3
$\begingroup$

I was just reading the question Why can't $\psi(x)=\delta(x)$ in the case of a harmonic oscillator? The accepted answer says that $\psi(x)=\delta(x)$ is a mathematically valid state, though it's not physically possible because delta distributions aren't normalizable. If it's possible, I'd like to ignore the fact that it's un-physical, and instead try to fill in the holes in my understanding of the mathematical model.

The answer concludes that the expectation value of the Hamiltonian of the system is infinity, which makes sense to me because it follows from the proposition that $\left<p^2\right>$ is infinity.

However, I see this as a contradiction of my impression that quantum mechanical systems at high energies are easily approximated to classical behavior. For instance, with a particle in a square well, the position probability distribution at high energies is approximately constant across the whole allowed region, which is exactly what classical physics predicts for the same energy.

Going back to the case of harmonic oscillators, I think that a classical model with zero energy would have a position distribution of $\delta (x)$, because the particle's localized in one exact place and it isn't oscillating or moving in any way. So this tells me that a quantum harmonic oscillator with an extremely high energy behaves like a classical zero-energy system, which just generally sounds wrong.

Is it wrong/meaningless to make such comparisons of a zero-energy classical particle to a quantum harmonic oscillator with $\psi(x)=\delta(x)$? Alternatively, is this apparent contradiction actually logically justifiable?


I think that there's something fishy in my comparison when I say that a classical system shows a position distribution of $\delta(x)$, since the momentum distribution is also a delta function (with a spike at $p=0$). However, the QM handling of momentum would predict a completely different state. But I can't build more out of that reasoning to see if it's relevant.

Furthermore, the lowest energy state of a QM oscillator is nonzero, so there is no valid QM counterpart of a classical oscillator with $\psi(x)=\delta(x)$. However, this holds true for the particle in an infinite well too, so I would expect a symmetry whereby in both cases, high-energy QM descriptions are similar to classical mechanics.

$\endgroup$
  • $\begingroup$ For reference, the answer mentioned in the bounty description is physics.stackexchange.com/a/457479. Sorry, I forgot that detail. $\endgroup$ – user191954 Feb 2 at 12:43
  • 1
    $\begingroup$ It is not correct to assume that a particle with high energy can always be described by classical mechanics. Consider for example two states $|\psi_1>$ and $|\psi_2>$, in which particle has a high energy. Any superposition of them is a valid state in quantum mechanics, but does not make sense in classical mechanics. $\endgroup$ – atarasenko Feb 2 at 13:28
  • 1
    $\begingroup$ By the way, $\psi(x)=\delta(x)$ does not satisfy the Schrödinger equation, that is why is cannot be the ground state. $\endgroup$ – atarasenko Feb 2 at 13:38
  • $\begingroup$ @atarasenko Your first comment would make a very good addition to your answer (or perhaps it's a second answer). It's a good counterexample to show that the comparison isn't valid. $\endgroup$ – user191954 Feb 2 at 13:40
  • $\begingroup$ I have added info from comment to the answer $\endgroup$ – atarasenko Feb 2 at 13:55
5
+50
$\begingroup$

The most classical-like states of harmonic oscillator are Coherent states of harmonic oscillator, not $\delta(x)$. The reason is that coherent state has balanced uncertainty of coordinate and momentum: $$\Delta x =\sqrt{\frac{\hbar}{2m\omega}}; \Delta p = \sqrt{\frac{m\hbar\omega}{2}},$$ and in the classical limit ($\hbar\rightarrow 0$) $\Delta x \approx 0$, $\Delta p \approx 0$.

In a state with $\phi(x)=\delta(x)$ the momentum uncertainty $\Delta p=\infty$, it does not tend to zero.

UPDATE:

It is not correct to assume that a particle with high energy can always be described by classical mechanics. Consider for example two states $|\psi_1\rangle$ and $|\psi_2\rangle$, in which particle has high energy. Any superposition of them is a valid state in quantum mechanics but does not make sense in classical mechanics.

$\endgroup$
5
$\begingroup$

Your attempts to take a classical limit don't really make sense. By the nature of classical mechanics, outside of statistical mechanics, there is no such thing as a "position distribution" of a classical system. A classical system always has definite position and momentum, so its position and momentum "distributions" would be $\delta(x)$ and $\delta(p)$, respectively. That is true completely regardless of the specific state of the classical system or any other property of the system, so it is not a useful approach to think about what quantum system corresponds to this.

$\endgroup$
  • $\begingroup$ 2. There does exist a position distribution in a classical system, at least for bound states. You just have to average over period of motion. Then quantum position distribution will go to a classical one in the appropriate limit. $\endgroup$ – Ruslan Feb 2 at 17:56
  • $\begingroup$ @Ruslan I don't understand how "averaging" over a period of motion is supposed to yield a distribution. Do you mean you want to uniformly randomly sample the position of the particle in time and that for periodic motion the probability distribution of the classical position in that case is the same as that of some quantum state? What sort of quantum state? (It certainly can't hold for all of them, and it's not a priori clear how to take the classical limit of a state) $\endgroup$ – ACuriousMind Feb 2 at 18:01
  • $\begingroup$ Yeah, I was sloppy in formulating this. I meant exactly this. And I think it's not the limit of a state ­— rather a limit of a sequence of states. E.g. choose a set of conserved quantities to unambiguously define a classical state. Now construct its "closest" quantum counterpart (smoothing all the exact values to have reasonable uncertainties, depending on $\hbar$). Then take the limit $\hbar\to0$. I do admit that I started from classical "target" state here though, and I do agree that not every quantum state can be meaningfully/uniquely taken to a classical limit. $\endgroup$ – Ruslan Feb 2 at 18:08
  • $\begingroup$ I find the first point in this answer to be extremely disingenuous. It is indeed correct to point out that $\delta(x)$ is not an element of $L_2(\mathbb R)$. However, it is utterly misleading to give the impression that there are no ways to fix that (when they absolutely do exist) and paper over any mention of those solutions. This is particularly notable given that ACM has explicitly acknowledged that they're aware of those solutions and it's only being dismissed out of personal preference. $\endgroup$ – Emilio Pisanty Feb 3 at 0:28
  • $\begingroup$ @EmilioPisanty Using a rigged Hilbert space does not mean it is valid to apply the standard version of the uncertainty principle to the $\lvert x\rangle$ objects. In fact, you're not even justified to apply it to all states in $L^2(\mathbb{R})$ because $x$ and $p$ are only defined on dense subsets and the r.h.s ($[x,p]$) only exists on the intersection of the subsets upon which they are defined. All using a rigged Hilbert space does is allow you to rigorously state what sort of object $\lvert x\rangle$ is, it does not allow you to act as if it is a state. $\endgroup$ – ACuriousMind Feb 3 at 0:35
1
$\begingroup$

Your identification of classical mechanics as the high energy limit of quantum mechanics is mistaken. I can give a number of examples. Consider the high energy harmonic oscillator (fock/number) state

$$ |\psi \rangle = |n=1000\rangle $$

The $1000^{th}$ energy eigenstate of the harmonic oscillator. The amplitude of motion is well defined, however the phase is completely undefined. The oscillator could appear at any position and with any momentum consistent with $X^2 + P^2 = 1000$ (in the appropriate units).

Also consider the cat state

$$ |\alpha = 1000 \rangle + |\alpha = -1000\rangle $$

Where this is a superposition of two high energy coherent states. In this case the system is in a macroscopic superposition of two states which is highly non-classical.

These are two examples of quantum states which are high energy but non-classical.

I think it is likely true that generically low energy quantum states (states close to the ground state) are going to always be non-classical. Thus, high energy might be a necessary condition for a state to look classical but it certainly not a sufficient condition.

The state you are considering is $\langle x|\psi \rangle = \psi(x) = \delta(x)$. It is incorrect to say that this state has high energy. It is not an energy eigenstate and, as can be seen by a Fourier transform, it is a superposition of states with all possible energies. That is, it has low energy and high energy contributions. It has energy at all levels! This is part of why it is a highly non-classical state. I'll just add here that the reason this state is unphysical is because this state requires infinite energy to prepare and that is not available. However, you could prepare a state which has uniform occupation of all energy levels up to energy $E_0$. In this case the wavefunction would be a narrow function approaching a dirac delta. It would have width proportional to $\frac{1}{E_0}$, so the higher energy you can get to the closer you can approximate the $\delta$ function.

Now the question of a classical dirac delta function. Yes, a classical system in ANY deterministic state (ignoring thermal/stochastic states) will show up as a dirac delta in phase space. It will have both well defined position and momentum. One can see that the energy can be calculated to be $E = X^2 + P^2$ (again with some scalings) and importantly that the energy is well-defined and finite.

The big difference is that quantum mechanically it is impossible to have a delta function in both position and momentum. It simply isn't a valid state within Hilbert space. One way of seeing this is because position and momentum are Fourier transform pairs and the dirac delta function is not its own Fourier transform. This is the main statement of the wavefunction version of the uncertainty principle.

Said yet another way, the ground state classical oscillator has $X=0$ and $P=0$. It is not moving. It has no energy. A quantum oscillator with $\psi(x) = \delta(x)$ has $X=0$ but it has ALL possible momenta. Meaning it is simultaneously not moving (analogous to the classical oscillator) and moving infinitely fast in all directions. This is very different from the classical oscillator.

I'm saying a lot here and I feel like I'm repeating myself so perhaps it would be best for you to ask any clarifying questions.

$\endgroup$
  • $\begingroup$ "That is, it has low energy and high energy contributions. It has energy at all levels! " => so the "average" energy would be very high (infinite). So the delta function state can be regarded as a high energy state. But I like your example of Fock state with large quanta as a counterexample. $\endgroup$ – wcc Feb 4 at 22:47
  • $\begingroup$ Yes, the delta function state does have an "average" energy of infinity, I don't disagree. I chose to phrase it in the somewhat strange way that I did to connect better to the OPs hope of connecting $\psi(x) = \delta(x)$ to a classical state. Thought of in phase space the spatial wave function $\psi(x) = \delta(x)$ looks like the classical state the OP is imagining. The momentum distribution looks something like $\phi(p) = 1$. So it does have a component at 0. $\endgroup$ – jgerber Feb 5 at 1:23
  • $\begingroup$ To that end the quantum state $\psi(x) = \delta(x)$ "contains" a component which "looks like" the classical state the OP is imagining. However, it also "contains" components of higher energy as well. This is what I was trying to emphasize. $\endgroup$ – jgerber Feb 5 at 1:23
1
$\begingroup$

Emilio Pisanty's answer sets $\hbar = 1$, which conceals some subtleties regarding the classical limit. With explicit $\hbar$, the Fourier transform contains $e^{i px/\hbar}$, which makes it hard to interpret the meaning of the classical limit $\hbar \to 0$. It's easier to see what's going on if you consider a Gaussian wavepacket with a position-space spread $\Delta x$ instead. You'll find that the spread in momentum space $\Delta p \propto \hbar / \Delta x$. So if we take $\Delta x \to 0$ while holding $\hbar$ fixed to get a $\delta$-function position wave function, as Emilio does, then we find that the kinetic energy goes like $(\Delta p)^2 \propto \hbar^2 / (\Delta x)^2$ and diverges. Physically, of course, $\hbar$ is fixed, so this is usually the logical thing to do. But in the classical limit it's more useful to think of taking $\hbar \to 0$ before we take $\Delta x \to 0$. In this case we get a "deterministic" classical particle with both $\Delta x = \Delta p = 0$, which can have zero energy.

That's all somewhat heuristic, because we should really be taking dimensionless ratios to zero. The more rigorous way to do it is to consider the characteristic length and momentum scales inherent to the particular oscillator potential, and think of the wave-packet spreads relative to those scales. You'll find that the the classical limit really corresponds to regimes in which both the position-space and momentum-space spreads are much smaller than those characteristic scales, so that the particle is almost perfectly localized in both position and momentum space. In this case we find that the "classical energy" turns out to be much larger than the zero-point energy $\propto \hbar \omega$, so the latter can be neglected. That's the rigorous way to justify whether to take $\Delta x \to 0$ or $\hbar \to 0$ first; the correct choice reflects the relative position spread of your wavefunction to the relevant length scale set by the Hamiltonian.

$\endgroup$
  • $\begingroup$ Hmmmm. It does seem that that answer had an inconsistent use of $\hbar$; it's been edited to fill in the missing $\hbar$s. I don't think it affects the question or the core of this answer, but you may want to rephrase a bit. $\endgroup$ – Emilio Pisanty Feb 3 at 3:27

Your Answer

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