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?

  • $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$ – Qmechanic Jun 29 '17 at 19:17
  • $\begingroup$ @Qmechanic I'm specifically interested in understanding the steps of Schwartz's book. In particular, how did it get the second equality from the first and at which step did he use $[H,\phi_0]=-i\partial_t\phi_0$? And also what happened to the integral over $d^3\textbf{p}$? $\endgroup$ – SRS Jun 30 '17 at 12:30
  • $\begingroup$ He utilized (ii) to derive/confirm (i), used in the first step. He then performed the momentum integral to derive the Fourier transform, the second step. $\endgroup$ – Cosmas Zachos Jun 30 '17 at 12:40
  • $\begingroup$ @CosmasZachos Is this momentum integral obvious? Because the RHS (2.85) has a minus sign that is not present in the definition of $\phi_0(\mathbf{x},t)$. $\endgroup$ – Dwagg Dec 25 '17 at 3:05
  • $\begingroup$ @CosmasZachos I agree, but I don't see how the momentum integral works out. I agree $\int \frac{d^3 p}{(2\pi)^3} \frac{\sqrt{p^2+m^2}}{\sqrt{2\omega_p}} a_p e^{-ipx} = \sqrt{m^2-\nabla^2} \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} a_p e^{-ipx} $. But the second term has a minus sign and an $e^{+ipx}$ instead of $e^{-ipx}$, which is the F.T. Do these sign differences cancel to give us $\phi_0(x,t)$ using some property of F.T.? $\endgroup$ – Dwagg Dec 25 '17 at 14:31

(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 $|{\bf p}|<< m^2$. 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...

  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.

  • $\begingroup$ At which step did he use the information (i) and (ii)? @ACuriousMind $\endgroup$ – SRS Jun 29 '17 at 18:57

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,


from your definition of the position space wavefunction.

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

  • $\begingroup$ Your answer does not specify at which step he used (ii) $[H,\phi_0]=-i\partial_t\phi_0$. Moreover, why does the integral over $d^3\textbf{p}/(2\pi)^3$ go away? @BobKnighton $\endgroup$ – SRS Jun 30 '17 at 12:25
  • $\begingroup$ The integration over momentum went away by pulling out the spacial derivative and noticing the definition of the fourier transform of the field is what's left. Also I don't see a need for the commutator if the first step is already complete. $\endgroup$ – Bob Knighton Jun 30 '17 at 13:52
  • $\begingroup$ @BobKnighton I agree that the operator $\sqrt{m^2 - \nabla^2}$ acts as $\sqrt{m^2 + \vec p^2}$ in momentum space, but the thing its acting on is not the quantum field operator is it? Because of the minus sign in (2.85)? $\endgroup$ – Dwagg Dec 25 '17 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.