7
$\begingroup$

In classical mechanics, a particle confined to move along only the $x$-direction can be fully described by a 2-tuple $(x_1,p_1)$ in phase space. In this case, the phase-space is clearly 2-dimensional. Similarly, a particle free to move in three spatial directions can be fully defined by a 6-tuple in phase space $(x_1,y_1,z_1,p_{x,1},p_{y,1},p_{z,1})$. Now phase space is 6-dimensional. More generally, if we have a system of $N$ point particles, each free to move in 3 directions, then the phase space of the system will have a dimension of $6N$.

If we now move over to quantum mechanics, and we examine an isolated particle free to move in 3 dimensions (confined to a box say), phase space appears to require only 3 dimensions to fully specify the state of the system (one dimension for each quantum number so $(n_x,n_y,n_z)$ defines a point in phase space) as opposed to 6 dimensions in the classical case (where $(x_1,y_1,z_1,p_{x,1},p_{y,1},p_{z,1})$ is a point in phase space). Is this true? It seems to me like this reduction in dimensions could be the result of momentum being undefined when the position is well-defined and vice versa. Is my thinking correct or am I totally off the mark?

I ask this because my lecturer has said that in classical mechanics, the energy of the state of a system represented by the phase point $(q^1,...q^f,p_1,...,p_f)$ is given by a function of the form $$E=E(q^1,...q^f,p_1,...,p_f).$$ If we then wish to introduce some QM, we divide phase space into cells of phase volume $h^f$, each cell corresponding to one state, then the number of allowed states in the phase space volume $dV=dq^1...dq^fdp_1...dp_f$ will be $\frac{dq^1...dq^fdp_1...dp_f}{h^f}$. Now this result makes sense to me from the point of view in classical mechanics since in this domain, for each $q^i$ we have a corresponding $p_i$. But in quantum mechanics there is no corresponding $p_i$ so should the number of allowed states not then be $\frac{dq^1...dq^f}{h^f}$ in QM instead of the classical case where we have $\frac{dq^1...dq^fdp_1...dp_f}{h^f}$?

Any help on this would be most appreciated!

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ I get the impulse but I don't get the argument that you only need three quantum numbers to describe the quantum system. You need only three quantum numbers to describe a system if the system is assumed to be an eigenstate, not a generic system. Am I missing something? $\endgroup$
    – user87745
    May 13, 2021 at 18:22

2 Answers 2

Reset to default
11
$\begingroup$

The most meaningful language to compare classical mechanics to quantum mechanics is the phase-space formulation of the latter, based on an invertible map between Hilbert space and phase-space: you want to compare apples with apples.

In this formulation, a particle is represented not by a mere point in phase space, but, instead, by a constrained distribution in phase space, a Wigner function. QM forces it to be "fuzzy" and somehow distributed over cells of size never smaller than roughly ℏ, which gives you the same dqdp/ℏ measure in state counting. This spreading out is a consequence of the uncertainty principle, which is embedded in the formulation, automatically, and prevents sharp trajectories. A particle is a profile distribution, usually centered on the classical point in phase space. (Even outside phase space, an x-wavefunction is much "larger" than a point it is centered at and its velocity.)

So, you might think, as you possibly did, that QM has lost information vis-a-vis classical mechanics, but it is the other way round. The classical limit has higher, not lower, entropy than the original QM description. (You know that: think of the ℏ-dependent info suppressed in the classical limit. Think of quantum computing.) In a sense, the momentum information of the classical trajectory is embedded in the x-spread of the Wigner function of its quantum progenitor, which the formalism correlates to its p-profile in an eerie way, thus ensuring the uncertainty principle prevails.

This is all words, and won't make much sense before you do some simple illustrative calculations, for which it serves as a trail map. A free particle, a linear oscillator mode, a Morse potential state, etc... might illustrate the point. There are nice YouTube videos illustrating the evolution of such states. (I also have a video there, "Deformation Quantization: Quantum Mechanics Lives & Works in Phase-Space".)

Improve this answer
$\endgroup$
4
  • $\begingroup$ Might be worth mentioning that algebraic quantum field theory promotes this to an axiom (called the nuclearity property). Roughly, it says that the part of the Hilbert space consisting of states with energy below a given upper bound and which look enough like the vacuum state outside a given volume must be "essentially finite dimensional". $\endgroup$
    – Chiral Anomaly
    May 13, 2021 at 17:20
  • 1
    $\begingroup$ @ChiralAnomaly. Thanks; it's a bit outside my comfort zone. I'm just dishing out the standard consensus in the street... Perhaps you could expand it to an answer yourself? $\endgroup$
    – Cosmas Zachos
    May 13, 2021 at 17:26
  • 1
    $\begingroup$ Buchholz, D., & Jacobi, P. (1987): "On the nuclearity condition for massless fields", Letters in mathematical physics 13 (4), 313-323 seems to be the mother of them, but distinctly beyond my comfort zone... $\endgroup$
    – Cosmas Zachos
    May 13, 2021 at 17:38
  • 1
    $\begingroup$ I don't know enough about the mathematical intricacies of the nuclearity condition(s) to expand it into an answer, but the paper you linked is representative. $\endgroup$
    – Chiral Anomaly
    May 14, 2021 at 1:42
0
$\begingroup$

One must be careful to specify the correct phase space for a quantum system. The phase space for a particle in a box is not exactly the same as the phase space for a free particle (there's a rigorous discussion here).

The phase space for a quantum harmonic oscillator is the same as for a classical harmonic oscillator; it has an $x$ coordinate and a $p$ coordinate. This phase space describes things like electromagnetic fields in quantum optics. Even though there is a single quantum number describing the nergy eigenstates, the phase space is still two-dimensional.

As another example, the phase space for a 2-level atoms is a sphere (the Bloch sphere). This has two angular degrees of freedom. This is true even though, again, there is only one quantum number involved. Phase space dimension and number of quantum numbers are different things.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.