{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW3 - PHY4221 Quantum Mechanic I(Fall term of 2004 Problem...

This preview shows pages 1–3. Sign up to view the full content.

PHY4221 Quantum Mechanic I (Fall term of 2004) Problem Set No. 3 (Answers) Question 1: (a): You could see from the time-depended Schrodinger equation (1) that, although the potential is changed suddenly, the wave function should be continuous as a function of time t . 2 2 2 2 i V t m x ψ ψ ψ = - + h h (1) As a result, immediately after the motion of the wall of this potential well, the wave function of this system stays undisturbed. For our two systems, we have been familiar with their eigenfunctions: ( 29 ( 29 2 2 2 2 2 sin 1, 2, 3, 2 n n n x n x E n L L mL π π ψ = = = ××× h (2) ( 29 ( 29 2 2 2 2 1 sin 1, 2, 3, 2 8 n n n x n x E n L L mL π π ψ = = = ××× h (3) Now let us analyse the condition we are facing: At first, the system is in its stable state (and particle in ground state); then suddenly we move the right wall of this potential well rightward (in an assumed very large velocity so that we do not consider the process of moving it). Once the wall is moved, the system changes into another, even though its wave function has not been changed as well ; as a result, immediately after the motion of the wall, we would better do our consideration using the quantities of the new system. Before the change of the potential wall, the particle is in its ground state simply given by: ( 29 [ ] 2 2 2 2 sin 0, 2 before before x x E x L L L mL π π ψ = = h (4) The eigenfunctions in the new system is given in equation (3) and the realistic wave function is given by equation (4); so we could expand equation (4) using equation (3)’s as: ( 29 ( 29 ( 29 2 1 sin sin 2 immediately after before n n n n n x n x x x c x c L L L L π π ψ ψ ψ = = = = (5) Then it is easy to obtain the coefficients: ( 29 ( 29 ( 29 2 0 4 2 sin 2 4 2 1 sin sin 2 2 2 2 L n before n n n x n x c dx x x dx L L L L for n π π π π ψ ψ - - = = × × = = (6) So obviously the most probable result is when n =2, that is to say, the measurement will give the following result most probably: 2 2 2 2 2 1 2 2 2 n n E p c mL π = = = = h (7)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(b): Obviously, the next most probable result is given by: 2 2 2 1 2 2 32 1 8 9 n n E p c mL π π = = = = h (8) (c): if you use the above result to calculate the energy expectation value, you will find yourself confronted with an infinite series and thus the calculation becomes difficult to carry out. Thus you should try another method. In fact, because the wave function is not influenced at all, we conclude that the motion of the potential wall doesn’t ever influence the particle, and thus the expectation value of the energy of the particle does not change. So we have this expectation value as: 2 2 2 2 immediately after before before E E E mL π = = = h (9) In equation (9) the 2 nd ‘=’ is due to the fact that before changing the potential the particle is in its ground state, not a mixed state.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

HW3 - PHY4221 Quantum Mechanic I(Fall term of 2004 Problem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online