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This problem might sound a bit trivial at face value but to me is sorta hard to grasp if not paradoxical!

We know that the energy gap between the quantum state of $n$ particles and $n+1$ particles is finite. Given the fact that quantum field are beyond any notion of waves or particles (and they are there to solve the wave-particle duality problem), how is it possible to visualize that: to excite a quantum field to another higher state, one needs only a finite amount of energy regardless of the volume!?

In fact in both cases of finite and infinite volume (consequently standing and propagating waves) the energy gaps are independent of the volume! How is it possible to excite an open-ended field at all points of space with a finite amount of energy? Or how is it possible to excite a standing quantum field (like the primitive instance of a quantized string in the 3 man paper by Jordan) by adding a fixed amount of energy that doesn't depend on the volume (of the black body or string or whatever capacitor)?

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  • $\begingroup$ It is not true in general that the energy gap is independent of volume in a finite system (and the statement is not terribly meaningful in an infinite system). Typically increasing the size of the system will decrease the size of the gap. $\endgroup$ Dec 29, 2020 at 13:37
  • $\begingroup$ Isn't the difference between the n+1 and n particles in the EM field excitations merely: h $\omega$ which is pretty much finite? $\endgroup$ Dec 29, 2020 at 13:39
  • $\begingroup$ The energy to add 1 photon is $\hbar\omega$, so the energy gap is determined by the smallest allowed $\omega$. This is determined by the longest allowed wavelength, which is determined by the size of the system. For photons the minimum $\omega$ tends to zero in the limit of infinite volume (and so infinite maximum wavelength). $\endgroup$ Dec 29, 2020 at 13:43
  • $\begingroup$ True, but what if I consider a certain nonzero frequency, and try to increase the number of such photons by one. How much energy do I need to do so? $\endgroup$ Dec 29, 2020 at 13:48
  • $\begingroup$ Specially that in the case of infinite volume there's no standing wave at all and all waves propagate. So there's no inverse relation between spatial dimensions of a hypothetical container and the frequency. $\endgroup$ Dec 29, 2020 at 14:00

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This question has nothing to do with quantum mechanics. The same "paradox" happens in classical field theory, where things are easier to visualize. Consider for example the electromagnetic field in $\mathbb R^3$. How can we excite this field with finite energy? Well, we make an excitation that is localized in space.

Recall that the energy of the electromagnetic field is $$ E=\int_{\mathbb R^3}\boldsymbol E^2+\boldsymbol B^2 $$

If $\boldsymbol E,\boldsymbol B$ are non-zero in a compact region (or if they decay sufficiently fast at infinity), then $E$ is finite, even if the total integration region is infinite.

In QFT the energy is given by an identical expression, $$ E=\int_{\mathbb R^3}\mathcal H(\varphi) $$ where $\mathcal H$ is the Hamiltonian density, a function of all the fields of the theory. Just like in the classical case, a finite-energy field configuration has fields of compact support (or that vanish sufficiently fast at infinity).

Please do keep in mind that not every field configuration satisfies $E<\infty$. There are important situations where this is violated. For example, solitons have finite energy and are topologically protected, because their decay involves configurations that have infinite energy. So keep in mind that not every field configuration has finite energy, but those that don't are not physical and cannot be realized in the real world. A more down-to-earth example of a configuration with infinite energy is the case of plane waves, which are constant in space (so they do not decay). Again, this field configuration is not physical and cannot be realized in the real world. There are no plane waves in nature.

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  • $\begingroup$ Finally someone understands the problem. Thanks. But one more point is: the plane waves have a definite number of particles(the mode is determined unequivocally). How about the case of a wave-packet? $\endgroup$ Dec 31, 2020 at 19:56
  • $\begingroup$ And can I interpret the situation this way: I locally add $h \omega$ amount of energy and the field configuration needs some large amount of time(physical infinity) to reach the asymptotic state in which the mode is n+1. And the field configuration approaches an asymptotic plane wave shape at the end. And since the volume is infinitely large, the overall energy density won't be changed. $\endgroup$ Dec 31, 2020 at 20:08
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    $\begingroup$ @BastamTajik Wave-packets may or may not have a well-defined particle number. For example, in the case of electrons you have the electric charge, which is quantized, so a wave-packet will have a well-defined particle number. In the case of photons there is no conserved charge, so the particle number is not defined. A wave-packet of photons is a coherent state, it has no definite number of photons. $\endgroup$ Dec 31, 2020 at 21:51
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If I understand your question properly, you are concerned that (1) the Hamiltonian is given by a Hamiltonian density which we integrate over all space, and (2) that an energy eigenstate (in an infinite system) is a plane wave which exists everywhere with the same amplitude, differing from point to point only by a phase factor. Therefore, either the Hamiltonian density acting on this state gives zero or something nonzero, corresponding to an excitation with either zero energy or infinite energy.

The answer to this is basically the same as the answer to a similar answer in elementary QM: In an infinite, free system, energy eigenstates are non-normalizable and hence unphysical.

We talk about momentum eigenstates $a^\dagger_\mathbf p|0\rangle$ in essentially the same way as we talk about them in 1D scattering problems from our first course in QM. They're useful tools, but all physical states will be of the form

$$|\psi\rangle = \int\frac{d^3p}{(2\pi\hbar)^3} f(\mathbf p) a^\dagger_{\mathbf p} |0\rangle$$

for some nicely-behaved function $f$. Expressing this state in terms of the spatial creation operators (i.e. the field operators) would yield

$$|\psi \rangle = \int d^3 x \ \tilde f(\mathbf x) \phi^\dagger(\mathbf x)|0\rangle$$

with $\tilde f$ the Fourier transform of $f$. As a result, all physical states will have some finite spread in momentum as well as some degree of spatial localization, as $\tilde f$ should be square-integrable. If you apply the Hamiltonian operator to this state, you will find that

$$H|\psi\rangle = \int\frac{d^3p}{(2\pi\hbar)^2} f(\mathbf p) \omega_\mathbf p a^\dagger_\mathbf p|0\rangle$$ $$\implies \frac{\langle \psi|H|\psi\rangle}{\langle \psi|\psi\rangle} = \frac{\int \frac{d^3p}{(2\pi\hbar)^3} |f(\mathbf p)|^2 \omega_p}{\int\frac{d^3p}{(2\pi\hbar)^3} |f(\mathbf p)|^2} \geq \omega_0 = mc^2$$

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  • $\begingroup$ I guess you haven't understood the question properly or perhaps I'm missing some point in your answer. Given to different eigenstates of the Hamiltonian say, n particle of momentum p and n+1 particle of momentum p, in order to excite the former one to the latest one, one should add merely $h {\omega}_p$ amount of energy to the n particle state. The question is that: Given the infinite extent of the quantum fields how is it possible to excite the field at all points of the space with a finite amount of energy. $\endgroup$ Dec 31, 2020 at 17:29
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    $\begingroup$ @BastamTajik I think I understood your question. The answer is that momentum eigenstates arent truly eigenstates of the Hamiltonian because they aren’t physical states at all. Real states consist of superpositions of momentum eigenstates, and therefore are not of infinite spatial extent. This is exactly the same as in elementary QM. Please read my answer again. $\endgroup$
    – J. Murray
    Dec 31, 2020 at 17:42
  • $\begingroup$ So you mean the contradictory observation is right but the way out is to resort to wave packets? $\endgroup$ Dec 31, 2020 at 17:47
  • $\begingroup$ @BastamTajik From my perspective I see no contradictory observation. Your trouble is in trying to calculate the energy by applying the Hamiltonian density operator to a plane wave and integrating over all space. Since plane waves are unphysical and not elements of the Fock space, there’s no reason to think this will give a meaningful answer. $\endgroup$
    – J. Murray
    Dec 31, 2020 at 17:51
  • $\begingroup$ @BastamTajik If you go back to elementary QM and try to calculate $\langle p|H|p\rangle$ for the free particle Hamiltonian, you will also find the result to be infinite (proportional to $\delta(0)$) for what is ultimately the same reason. $\endgroup$
    – J. Murray
    Dec 31, 2020 at 17:54

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