Question About Quantum Numbers

[Aero51]

I have a question about a specific passage in Hill's book on statistical thermodynamics. I have decided to teach myself the subject, but with having taken any classes in modern physics sometimes I encounter foundational material that I have no experience with. This question should be pretty straight forward to someone with more experience:

In the beginning of chapter 4, Hill starts to describe an ideal monoatmoic gas with three translational degrees of freedom. The gas is confined into a container with volume L³. The formula for possible energy states is given:

\epsilon_{l_x l_y l_z} = h^2 ({l_x}^2 + {l_y}^2 + {l_z}^2)/(8 mL^2)

where l_x, l_y, l_z are "quantum numbers".

This brings me to my first question:
What is a quantum number in the context of this problem? Does it represent a coordinate or a particle with a specific energy state indexed by l_x, l_y, l_z?

Second:
What does "h" represent? Judging by the equation I would guess it is the momentum in Newtonian mechanics so the units cancel out appropriately.

In addition Hill describes what is called l_xl_yl_z space. Would this simply be the set of all molecules who have accessible quantum states described by the quantum numbers above?

Thanks

 

[lugita15]

This brings me to my first question:
What is a quantum number in the context of this problem? Does it represent a coordinate or a particle with a specific energy state indexed by l_x, l_y, l_z?

The quantum numbers l_x, l_y, l_z index the energy states. It's important to note, though, that different triplets (l_x, l_y, l_z) can give rise to the same value of energy. That's known as degeneracy.

Second:
What does "h" represent? Judging by the equation I would guess it is the momentum in Newtonian mechanics so the units cancel out appropriately.

It's Planck's constant, and it has units of angular momentum, i.e. kg m²/s.

In addition Hill describes what is called l_xl_yl_z space. Would this simply be the set of all molecules who have accessible quantum states described by the quantum numbers above?

Each point in this space is a particular quantum state, indexed by the three quantum numbers.

 

[Aero51]

The quantum numbers lx, ly, lz index the energy states. It's important to note, though, that different triplets (lx, ly, lz) can give rise to the same value of energy. That's known as degeneracy.

I always thought degeneracy was the instance where all energy states have an equal probability of being accessed. I assume you are describing a degenerate case between two molecules and not the whole system. Also, how do the quantum numbers relate to the spatial coordinate of a specific molecule? In other words, how do the quantum numbers correlate to the energy distribution inside the volume?

On a side note the difference between an energy level and energy state is that the former represents a magnitude while the latter is a classification (rotational, transnational, etc)?

 

[unchained1978]

Degeneracy arises when different quantum numbers give rise to the same energy. For the particle in a box, the state n_{x}=2 , n_{y}=1, n_{z}=2 has the same energy according to the formula as the state n_{x}=1 , n_{y}=2, n_{z}=2 even though they're different states. The quantum numbers do not correspond to the spatial coordinates of a particle, they describe a representation of the state. The principal quantum number n determines which energy level a particle is in, the azimuthal number l describes the orbital angular momentum of the particle and therefore which orbital the particle is in (in the case of a hydrogen atom) and the spin quantum number m_{s} describes which way the spin of the particle is oriented.

 

[Aero51]

Ok, what would properties such as position and momentum be described as if they are not states as described above. How are they represented in quantum mechanics?

 

[Rap]

Hill is describing a single particle confined to a cube of side L. The particle is described in QM by a wave function \Psi(x,y,z,t), which describes what you know about the particle. The Heisenberg uncertainty principle says you cannot know everything about the particle all at once, and the wave function and the way you calculate with it makes sure that this always holds true. For example, you cannot know the position of the particle and its momentum at the same time. You know the particle is inside the box somewhere, so that means you cannot know the momentum of the particle [px,py,pz] exactly, but you can know the absolute value of the momentum components |px|, |py|, |pz|. It turns out that if you know the absolute value of the momentum components and that the particle is inside the box, there is a wave function that you can write down which describes this situation. It turns out that not all momenta are allowed. The only momenta that are allowed are [px,py,pz] where |p_x|=h l_x/2 L etc. for y and z, where lx=1,2,3...m etc. So you can label these particular wave functions by their lx, ly, and lz. Those three positive integers are all you need (along with L) to write down this special type of wave function describing the situation inside the box. In other words, \Psi(x,y,z,t) is any wave function that might describe the particle in the box, but the special wave functions (called momentum eigenfunctions) can be written down as \Psi_{l_x,l_y,l_z}(x,y,z,t) and these are explicitly known functions.

The energy of the particle is (|px|^2+|py|^2+|pz|^2)/2m ( e.g. mv^2/2=(mv)^2/2m) and that's where the energy relationship you wrote down comes from.

When they talk about lx, ly, lz space, just think of a 3 dimensional space, with points on the positive integer triplets. So [1,2,5] is in that space, but [1,1.72,5] is not, and [-1,2,5] is not.

 

[eudo]

One thing that I think is missing from the previous responses is this: You need to pick up a good intro to quantum mechanics book. The real problem is that you're trying to run without learning to walk first. That's a painful and inefficient way to go about things. It's like trying to teach yourself calculus without learning trigonometry. I'm guessing that as you get more and more into your book you're going to run into more cases like these.

A quantum particle in a box problem is solved pretty early on in any quantum 101 book, and your thermo book is clearly assuming you know this stuff cold.

So trust me, your life will be a lot easier and more enjoyable if you pick up that quantum book now.

 

[Rap]

One thing that I think is missing from the previous responses is this: You need to pick up a good intro to quantum mechanics book. The real problem is that you're trying to run without learning to walk first. That's a painful and inefficient way to go about things. It's like trying to teach yourself calculus without learning trigonometry. I'm guessing that as you get more and more into your book you're going to run into more cases like these.

A quantum particle in a box problem is solved pretty early on in any quantum 101 book, and your thermo book is clearly assuming you know this stuff cold.

So trust me, your life will be a lot easier and more enjoyable if you pick up that quantum book now.

I agree, if you know little quantum mechanics, you need to get up to speed on that before tackling the quantum approach to thermodynamics.

 

[Aero51]

I was hoping that I could pick up some of the material as I went along, but it is clear that won't be sufficient. I'm thinking about auditing a modern physics class at my university, then continuing with stat thermo once a firm foundation has been developed.

 

[unchained1978]

Positions and momenta are represented by Hermitian operators in a hilbert space. Hermitian operators are a generalization of matrices such that A^{\dagger}=A, which gives rise to real eigenvalues and hence observable quantities (such as position, momentum, etc...). This will make much more sense if you've taken a course on linear algebra. Essentially, and operator is a matrix such that Ax=ax where A is your matrix in question, and a is the eigenvalue of that operator. The x's are the eigenvectors for this operator. For example, given a wavefunction \psi(x,t)=A_{0}e^{i\textbf{k} \cdot \textbf{x}-\omega t}, we can treat this wave function as an "eigenvector" of the momentum operator \hat p =-i\hbar\vec\nabla so that \hat p \psi(x,t)=\vec p \psi(x,t) given the deBroglie relations \vec p=\hbar \vec k. The position operator obeys a similar relationship.