As for the evolution, notice that if QM were to describe particle annihilation, it should also be able to describe particle creation because it is time-reversible. However, you can't give a wavefunction a interpretation of two particles at once. It only tells you how to find a single particle.

From a more mathematical point of view, saying that $\psi = 0$ at any time (e.g., after annihilation) determines the whole evolution of the state. Notice that evolving the Schrödinger equation to any time with $\psi(0,x) = 0$ can only lead to $\psi(t,x) = 0$. Hence, wanting to describe the state of the system with a single wavefunction will also prevent one from getting $\psi(t_0,x) = 0$ at any particular time $t_0$, unless the wavefunction vanishes at all times.

To actually describe particle creation and annihilation, one must use different formalisms, such as second-quantization and quantum field theory. While this keeps the main ideas of QM (states are vectors on a Hilbert space, observables are operators, etc), you no longer deal with wavefunctions, and the states now live in a more complicated Hilbert space known as Fock space. This space does have states corresponding to different particle numbers and it should be mentioned that even now that particle annihilation is possible, the vacuum state (no particles) is not $\psi = 0$, but rather a non-vanishing vector corresponding to zero particles.

Níck Aguiar Alves
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You are correct that in order to represent a valid physical state, a wavefunction $\psi(x)$ must be normalizable and not just square-integrable. In practice, people tend to (sloppily) use those two terms interchangeably, because the only functions that are square-integrable but not normalizable are those that are zero almost everywhere - and since the Hilbert space of physical states identifies together functions that are equal almost everywhere, all of these functions correspond to a single vector (not state!) $|0\rangle$ in the Hilbert space. (Here the notation $|0\rangle$ represents the additive identity vector in the Hilbert space, not the ground state of any Hamiltonian.)

So the space of square-integrable functions does indeed include unphysical wavefunctions, but they all only correspond to a single unphysical vector in the Hilbert space. So it's almost correct to say that all (really should be "almost all") square-integrable wavefunctions represent a valid physical state.

tparker
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