I am reading "An introduction to quantum field theory" by Peskin and Schroeder and I am confused. I appreciate your help.

Here's the context to my question: In chapter 2, the book introduces quantum field theory by first talking about classical field theory and Klein-Gordon equation. The book expands a typical classical field in momentum space as,

\begin{equation} \phi(x, t)=\int\frac{d^3p}{(2\pi)^3}e^{ipx}\phi(p, t). \end{equation}

The classical Klein-Gordon equation in momentum space becomes (equation (2.21)),

\begin{equation} \left[\frac{\partial^2}{\partial t^2} + (|p|^2 + m^2)\right]\phi(p, t) = 0, \tag{2.21} \end{equation}

which is the equation of motion of a (classical) simple harmonic oscillator with frequency $\omega_p=\sqrt{|p|^2 + m^2 }$. The book then claims that the solution to quantum harmonic oscillator is well known (equation (2.23)),

\begin{equation} \phi = \frac{1}{\sqrt{2\omega}}(a + a^{\dagger}).\tag{2.23} \end{equation}

Where $a$ is the ladder operator. If I understand the book correctly, it finally claims that if we use the solutions of quantum harmonic oscillator as the solution to Klein-Gordon equation in momentum space, i.e. $\phi(p,t) = \frac{1}{\sqrt{2\omega_p}}(a_p + a_p^\dagger)$, we obtain the solution to the quantum Klein-Gordon equation. Such solution satisfies the commutation relation automatically so I can see where the author is going with this idea. The solution to quantum Klein-Gordon equation is then (equation (2.25)),

\begin{equation} \phi(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_pe^{ipx} + a^\dagger_pe^{-ipx})\tag{2.25} \end{equation}

Here's my question: Claiming equation (2.23) as the solution to the equation (2.21) is a bit non-rigorous, don't you think? Put it this way, how can we prove that,

\begin{equation} \left[\frac{\partial^2}{\partial t^2} + (|p|^2 + m^2)\right]\frac{1}{\sqrt{2\omega_p}}(a_p + a_p^{\dagger}) = 0 ??? \end{equation}

The disconnect I feel here stems from the fact that although expanding $\phi$ as ladder operators is useful in finding the eigenstate (and eigenvalue) to Hamiltonian of harmonic oscillator, equation (2.21) is just a differential equation that, at first glance, isn't about finding eigenstate. Operation-wise, finding eigenstate is so much different from solving differential equation that I think a mathematical proof is needed to before we can claim that the solution to quantum harmonic oscillator is also a solution to the equation of motion of the classical harmonic oscillator. I would like to ask kindly if you can prove a proof to my final equation to fill in the gap in knowledge I current have while reading the book.

  • $\begingroup$ The equation you're trying to prove isn't true. Equations of motion and boundary conditions being enough to determine everything is unique to classical physics. $\endgroup$ Jul 22, 2022 at 22:20

1 Answer 1


Please first see this reference about solving the quantum simple harmonic oscillator with raising and lowering operators.

Next, realize that we are dealing with a free field theory, and the Hamiltonian (which governs the dynamics) is the free field Hamiltonian: $$ H = \sum_\vec p \omega_p a^\dagger_{\vec p }a_{\vec p}\;, $$ which is clearly a sum over individual simple harmonic oscillator Hamiltonians (one for each $\vec p$ value). (To put is another way the free field is just a bunch of uncoupled simple harmonic oscillators.)

In quantum mechanics, the time-dependence of an operator in the Hamiltonian picture (or, once we add interactions, we will call this the interaction picture) is given by: $$ a_{\vec p}(t) = e^{-iHt}a_{\vec p}e^{iHt} $$

Taking the time derivative we find, as usual: $$ \dot a_{\vec p} = -i[H, a_{\vec p}(t)] = i\omega_{\vec p}a_{\vec p}(t)\;. $$

Take a second time derivative to find: $$ \ddot a_{\vec p} = -\omega_p^2 a_{\vec p}\;. $$

Or, using $\omega_{\vec p}^2 = |\vec p|^2 + m^2$ and rearranging: $$ \ddot a_{\vec p} + (|\vec p|^2 + m^2)a_{\vec p} = 0\;, $$ just like we want.

I am quite sure you can work out the analogous relation for $a^\dagger_{\vec p}$ on your own.

  • 1
    $\begingroup$ Amazing! I completely forgot these dynamic equations while I was reading. $\endgroup$ Jul 22, 2022 at 22:46
  • $\begingroup$ I hate it when that happens ;) $\endgroup$
    – hft
    Jul 23, 2022 at 0:49

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