What is the temperature of a quantum particle in a box?

Some simple examples in textbooks include simple 1D systems such as particle in an infinite potential well or in harmonic oscillator potential.

It is also said that at absolute temperature of the particle is zero in the ground state.

My question is how temperature enters this quantum picture? because we never started with anything related to temperature, where does it come from?

• Temperature is introduced as a Lagrange multiplier in statistical physics. If all particles occupy the lowest state, it is 0 K. – user26143 Nov 29 '13 at 13:47
• Temperature is a statistical phenomenon. If you have only one harmonic oscillator in a definite state, it's stretching things to say it has a temperature. If you have a large number of harmonic oscillators in the ground state, they have temperature zero. If you have a large number of harmonic oscillators in the first excited state, this isn't an equilibrium distribution so, strictly speaking, they don't have a temperature. – Peter Shor Nov 29 '13 at 14:41

Let a quantum system with Hilbert space $\mathcal H$ and hamiltonian $H$ be given.

If the system is in equilibrium with a heat bath at temperature $T$, then the system is in a so-called mixed state and is modeled by a linear operator on $\mathcal H$ instead of as a vector in $\mathcal H$. This operator is called the density matrix (or density operator). For the specific situation in which the system can exchange energy with the heat bath with which it is in equilibrium but is otherwise closed, the system is said to be described by the canonical ensemble, and its density operator is given by \begin{align} \rho(\beta) = \frac{1}{Z(\beta)}e^{-\beta H}, \qquad Z(\beta) = \mathrm{tr} (e^{-\beta H}), \qquad \beta = \frac{1}{kT} \end{align} When we say that a single particle in a box is at temperature $T$, we mean that its state is given by the density operator defined above where $H$ is the Hamiltonian for the single particle in a box.

You might then ask what this particular density operator has anything to do with temperature the way we usually understand it and why we need such a beast to model systems with temperature. It would take far to long to explain that here, but that's something that is explained in books on quantum statistical mechanics. If you plan to explore this stuff further, you might find the following physics.SE posts helpful:

States versus ensembles in quantum mechanics

(Canonical) Partition function - what assumption is at work here?

Why is (von Neumann) entropy maximized for an ensemble in thermal equilibrium?

Addendum. I showed in the following answer why a system in the canonical ensemble occupies its ground state at zero temperature:

https://physics.stackexchange.com/a/65167/19976

• So the above replies are wrong? – Revo Nov 29 '13 at 18:45
• @Revo No they're actually both consistent with this response. You can, for example, show that at zero temperature, the system will occupy its ground state (provided it is non-degenerate, otherwise it's a little more subtle than that). See the addendum for a post with more details on this fact. None of Peter's remarks are inconsistent with what I wrote either as far as I can tell. – joshphysics Nov 29 '13 at 18:59