quantum field theory - Why is there a minus sign in this wave equation derivation?

My book on quantum mechanics suggests a derivation of the wave equation

$$\left(\Delta - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \right) \psi(\bar{r},t) = 0$$

from the photon energy-impulse relation

$$E^2 = c^2p^2$$

using the substitutions

$$\bar{p} \to -i\hbar \nabla$$ and $$E \to i \hbar \frac{\partial}{\partial t}$$

I suppose the $\hbar$s are there for the non-photon generalisation, but is there a reason for the minus sign in the $p$ substitution? Removing it would yield the same result.

Answer

The relative sign is not just a convention. Once you decide that $E$ is represented by $i\hbar \partial/\partial t$, there must be a minus sign in the formula for $p$, namely $p=-i\hbar \partial / \partial x$. Or vice versa.

First of all, there has to be $i$ or $-i$ in all the formulae because $\partial/\partial x$ is an anti-Hermitian operator (because of the minus sign in the integration by parts) and we need Hermitian operators (which has real, measured eigenvalues) for the energy, momentum, and others. What about the signs?

The only sign convention that was chosen in the early days of quantum mechanics was one for the energy; indeed, one could have replaced $i$ by $-i$ in that equation because $i$ and $-i$ play the same algebraic role: exchanging $i$ and $-i$ is an "outer automorphism" of complex numbers. But once this sign is fixed, all other signs are fixed, too. That includes minus signs in momentum, angular momentum, gauge transformations, Schrödinger's picture, Heisenberg's picture, Feynman's path integral, and any other formula of quantum mechanics. There is only one way to define quantum mechanics given a classical limit (with its sign conventions) we need to get.

The relative minus sign in $p,E$ may become invisible if you only act with second derivatives - squared momentum, squared energy - but it is visible if you act with first powers of the operators.

De Broglie's wave - or a wave associated with a particle - is proportional to

$$\exp \left[ \frac{i}{\hbar} (\vec p\cdot \vec x-Et)\right]$$ Note that if you differentiate with respect to $x$ and $t$, and multiply the result by $i\hbar$, you get $-p$ and $E$, respectively. (Omit the vector signs if you want just one-dimensional space with one $x$ and one $p$.)

The relative sign between $\vec p\cdot \vec x$ and $Et$ in de Broglie's formula above is physically necessary because only $Et - \vec p \cdot \vec x$ is the correct Lorentzian inner product of the vectors $(E,\vec p)$ and $(t,\vec x)$: the relative minus sign comes from the opposite signs of space and time in the signature of spacetime. Note that the doubly relative sign in $(E,\vec p)$ and $(t,\vec x)$ can't be flipped because in relativity, $\vec pc^2/E$ is the velocity $\vec v$.

The argument above is for a relativistic interpretation of $E$. However, the non-relativistic kinetic energy is defined just as a shifted relativistic energy,

$$E_{\rm nonrel} = E_{\rm rel} - mc^2$$ so the coefficient (and sign) in front of $E$ remains unchanged and $\vec p$ is totally unchanged. One may also design related arguments analyzing where the de Broglie wave is moving. For it to be moving in the right direction, there has to be a minus sign in $Et-px$. It's related to the fact that the shape of objects moving by velocity $v$ only depends on $x-vt$ because $x=vt$ (without a minus sign, which becomes a minus sign if you move both terms to the same side) is the equation for their center of mass.

The hypersurfaces of constant phase of the de Broglie wave are orthogonal to the world lines of the particles which dictates the sign. After all, the waves' maxima and minima should be moving in the same direction as the particles themselves.