Angular dependence in QM: Why is it present in the hydrogen atom?

  • Thread starter Thread starter Repetit
  • Start date Start date
  • Tags Tags
    Angular Qm
Click For Summary
In the hydrogen atom, eigenstates exhibit angular dependence despite the spherically symmetric potential, as seen in orbitals like 2p compared to s orbitals. This angular dependence arises from the initial conditions of motion, which can vary even when the force is central and symmetric. The conservation of angular momentum allows for non-spherical motion, as illustrated by a comet's elliptical trajectory when thrown in a specific direction. Thus, while the potential in the Schrödinger equation is symmetric, the solutions can reflect less symmetry due to these initial conditions. Understanding this distinction is crucial for interpreting angular dependence in quantum mechanics.
Repetit
Messages
128
Reaction score
2
When calculating eigenstates in the hydrogen atom one finds plenty of eigenstates with angular dependence. The s orbitals are spherically symmetric, but an orbital like 2p is not, there is some angular dependence through the spherical harmonics. But why is there angular dependence? It is a totally spherically symmetric problem so how can there be? Certainly, starting from the nucleus and going out in one direction should not be any different from going out in any other direction. How is the angular dependence to be interpreted?
 
Physics news on Phys.org
Recall that most planets have elliptical orbits while undergoing a central force, the sun's gravity.
Regards,
Reilly Atkinson
 
Repetit said:
When calculating eigenstates in the hydrogen atom one finds plenty of eigenstates with angular dependence. The s orbitals are spherically symmetric, but an orbital like 2p is not, there is some angular dependence through the spherical harmonics. But why is there angular dependence? It is a totally spherically symmetric problem so how can there be? Certainly, starting from the nucleus and going out in one direction should not be any different from going out in any other direction. How is the angular dependence to be interpreted?
Consider a completely classical situation
Even if the force is spherically symmetric, the initial conditions of the motion don't have to be!

If the force is a central force, the angular momentum will be conserved. But the motion is not necessarily spherically symmetric since you can choose whatever initial conditions you want!

Consider tossing a comet at some arbitary direction in the solar system (and assume it doe snot have enough energy to escape the potential). It will move following an ellipse which is obviosuly not spherically symmetric.

Again the key point is that you must consider both the symmetry of the force and the symmetry of the initial conditions.
 
rephrasing the last post in "quantum terms", even thought the potential which appears in the schorodinger equation is spherically symmetric the solutions to the schrodinger equation are not necessarily spherically symmetric.
 
kdv said:
Consider a completely classical situation
Even if the force is spherically symmetric, the initial conditions of the motion don't have to be!

If the force is a central force, the angular momentum will be conserved. But the motion is not necessarily spherically symmetric since you can choose whatever initial conditions you want!

Consider tossing a comet at some arbitary direction in the solar system (and assume it doe snot have enough energy to escape the potential). It will move following an ellipse which is obviosuly not spherically symmetric.

Again the key point is that you must consider both the symmetry of the force and the symmetry of the initial conditions.


Great! The solutions of an aquation may have less symmetry than the equation.
 
jiadong said:
Great! The solutions of an aquation may have less symmetry than the equation.

Welcome to PF, Jiadong. Just a heads-up, though - you might want to check the date (upper-left corner of the post) before responding; you've answered a few that are a bit old.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
979
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K