Where does phi go in the probability density for hydrogen?

In summary, the spherical harmonic part of the analytical solution for the wavefunction of a hydrogenic electron with quantum numbers n, l, and m does not affect the probability density and only contributes to the phase of the wavefunction. However, it is important in other respects, such as representing symmetries and providing a way to visualize the phase of the wavefunction.
  • #1
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The analytical solution for the wavefunction of a hydrogenic electron with quantum numbers n, l and m has a spherical harmonic part that involves theta and phi (in spherical coordinates). I was looking in Griffiths, and the spherical harmonics part only has phi as exp(i m phi) where i is the imaginary unit, m is the magnetic quantum number and phi is phi (sorry I didn't use TeX). Phi doesn't show up in any other way, than attached to an i in an exponential...

So the complex conjugate of the wave function multiplied by the wave function itself should kill all the terms with phi. Am I correct?

How does phi factor into the probability density? Isn't it removed in taking psi*psi? I am trying to sketch the probability densities of hydrogen's first few wavefunctions.

Thank you for any help.
 
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  • #2
Indeed. [tex]\varphi[/tex] only affects phase, and not amplitude, but it is important in other respects. Sometimes it is useful to sketch the phase of the wavefunction using colour.
 
  • #3
You're right, all eigenstates have the form [itex]\psi = A(r,\theta) e^{i m \phi}[/itex], so that:

[tex] \psi^* \psi = |A(r,\theta)|^2[/tex]

which doesn't depend on [itex]\phi[/itex]. This is good: just as an energy eigenstate has probabilities that don't depend on time, and a momenutm eigenstate (e^ipx) has probabiliites that don't depend on position, an angular momentum eigenstate should have probabilities that don't depend on [itex]\phi[/itex] (this is related to Noether's theorem, a far reaching result connecting conserved quantities like momentum and energy to symmetries like rotation and time translation).

On the other hand, if we have a state that is not an angular momentum eigenstate, such as:

[tex] \psi= A(r,\theta) e^{im\phi} + B(r,\theta) e^{in\phi} [/tex]

we get:

[tex] \psi^* \psi = |A(r,\theta)|^2 + |B(r,\theta)|^2 + 2Re \{ A B^* e^{i(m-n)\phi}\}[/tex]

or, assuming A and B are real:

[tex] \psi^* \psi = A(r,\theta)^2 + B(r,\theta)^2 + 2AB\cos((m-n)\phi)[/tex]

which should reming you of the oscillation between two energy eigenstates.
 
Last edited:

1. Where does phi go in the probability density for hydrogen?

In the probability density for hydrogen, phi is typically represented as the wave function or orbital function. It describes the spatial distribution of the electron around the nucleus.

2. What is the significance of phi in the probability density for hydrogen?

Phi plays a crucial role in determining the probability of finding an electron at a specific location around the nucleus. It is a mathematical representation of the electron's behavior in the hydrogen atom.

3. How is phi related to the energy levels in the hydrogen atom?

The value of phi at a given point around the nucleus is directly related to the energy level of the electron in the hydrogen atom. Higher energy levels have a larger amplitude of phi and a greater probability of finding the electron at a greater distance from the nucleus.

4. Can phi be used to predict the exact location of the electron in the hydrogen atom?

No, phi cannot be used to predict the exact location of the electron in the hydrogen atom. It only provides a probability distribution of where the electron is likely to be found.

5. How does phi change as the electron transitions between energy levels in the hydrogen atom?

As the electron transitions between energy levels, the amplitude of phi changes, indicating a different probability distribution of finding the electron at different distances from the nucleus. The shape of phi also changes as the electron moves to a different orbital with a different energy level.

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