As someone who has just finished my first year of undergrad, I don't really have any experience with QFT but have some experience with quantum mechanics and the math behind it. I have taken college courses on Newtonian mechanics, special relativity, and electromagnetism.

In the "Fields" episode of his "Biggest Ideas In The Universe" series on YouTube, Sean Carroll says that classical fields have a configuration $\Phi(\vec{x})$ which depends on the position $\vec{x}$.

He then goes on to say that quantum fields are described by a wave function $\Psi(\Phi(\vec{x}))$ which is the probability amplitude of the quantum field being in the configuration $\Phi(\vec{x})$.

Next, he considers a non-interacting (And also I guess he's assuming it's non-relativistic) "free" quantum field and says that you can use Fourier analysis to decompose the field configuration $\Phi(\vec{x})$ into its constituent frequencies or modes. Each mode is described by its wave vector $\vec{k}$ where the wavelength $\lambda = \frac{2\pi}{|\vec{k}|}$ So that the mode can be labeled as $\Phi_{\vec{k}}(\vec{x})$.

The kinetic energy of each mode is $K = \frac{1}{2}(\frac{\ d \Phi_{\vec{k}}}{\ dt})^2$ and the potential energy is $V = \frac{1}{2}m^2(\Phi_{\vec{k}})^2$ where $m$ is the mass of the particles described by the field.

In general, the energies of a mode are proportional to $h^2$ where $h$ is the amplitude of the mode and a wave function of the mode $\Psi(\Phi_{\vec{k}}(h))$ is the probability amplitude of the mode $\Phi_{\vec{k}}$ having the amplitude $h$. However, the energy levels of the mode are quantized (like in the quantum harmonic oscillator) so that there is a wave function for each energy level $n$ labeled $\Psi_n(\Phi_{\vec{k}}(h))$ for $n = 0$ to $\infty$.

Sean Carroll's Diagram: enter image description here

Finally, he says that an interpretation of the $n$th energy state $\Psi_n$ for the mode $\Phi_{\vec{k}}$ is as follows:

Since $\vec{k}$ represents the momentum of the mode, $\Psi_n$ is the probability amplitude of measuring $n$ particles with momentum/wave number $\vec{k}$.

The Question

It seems that this description basically allows you to find the probability of having $n$ particles with momentum/wave number $\vec{k}$. It tells you the probability amplitude of having $n$ particles with the same momentum but it doesn't tell you the probability amplitude of the positions of each of those $n$ particles.

My question is how can we find the probability of measuring a particle associated with the quantum field at some position $\vec{x}$ given that we know the probability of having $n$ particles with momentum / wave number $\vec{k}$. Would this require a different explanation than what Sean Carroll gave? Or can it be explained by continuing off of his explanation?

His lecture was a more casual explanation of quantum field theory with only some math and not going into too much detail. I was hoping for a similar casual explanation of my question with some math that builds off of his. Also, if there is anything in his explanation that is outright wrong and could be made more precise without overly complicating things please let me know of that as well.

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    $\begingroup$ Is the question specifically asking about relativistic (Lorentz-symmetric) QFT? That makes a big difference. $\endgroup$ Jul 25, 2020 at 19:00
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    $\begingroup$ @ChiralAnomaly Since it is an intro to QFT video I'm assuming it's non-relativistic. It wasn't really an advanced physics lecture. $\endgroup$
    – mihirb
    Jul 25, 2020 at 19:01
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    $\begingroup$ @ChiralAnomaly I don't really have much experience with QFT but have some experience with quantum mechanics $\endgroup$
    – mihirb
    Jul 25, 2020 at 19:04
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    $\begingroup$ "his explanation was meant to be more of a casual explanation of quantum field theory" casual explanations are casual. This sort of thing is clear in actual explanations. As @ChiralAnomaly says the context matters etc etc $\endgroup$
    – user21299
    Jul 25, 2020 at 23:11

1 Answer 1


Always remember that the point of view of a single individual may contain certain biases that can skew one's understanding. As a general principle, always gather as much information from diverse sources to get a better picture.

I don't want to say anything negative about Sean Carroll's video, but perhaps he is making the situation a bit more complicated than it needs to be in an attempt to be as general as possible.

Although the wave function of a multi-particle state can be rather complicated, the wave function for a single particle is essentially the same as that of the classical field. In other words, if the field $\Phi(x)$ in the classical case obeys the equations of motion, then in the quantum case for a single particle $\Phi(x)$ would be the wave function (barring some dimensional changes) that would obey the same equations of motion. The same equations of motion should govern both the classical field and the quantum wave function.

This can be confirmed experimentally. One can use a small detector to measure the classical optical field at different locations $x$. In the quantum case, one can use a single photon detector to measure the wave function as a function of $x$. If the initial conditions that produce these fields are the same, then the functions would be the same. In both cases, the measurements give us $|\Phi(x)|^2$. In the classical case, it is interpreted as a local intensity and, in the quantum case, it is a probability, but other than that they are the same.

For multiple particles, there are more possibilities. What makes it complicated is the fact that the different particles can have different wave functions and, on top of that, one can have superpositions of different such combinations. However, such complicated state are difficult to produce. Usually, all particles in a multi-particle state would have the same wave function. In that case, it is very much like the classical case. In fact, in the limit of a large number of particles with superpositions of different numbers of particles all having the same wave function, the state can essentially be considered a classical state. That is what one has with a coherent state.

  • $\begingroup$ Thanks for the answer! I'll keep in mind to keep to not take Carroll's explanation as the only possible interpretation and look at other sources on QFT as well. $\endgroup$
    – mihirb
    Jul 26, 2020 at 5:22

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