Why Can the Hamiltonian Be Split for a Stationary State?

Click For Summary
SUMMARY

The discussion centers on the separation of the Hamiltonian operator for a stationary state in quantum mechanics, specifically the Hamiltonian defined as H = H_x + H_y = -\frac{\hbar^2}{2m}(d^2/dx^2 + d^2/dy^2). The user inquires why the Hamiltonian can be split when the stationary state is expressed as ψ(x,y) = f(x)g(y). A key insight provided by a participant is that upon inserting this form into the time-independent Schrödinger equation and manipulating the resulting equation, one can derive that both sides must equal a constant, leading to the separation of variables.

PREREQUISITES
  • Understanding of the time-independent Schrödinger equation
  • Familiarity with Hamiltonian mechanics
  • Knowledge of separation of variables in differential equations
  • Basic concepts of quantum mechanics and wave functions
NEXT STEPS
  • Study the time-independent Schrödinger equation in detail
  • Learn about Hamiltonian operators in quantum mechanics
  • Explore the method of separation of variables in differential equations
  • Investigate stationary states and their implications in quantum systems
USEFUL FOR

Students of quantum mechanics, physicists exploring Hamiltonian systems, and anyone interested in the mathematical foundations of wave functions and their applications in quantum theory.

Niles
Messages
1,834
Reaction score
0

Homework Statement


Hi all.

I have a Hamiltonian given by:

[tex] H = H_x + H_y = -\frac{\hbar^2}{2m}(d^2/dx^2 + d^2/dy^2).[/tex]

Now I have a stationary state on the form [itex]\psi(x,y)=f(x)g(y)[/itex]. According to my teacher, then the Hamiltonian can be split up, i.e. we have the two equations:

[tex] H_x f(x) = E_xf(x) \qquad \text{and}\qquad H_y g(y)=E_yg(y).[/tex]

I can't see why this must be true. Inserting in the time-independent Schrödinger-equation doesn't give me these expressions. What am I missing here?

Thanks in advance.


Niles
 
Physics news on Phys.org
If you put it in the time-indep Schrödinger equation, and then if you divide both sides by f(x)g(x), and then if you take one of the two terms on the left hand side to the right hand side, you get : a LHS that is a function of x only, and RHS that's function of y only. This can only be true if both sides are equal to some constant. Try to take it from there.
 
Ahh, yes. I see.

Thanks for that.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K