Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Zurek 2001 is a review article on decoherence in quantum mechanics. Equation 5.36 on p. 24 gives an estimate of the decoherence time, which I'll paraphrase as follows:

$ \frac{t_D}{t_R} = \left(\frac{\lambda_T}{x-x'}\right)^2 $ .

My attempt to interpret the meaning of the variables is:

$t_D$ is the decoherence time

$t_R$ is the thermal relaxation time for the object, which isn't perfectly isolated from its environment

$\lambda_T$ is the de Broglie wavelength for the mass of the whole object at an energy corresponding to the temperature of the environment

$x-x'$ is the difference in position between two states of interest, i.e., $t_D$ is going to be the time-scale for the exponential decay of the off-diagonal element of the density matrix corresponding to $x$ and $x'$

The result is that for a macroscopic object such as Schrodinger's cat at room temperature, we have $t_D/t_R\sim10^{-40}$.

Question 1: Am I correctly interpreting the meaning of the variables?

Question 2: Can anyone give a hand-waving argument for why we have this particular dependence on the variables?

Zurek arrives at this estimate after a very long derivation that to me is not very understandable. All I've been able to accomplish is some plausibility arguments:

  1. In the classical limit we expect the ratio inside the parens to go to zero, and it does, since the de Broglie wavelength goes to zero.

  2. Coherence is more difficult to maintain at macroscopic scales, where $|x-x'|$ is large.

  3. The result is positive, since the exponent is even.

Zurek, "Decoherence, einselection, and the quantum origins of the classical,"

share|cite|improve this question
Thermal relaxation time? I havent heard of that term before. Is that the same as time scale for equilibration – Prathyush Aug 21 '13 at 10:21
up vote 2 down vote accepted

The equation (5.36) in the (newest version) of the paper is:

$$ρ_S (x, x′ , t) = ρ_S (x, x′ , 0) e^{ −γt\left(\frac{x−x'}{\lambda_T}\right)^2} \tag{1}$$

where $\rho_S$ should be the off-diagonal term in the density operator, $\gamma$ the relaxation coefficient and $\lambda_T=\frac{\hbar}{\sqrt{2Mk_BT}}$ is the thermal de Broglie wavelength of the system. The equation follows from their master equation(which they derived earlier in the paper), in the high-temperature limit.

Master equation: $\dot \rho_S=-\frac{i}{\hbar}\left[ H_{ren},\rho_S\right]-\gamma(x-x')(\partial_x-\partial_{x'})\rho_S-\frac{\gamma}{\lambda_T^2}(x-x')^2\rho_S $

where $H_{ren}$ is the renormalized Hamiltonian.

In the macroscopic limit (i.e. when $\hbar$ is small compared to other quantities with dimensions of action, such as $2M k_B T (x − x′ )^2$ in the last term) the high-temperature master equation is dominated by:

$$\dot \rho_S(x,x',t)= -\gamma \left( \frac{x-x'}{\lambda_T}\right)^2\rho_S(x,x',t)$$

Solve this, and equation $(1)$ follows smoothly. Now, the exponential decay of the off-diagonal terms makes it plausible to define a decoherence timescale $\tau_D$:

$$\tau_D=\gamma^{-1}\left(\frac{ \lambda_T}{x-x'}\right)^2$$

Now let's get back to O.P.'s questions; I think the answer to the first question is yes, the interpretation looks correct($\tau_R=\gamma^{-1}$ is the thermal relaxation time).

About the second question, I would be more than happy if the O.P. finds the above derivation enough hand-wavy! Else, there is a typical representation of what's going on in this other well known paper(around page 13) by the same author. The reasoning is around the lines of having Gaussian distribution wave packets and doing numerical simulations and seeing what happens.

share|cite|improve this answer

protected by ACuriousMind yesterday

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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