Home » Ask & Discuss » Physics. » Modern Physics « Back to Discussion

Modern Physics

Cool goIITian

Joined:	5 Mar 2007
Post:	31

30 Mar 2008 11:09:04 IST

0 People liked this

354

Quantum mechanics

None

Use the Hamiltonian Operator to write the Schrodinger equation in two dimensions and calculate the expectation value for position.

Salute is promised

Share this article on:

Comments (3)

Krishna Gopal Singh

Forum Expert

Joined: 29 Dec 2006

Posts: 5153

30 Mar 2008 11:31:07 IST

Like 0 people liked this

We need some kind of potential function to write Scrodinger equation. Please tell what is the kind of potential function and what is the particle whose schrodinger equation is to be written. Unless that is given how we can solve schrodinger equation

edison

Forum Expert

Joined: 19 Oct 2006

Posts: 7537

1 Apr 2008 15:40:31 IST

Like 0 people liked this

Sol) Schrodinger Equation in its most simple form can be written as

H = E ...(1)

Where H = Hamiltonian Operator

E = Energy Eigen Values

Hamiltonian Operator can be expressed as

H = T + V ...(2)

Where, T = Kinetic energy of the Particle

V = Potential energy of the particle

T = mv²/2 = p²/2m (where p = mv is momentum of the particle)...(3)

Now we want to express Hamiltonian H in the operator form so we substitute all the terms involved with their respective operators.

Since p = p_x = -i h' d/dx (for one dimension)

Therefore, p² = -(h'²/2m)(d²/dx² + d²/dy²) (as the

equation is for two dimensions) ...(4)

From eqns (1), (2) (3) and (4) we obtain

[-(h'²/2m)(d²/dx² + d²/dy²) + V]

= E

or [-(h'²/2m)(d²/dx² + d²/dy²) + V]

= ih'd/dt

, is the

required Schrodinger equation

Expectation value for position

Since no boundary conditions are mentioned in the problem and it is not specified whether the particle is free or under the influence of certain potential thus I make following assumptions

Assumming boundary to be -L/2

L/2

And -L/2

L/2

The normalized solution of above equation is

= L^-1.e^ik.r

This wave function is normalized now.

Here expectation values need to be evaluated separately for x and y.

Now expectation value for position is given by

<x> = _{[ -L/2]}

^[L/2] x I

I² d²x,

Substitute the value of

and to evaluate the expectation value, we obtain

<x> = 0

Similarly <y> = 0

edison

Forum Expert

Joined: 19 Oct 2006

Posts: 7537

1 Apr 2008 15:55:17 IST

Like 0 people liked this

Here h' means h/2pi (as usually it is represented by h cross)

h = plank's constant

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team