Statistical operator of hydrogen atom

You're trying to write the density matrix in terms of a sum of pure states, each with some probability. But there are some errors in your notation, and you're missing some quantum numbers. Also, the probability is not associated with each state, but with each set of quantum numbers. So the correct expression should be:\hat{\rho} = \frac{1}{P} \sum\limits_{n = 2}^{2} \left(\sum\limits_{l = 0}^{1} \left(\sum\limits_{m_{l} = -l}^{l} \left(\sum\limits_{m_{s} = -1/2}^{1/2}|
  • #1
sunrah
199
22

Homework Statement


Individual hydrogen atoms have been prepared in the energy state n = 2. However, nothing is known about the remaining quantum numbers. Fine structure and all corrections can be ignored.

What is the micro-canonical statistical operator.

Homework Equations



[itex] \hat{\rho_{mc}} = \frac{1}{\Omega (E)}\delta(E-\hat{H})[/itex]

[itex] \hat{H} = -\frac{\hbar^{2}}{2m_{p}}\Delta_{p} -\frac{\hbar^{2}}{2m_{e}}\Delta_{e} - \frac{e^{2}}{4\pi \epsilon_{0}}\frac{1}{\left|r_{p} - r_{e}\right|}[/itex]

[itex] \Omega (E) = Tr( \delta(E-\hat{H}) )[/itex]

The Attempt at a Solution



I don't understand this. If the atoms are all in state n = 2 then with have a system in a pure state: therefore the probability [itex]p_{i} = \frac{1}{\Omega (E)} = 1[/itex] and [itex]\rho = |2\rangle \langle 2|[/itex].

is that so?
 
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  • #2
All of the atoms have the same energy determined by the quantum number n = 2. But that doesn't mean that all the atoms are in the same quantum state. Consider the "remaining quantum numbers" and list all the possible quantum states that have this energy.
 
  • #3
TSny said:
Consider the "remaining quantum numbers" and list all the possible quantum states that have this energy.

n = 2
l = 0, 1
ml = -1, 0, 1
ms = -1/2, 1/2

so if the generalised state vector is [itex]|n,l,m_{l},m_{s}\rangle [/itex] then

[itex]\hat{\rho} = \frac{1}{P} \cdot \sum\limits_{i = 0}^{1} \left(\sum\limits_{j = 0}^{2} \left(\sum\limits_{k = 0}^{1}|2,i,j-1,k - \frac{1}{2}\rangle\langle k - \frac{1}{2}, j-1,i,2| \right)\right)[/itex]

where [itex]\frac{1}{P}[/itex] is the probability associated with each state, which I assume to be all equal, e.g. normal distribution (I think 1/12 but not sure )
 
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  • #4
When ##l = 0## what are the possible values of ##m_l##?

Otherwise, I think you're on the right track.
 
  • #5


Yes, that is correct. Since all the atoms are in state n = 2, the system is in a pure state and the probability of being in that state is 1. Therefore, the micro-canonical statistical operator would be the projection operator onto the state |2⟩, which is represented as |2⟩⟨2|. This means that the system is described by a single state and there is no need for a statistical operator.
 

1. What is a statistical operator?

A statistical operator is a mathematical tool used in statistical mechanics to describe the state of a physical system. It is a matrix that contains information about the probabilities of finding a system in different states.

2. How is the statistical operator used to describe the hydrogen atom?

In the context of the hydrogen atom, the statistical operator is used to describe the quantum state of the system, including the electron's position and momentum. It takes into account the probabilities of different energy levels and angular momentum states.

3. What is the significance of the statistical operator in quantum mechanics?

The statistical operator is a fundamental concept in quantum mechanics as it allows us to calculate the properties and behavior of a system based on its underlying probabilistic nature. It is used to calculate expectation values of observables and to describe the time evolution of a system.

4. How does the statistical operator of the hydrogen atom relate to the Schrödinger equation?

The Schrödinger equation is used to determine the wave function of a system, which contains all the information about the system's state. The statistical operator is then derived from the wave function and is used to calculate the probabilities of different states and observables.

5. Can the statistical operator be used for other atoms or molecules?

Yes, the statistical operator can be used to describe the state of any physical system, including atoms and molecules. It is a powerful tool in statistical mechanics and is used to analyze the behavior of a wide range of systems in physics and chemistry.

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