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Do states with infinite average energy make sense?

For the sake of concreteness consider a harmonic oscillator with the Hamiltonian $H=a^\dagger a$ and eigenstates $H|n\rangle=n|n\rangle$, $\langle n|m\rangle=\delta_{n,m}$. Then prepare a state $$|\psi_m\rangle = c \sum_{n=1}^\infty\frac1{n}|n^m\rangle,\qquad m\in\mathbb{N}.$$ This state is normalizable (with normalization constant $c^{-2}=\sum_{n=1} \frac1{n^2}$) and hence seems to be an OK quantum state. However, its average energy $$\langle\psi_m|H|\psi_m\rangle=c^2\sum_{n=1}^\infty n^{m-1}$$ is infinite for any $m\ge0$.

There is no direct contradiction here, but I have never thought about such states before and they seem puzzling. Do they make sense? Do they appear in theory/experiment? Does it take an infinite amount of energy to prepare such a state from the vacuum state?

Qmechanic
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Weather Report
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  • Possible duplicate: Expectation value of Hamiltonian? – Tobias Fünke Nov 22 '21 at 20:25
  • Wouldn't the Cauchy-Schwarz inequality imply that $H\left|\Psi_m\right>$ has an infinite norm and thus isn't in the domain of the inner product? So perhaps $\left|\Psi_m\right>$ is fine but calculating the expectation value as $\left<\Psi_m|H|\Psi_m\right>$ is invalid? – user7896 Nov 22 '21 at 21:23
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    It is impossible to resist saying that the sum has a nice regularized value: $\zeta(1-m)$. I make no claims that this regularization is physical – Sal Nov 22 '21 at 21:36
  • In position representation your state has fast decay for $x\to-\infty$ and a heavy tail $\sim x^{-1.5}$ for $x\to+\infty$ – Ruslan Nov 22 '21 at 23:43

3 Answers3

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If you adopt the frequentist perspective, the fact that $\langle E\rangle$ does not exist simply means that if you measure the energies of $N$ identically-prepared systems and average the results, then that number does not approach a well-defined limit as $N\rightarrow \infty$.

If the spectrum of the Hamiltonian is unbounded (as it very often is), then such states must exist in the Hilbert space, so they certainly appear mathematically. I personally don't find anything necessarily unphysical about them either; $\langle E\rangle \rightarrow \infty$ doesn't mean that any measurement returns infinity as a result, but rather that increasing the number of measurements tends to increase the average energy.

For example, consider a particle in a box of length $a$ with wavefunction $\psi(x) = 1/\sqrt{a}$. This corresponds to the position of the particle being completely unknown (other than the fact that it's in the box). If you compute $\langle E\rangle$ in this state, you will find that it diverges to infinity. The particle in a box is of course a mathematical idealization, but so is every model we ultimately use, and since this one is very plausibly a useful model I'm not so willing to write it off.

Your example is tricker than it seems since [$\psi$ does not satisfy the boundary conditions associated with the domain of $H$]

@ZeroTheHero raises an excellent point. $H$ (whose domain is twice weakly differentiable functions with Dirchlet boundary conditions) is self-adjoint (as all Hamiltonians must be), but it is correct that the $\psi$ I've written down is not in its domain. In this respect, we cannot write $\langle E\rangle = \langle \psi,H\psi\rangle$ simply because the right-hand side is illegal!

This does not necessarily mean that $\langle E \rangle$ is undefined, but it should make us suspicious. The correct thing to do is to expand $\psi$ in terms of the normalized eigenvectors of $H$: $$\psi = \sum_n c_n \phi_n \qquad H\phi_n = E_n\phi_n$$ which can always be done due to the self-adjointness of $H$. We then define $\langle E\rangle := \sum_n E_n |c_n|^2$. It's easy to see that this coincides with $\langle \psi,H\psi\rangle$ when $\psi\in\mathrm{dom}(H)$, but is in fact slightly more general, and it is this calculation which diverges to infinity. Indeed, if we naively calculate $\langle \psi,H\psi\rangle$ by differentiating $\psi$ twice, we get zero.

J. Murray
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  • OK, that makes good sense to me! In my case a state with the infinite average energy appears as a result of an evolution in an open quantum system. I'm trying to understand if this means that I used incorrect approximations along the way and in particular pumped an infinite amount of energy into the system, or that the resulting state can still be trusted to compute some observables (presumably not the energy). – Weather Report Nov 22 '21 at 20:36
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    @WeatherReport Unfortunately I'm not an expert in open quantum systems, so I'm not the right person to speak to your specific example. However, it's worth noting that even in my particle-in-a-box example, the probabilities of measuring the system to have any particular energy are still well-defined and all of those probabilities add up to one, even though $\langle E\rangle \rightarrow \infty$. Point being, I could imagine the introduction of an infinitely large thermal reservoir could produce infinite expectation values without making the state itself inherently useless. – J. Murray Nov 22 '21 at 20:55
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The infinite expectation value for the energy arises due to the fact that the harmonic potential extends to infinity. As a result one can keep going up the ladder of states indefinitely. Practically such an infinitely high potential cannot exist. After a certain energy level there are not going to be any bound states and you will be left with scattering states only.

So there is no contradiction and such unintuitive behaviour is a consequence of the unphysical nature of the potential. A similar thing happens with the infinite potential well where the first derivative of the wavefunction is discontinuous. Again theoretically no contradiction but physically unrealisable.

user7896
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  • Can't I play the same game with a free particle (where all states are scattering)? Prepare a state $|\psi\rangle=\int dE f(E) |E\rangle$ where $\langle \psi|\psi\rangle=\int dE,, |f(E)|^2$ converges but $\langle \psi|H|\psi\rangle=\int dE,, E|f(E)|^2$ does not? – Weather Report Nov 22 '21 at 19:20
  • Even in this case, as you point out in your question, you could never prepare such a state in the actual universe without access to an infinite amount of energy. – sasquires Nov 22 '21 at 19:39
  • The reason for intinite energy is the inclusion of the infinitely high energy eigenstates with finite probability to be found in them. One can play this game with any unbounded spectrum. Physically, however, this is not possible, as this answer correctly points out. – Roger V. Nov 22 '21 at 19:54
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    @RogerVadim well, infinite energy state has zero probability in my construction, so your description is a bit inaccurate. Inclusion of arbitrarily high energy eigenstates is not a problem by itself, consider a coherent state as a reasonable example. Moreover, if the states I am describing are really somehow banned by nature, that would imply that the state space does not really include all normalizable superpositions, which seems to be a non-trivial statement. – Weather Report Nov 22 '21 at 20:10
  • @WeatherReport You are confusing what is allowed/banned by nature and what is allowed/banned by physical theory / math. Also, my description is quite precise: for any energy $E < +\infty$ there is such a finite $\epsilon$ (dependent on $E$) that the probability of finding the system in this state is greater than $\epsilon$. – Roger V. Nov 22 '21 at 20:15
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    @RogerVadim this is subtle but I disagree. In particular, it is not obvious to me why my $\psi_m$ state is worse than a coherent state and should be considered unphysical? – Weather Report Nov 22 '21 at 20:42
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    @WeatherReport they say that "all models are wrong, but some of them are useful". Physics is an approximate description of nature created by humans. This applies to the coherent states as well - such a state created in a lab is only approximately coherent. – Roger V. Nov 22 '21 at 21:21
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To interpret $\langle E \rangle = \infty$ one has to distinguish between the possible values, the expected value and the most probable value of energy.

All eigenvalues $\{1,2,3,\dots\} =\mathbb N$ of the Hamiltonian are possible energies for the state you suggest (with that $m=1$ of course). Every single eigenenergy is a finite number, but there is no highest energy, the spectrum is unbounded from above.

This state, however, has the lowest energy as the most probable one, and the probability decreases with increasing energy. $$ (p_1,p_2,p_3,\dots) = \frac{1}{\sum_{n=1}^\infty \frac{1}{n^2}} \left(\frac{1}{1^2}, \frac{1}{2^2},\frac{1}{3^2},\dots \right) $$ So the lowest energies are those one gets every day upon measurement of energy in that state.

Now for the expectation (or average value) it is (in general) not necessarily a possible value nor the most probable. Furthermore the distribution for this state has an indefinitely large mean value (with $E_n = n$) $$ \langle E\rangle = \sum_{n=1}^\infty p_n E_n = \frac{\sum_{n=1}^\infty \frac{1}{n}}{\sum_{n=1}^\infty \frac{1}{n^2}} = \infty, $$ but what you get with measurement is always a finite number in the spectrum. It is still possible, (in theory at least,) by chance, to measure as high energy as you like, as long as you keep measuring particles in that state forever, since it is unlikely to get those low energies every day.

Physor
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