Solving the Quantum Mechanics of a Hydrogen Atom

[frustrationboltzmann]

Hello, I have a little problem understanding the quantum mechanics of a hydrogen atom.
Im troubled with the following question: before i measure the state of a (simplified: without fine-, hyperfinestructure) hydrogen atom, which is the right probability density of finding the electron? is it the density of the groud state (s) or the 2s, 3s...?

my assumption is the following:
the probability of a state to be occupied is given by the fermi dirac distribution for a certain temperature. so in order to get the probability density of the location of the electron before it is measured, I multiply each (normalized) probability density with the occupation probability of every eigenstate and sum over all those products. is this right or complete nonsense?

also if its right, how would I treat degenerated quantum states. my assumption. multiply every degenerated states by the probability of the fermi dirac distribution and divide by the number of degenerated states, since I assume that they all have the same probability.

this would be a mixed state, no?

I appreciate any help!
thank you very much

 

[vanhees71]

First of all you have to get the concepts right. In quantum theory you have to clearly distinguish between "states" and "observables". You cannot "measure states" in the usual sense but only observables.

The state of the electron is in general described by a statistical operator, i.e., a self-adjoint positive semidefinite operator with trace 1, $\hat{\rho}$.

Then observables are described by self-adjoint operators too. The possible outcome of measuring an observable $O$, represented by a self-adjoint operator $\hat{O}$ are the (generalized) eigenvalues of this operator. Then for each (generalized) eigenvalue $o$ of $\hat{O}$ there exists a orthonormal set of eigenvectors $|o,\alpha \rangle$ with $\langle o,\alpha|o',\alpha' \rangle=\delta_{oo'} \delta_{\alpha \alpha'}$. If there are eigenvalues in the continuum (which can also be the case for the parameter(s) $\alpha$) then the Kronecker $\delta$'s become Dirac $\delta$ distributions.

If the system is prepared in the state described by the statistical operator $\hat{\rho}$, then the probability (distribution) for finding the value $o$ when measuring $O$ is given by
$$\sum_{\alpha} \langle o,\alpha|\hat{\rho}|\langle o,\alpha \rangle.$$
For the hydrogen atom (at rest) you can use energy eigenvectors as a complete set. To fully determine the eigenvectors you can diagonalize the compatible operators $\hat{\vec{L}}^2$ and $\hat{L}_z$ as well as the spin $\hat{s}_z$ of the electron (the usual non-relativistic approximation).

Then you get energy eigenvectors for the discrete part of the spectrum (bound states)
$$|E_n,l,m,s_z \rangle, \quad E_{n} =-\frac{13.6 \; \text{eV}}{n^2}, \quad n \in \{1,2,3,\ldots \}, \quad l \in \{0,1,2,\ldots\}, \quad m \in \{-l,-l+1,...,l-1,l \}, \quad s_z \in \{-1/2,1/2 \}.$$
and generalized eigenvectors for the continuous part (scattering states)
$$|E,l,m,s_z \rangle \quad \text{with} \quad E>0, \quad l,m,s_z \quad \text{as for the bound states}.$$
The thermal-equilibrium state is given by the stat. op.
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H}), \quad \beta=1/(k T), \quad Z=\mathrm{Tr} \exp(-\beta \hat{H}).$$
The energy eigenstates are also eigenstates of the thermal-equilibrium state of course, and you can use the basis to evaluate all probabilities with the above given rules.

 

[frustrationboltzmann]

First of all you have to get the concepts right. In quantum theory you have to clearly distinguish between "states" and "observables". You cannot "measure states" in the usual sense but only observables.

The state of the electron is in general described by a statistical operator, i.e., a self-adjoint positive semidefinite operator with trace 1, $\hat{\rho}$.

Then observables are described by self-adjoint operators too. The possible outcome of measuring an observable $O$, represented by a self-adjoint operator $\hat{O}$ are the (generalized) eigenvalues of this operator. Then for each (generalized) eigenvalue $o$ of $\hat{O}$ there exists a orthonormal set of eigenvectors $|o,\alpha \rangle$ with $\langle o,\alpha|o',\alpha' \rangle=\delta_{oo'} \delta_{\alpha \alpha'}$. If there are eigenvalues in the continuum (which can also be the case for the parameter(s) $\alpha$) then the Kronecker $\delta$'s become Dirac $\delta$ distributions.

If the system is prepared in the state described by the statistical operator $\hat{\rho}$, then the probability (distribution) for finding the value $o$ when measuring $O$ is given by
$$\sum_{\alpha} \langle o,\alpha|\hat{\rho}|\langle o,\alpha \rangle.$$
For the hydrogen atom (at rest) you can use energy eigenvectors as a complete set. To fully determine the eigenvectors you can diagonalize the compatible operators $\hat{\vec{L}}^2$ and $\hat{L}_z$ as well as the spin $\hat{s}_z$ of the electron (the usual non-relativistic approximation).

Then you get energy eigenvectors for the discrete part of the spectrum (bound states)
$$|E_n,l,m,s_z \rangle, \quad E_{n} =-\frac{13.6 \; \text{eV}}{n^2}, \quad n \in \{1,2,3,\ldots \}, \quad l \in \{0,1,2,\ldots\}, \quad m \in \{-l,-l+1,...,l-1,l \}, \quad s_z \in \{-1/2,1/2 \}.$$
and generalized eigenvectors for the continuous part (scattering states)
$$|E,l,m,s_z \rangle \quad \text{with} \quad E>0, \quad l,m,s_z \quad \text{as for the bound states}.$$
The thermal-equilibrium state is given by the stat. op.
$$\hat{\rho}=\frac{1}{Z} \exp(-\beta \hat{H}), \quad \beta=1/(k T), \quad Z=\mathrm{Tr} \exp(-\beta \hat{H}).$$
The energy eigenstates are also eigenstates of the thermal-equilib state of course, and you can basis to evaluate all probabilities with the above given rules.

Thank you for this detailled answer. 1 question remains unclear. Why do I use the Boltzmann factor here and not the fermi dirac probability since electrons are fermions and its not a classical system?

 

[vanhees71]

Here you deal only with 1 electron, and for one particle you don't recognize whether it's a boson or a fermion.

In many-body physics the statistical operator for the grand-canonical potential is still
$$\hat{\rho}=\frac{1}{Z} \exp[-\beta (\hat{H}-\mu \hat{Q} ],$$
but this time $\hat{H}$ is the quantum-field theoretical Hamiltonian, and $\hat{Q}$ is some conserved charge (or in non-relativistic physics often the particle number).

Here the Bose or Fermi nature of the particle comes in naturally due to field quantization in terms of commutators or anticommutators, respectively.