Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How do we determine the energy density of a given system? I have seen that the density operator

$$\rho~=~\frac{\exp(-\beta \hat{H})}{\text{tr}(\exp(-\beta \hat{H}))}.$$

What does this mean exactly and how does it relate to the pure and unpolarized states of a system? For example, given a system of relativistic spin-1/2 particles, the completely unpolarized beam density is $\sigma~=~\frac{1}{2} \left| \uparrow\right>\left<\uparrow \right|+\frac{1}{2} \left| \downarrow\right>\left<\downarrow \right|$ or

$$\sigma~=~\left( \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array} \right)$$

Are the diagonal elements the probabilities that within the ensemble a particular particle will be found to be in that spin state? If so, how does this affect the energy density of the system? I am not entirely sure what the Hamiltonian would be in this case. But I think that spin-up and spin-down are eigenstates with eigenvalues $ \lambda_{\pm}=\pm \frac{\hbar}{2}$, so as a guess, since we are talking about spin-projections, that the energy eigenvalues are going to be proportional to $\lambda_{\pm}$

$$\displaystyle \rho~=~\frac{1}{\text{e}^{-\frac{\beta\hbar}{2}} + \text{e}^{\frac{\beta\hbar}{2}}} \left( \begin{array}{cc} \text{e}^{\frac{\beta\hbar}{2}} & 0 \\ 0 & \text{e}^{-\frac{\beta\hbar}{2}} \end{array} \right) $$

where $\beta$ is the inverse of temperature given by $\beta=\frac{1}{kT}$, however I am not sure what the constant of proportionality will be since I am assuming that the Hamiltonian will be proportional to $\sigma_z$ and not equal to it.

Would it be possible to see an example or have an explanation of what is happening? I am referring to Sakurai's Modern Quantum Mechanics and Quantum Mechanics by Bransden and Joachain.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Yeah, the example $\sigma=\frac{1}{2} \left | \uparrow \right \rangle \langle \uparrow|+\frac{1}{2} \left | \downarrow \right \rangle \langle \downarrow| $ you give describes a completely unpolarized ensemble for single spin-$1/2$ system, and the coefficients $1/2$ definitely represent the probabilities for the particle be in either up or down spin state.Now I will explain the meaning of the density operator $\rho =exp(-\beta H)/Z—(1)$ where $Z\equiv tr(exp(-\beta H))$ and its application to the single spin-$1/2$ system including its connections to the pure and unpolarized states.

Generally speaking, eqn(1) is the density operator for an equilibrium system $H$, its physical meaning becomes more clear if we rewrite it as $\rho =\frac{exp(-\beta E_1)}{Z}\left | 1 \right \rangle \langle 1|+\frac{exp(-\beta E_2)}{Z}\left | 2 \right \rangle \langle 2|+...$(you can prove this formula by yourself), where $\left | n \right \rangle $ are normalized energy eigenstates of $H$ with eigenvalues $E_n$. Here the coefficients $\frac{exp(-\beta E_n)}{Z}$ give the probabilities of the system being in state $\left | n \right \rangle $.

Now consider two extreme cases which are physically important:

(1)Low temperature limit$(\beta\rightarrow\infty)$ , $\rho =\frac{1}{D}\left | 1 \right \rangle \langle 1|+\frac{1}{D}\left | 2 \right \rangle \langle 2|+...+\frac{1}{D}\left | D \right \rangle \langle D|$, where $\left | 1 \right \rangle,\left | 2 \right \rangle,...,\left | D \right \rangle$ are the $D$ degenerate groundstates of $H$.

(2) High temperature limit$(\beta\rightarrow0)$, $\rho =\frac{1}{d}\left | 1 \right \rangle \langle 1|+\frac{1}{d}\left | 2 \right \rangle \langle 2|+...+\frac{1}{d}\left | d\right \rangle \langle d|$, where $d$ is the dimension of the system's Hilbert space, $\left | 1 \right \rangle,\left | 2 \right \rangle,...,\left | d \right \rangle$ are the energy eigenstates.

Now apply the above formulas to your example, let $H=\sigma_z$ be the Hamiltonian of the single spin-$1/2$ system. Then its energy eigenstates are just $\left | \uparrow \right \rangle $ and $\left | \downarrow \right \rangle $, its groundstate is simply $\left | \downarrow \right \rangle $(so here $d=2,D=1$). So for low temperature limit , we have $\rho= \left | \downarrow \right \rangle \langle \downarrow|$ (pure state you mentioned); and for high temperature limit, we have $\rho=\frac{1}{2} \left | \uparrow \right \rangle \langle \uparrow|+\frac{1}{2} \left | \downarrow \right \rangle \langle \downarrow| $(completely unpolarized states you mentioned).

Hope my answer is useful for you.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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