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In the book "Quantum Field theory and the Standard Model" by Matthew Schwartz, page 23-24, the position space wavefunction is defined as

$$\psi(x)=\langle 0|\phi(x)|\psi\rangle, \tag{2.82+2.83}$$

where $|\psi\rangle$ is any state in the Fock space. Then he uses the equations (i) $\partial_t^2\phi_0=(\nabla^2-m^2)\phi_0$ (i.e., the Klein-Gordon equation for the free massive scalar field $\phi_0(\textbf{x},t)$) and (ii) $[H,\phi_0]=-i\partial_t\phi_0$ to derive Eq. 2.85, in following 3 steps:

$$i\langle 0|\partial_t\phi_0(\textbf{x},t)|\psi\rangle=\langle 0|\int \frac{d^3\textbf{p}}{(2\pi)^3}\frac{\sqrt{\textbf{p}^2+m^2}}{\sqrt{2\omega_\textbf{p}}}(a_p e^{-ip\cdot x}-a^\dagger_p e^{ip\cdot x})|\psi\rangle\\ =\langle 0|\sqrt{m^2-\nabla^2}\phi_0(\textbf{x},t)|\psi\rangle.\tag{2.85}$$

This equation is used to successfully derive the Schrodinger equation in quantum mechanics for the state $\psi(\textbf{x},t)\equiv \langle 0|\phi_0(\textbf{x},t)|\psi\rangle$.

  1. The first equality follows from differentiation of $\phi_0(x)$ w.r.t $t$. How does the second equality follow from the first? How are the inputs (i) and (ii) are utilized to derive the second equality from the first?

  2. Is it true that the operator $$i\partial_t\sim\sqrt{m^2-\nabla^2}~?$$ If yes, what is the use of doing the middle step?

Cosmas Zachos
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3 Answers3

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(as a personal comment: I do not know why one should deal with complicated things like approximation of operators and so on when physics is evident...)

The only requirement is that

the energy content of the state must have values very small with respect to the mass of the particle.

In practice the support of the ${\bf k}$-Fourier transform of $\psi({\bf x}, t)$ must be sufficiently concentrated in a set where $|{\bf p}|<< m$. In this situation we my safely approximate $p^0 \simeq \frac{{\bf p}^2}{2m} + m$ (I assume $c= \hbar=1$). In this situation $$\psi({\bf x},t) = \langle 0|\phi_0({\bf x},t)|\psi\rangle \simeq \frac{e^{-imt}}{(2\pi)^{3/2}}\int d^3{\bf p} \: e^{i\left({\bf p}\cdot {\bf x} -\frac{{\bf p}^2}{2m}t\right)}\:\frac{\hat{\psi}({\bf p})}{2m}$$ Up to the factor $e^{-imt}$ that is a phase (even if time depeding) and can be omitted (actually it is responsible for Bargamann's superselection rule of mass), the obtained function evidently satisfies the standard free Schroedinger equation.

This approximation clearly does not hold for massless particles and this explains why photons do not (approximately) solve Schroedinger equation...

Valter Moretti
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  1. The step follows just by the usual relation that the Fourier transform of $p\cdot \phi(p)$ is $\partial_x \tilde{\phi}(x)$, i.e. multiplication by the variable becomes differentiation.

  2. There is no general relation between $\partial_t$ and $\sqrt{m^2-\nabla^2}$, since it is in general not even defined what $m$ is. The operators have to act on some specific given function to have any relation at all.

ACuriousMind
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If you can agree that the first step come from differentiation with respect to time, then the second step is simply noting that the operator $\sqrt{m^2-\boldsymbol{\nabla}^2}$ acts as $\sqrt{m^2+\textbf{p}^2}$ in momentum space. This is just a generalization of the fac that the operator $\boldsymbol{\nabla}$ simply acts as $i\textbf{p}$ in momentum space. They give the same result. So, no, there is no general relation between $\partial_t$ and $\sqrt{m^2-\boldsymbol{\nabla^2}},$ since they are independent differntial operators. The result is just a property of the Fourier transform. The end result is that,

$$i\partial_t\psi(x)=\sqrt{m^2-\boldsymbol{\nabla}^2}\,\psi(x)$$

from your definition of the position space wavefunction.

I hope this helps! Also, I hope you keep reading Schwartz. It's a fantastic book.

Bob Knighton
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