Degrees of freedom in Quantum Mechanics

Question

If we look at a particle in classical mechanics, the degrees of freedom increase as its size decreases like the degrees of freedom of an atom is more than that of molecule, and subsequently, the degrees of freedom of an electron is more than that of an atom. However, is the 'degrees of freedom' for a particle fixed in Quantum Mechanics? if so, what determines its degrees of freedom in Quantum Mechanics.

Answer 1

I think you are just confused about the words. Textbooks say quantum "particle" to mean "anything which happens to have the Hamiltonian we're talking about right now". Consider, for example, the harmonic oscillator Hamiltonian

$$H = \frac{\hat{p}^2}{2m} + \frac{1}{2} k \hat{x}^2 \, .$$

This Hamiltonian has two degrees of freedom: $x$ and $p$. It could apply to a single electron in one dimension under the influence of a parabolic potential. In that case we truly have a single "particle", and it really has two degrees of freedom: $x$ and $p$ (ignoring spin). However, the same Hamiltonian applies to an atom in a one dimensional crystal lattice (for small amplitude displacements). If we only consider the translational motion of the atom, then it too has two degrees of freedom: $x$ and $p$. In this case, the atom can be regarded as a "particle". However, if we perturb the atom with higher energy radiation we might excite one of its electrons into a higher state, or even cause that electron to leave the atom. If we're dealing with these higher energy processes, then we have to take into account the degrees of freedom internal to the atom, such as the electron degrees of freedom. In that case, the harmonic oscillator and its two degrees of freedom aren't enough to describe the system. So you see, the number of degrees of freedom is not something fixed in quantum mechanics, it depends on the system you are studying and the range of energies which perturb that system.

By the way, the quantum harmonic oscillator Hamiltonian applies even to a macroscopic $LC$ circuit under the right conditions. For example, if the circuit is made of superconducting metal and protected from too much external noise from radiation etc., then the electrons all sit in the superconducting ground state and can be ignored. In this case, only the current $I$ and voltage $V$ of the circuit are active degrees of freedom and again the Harmonic oscillator Hamiltonian applies:

$$H = \frac{1}{2}CV^2 + \frac{1}{2}LI^2 \, .$$

So you see, even an object as big as a circuit with a huge number of internal degrees of freedom including billions of atoms can be represented with a Hamiltonian with only two degrees of freedom in the right conditions.

Now that we said all of that, please note that whether or not a system is quantum mechanical isn't really the issue when you're deciding the number of degrees of freedom. What matters is what parts of the system are active. As CuriousOne says in the comments, the Earth can be regarded as a ball with six degrees of freedom: $x$, $y$, and $z$ components of position and momentum, but of course all of the $>10^{23}$ atoms in the earth have their own internal degrees of freedom. If you're calculating orbits, you don't care about those internal atoms and you ignore their degrees of freedom.