hw8 - 6-35 d Applying the momentum operator px to each of...

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

View Full Document Right Arrow Icon
6-35 Applying the momentum operator p x  i d dx to each of the candidate functions yields (a) p x A sin kx   i kA cos kx (b) p x A sin kx A cos kx i cos kx A sin kx (c) p x A cos kx iA sin kx i k A sin kx iA cos kx (d) p x e ik x a i ik e ik x a In case (c), the result is a multiple of the original function, since A sin kx iA cos kx iA cos kx iA sin kx . The multiple is i ik k and is the eigenvalue. Likewise for (d), the operation p x returns the original function with the multiplier k . Thus, (c) and (d) are eigenfunctions of p x with eigenvalue k , whereas (a) and (b) are not eigenfunctions of this operator. 6-37 (a) Normalization requires 1  2 dx  C 2 1 * 2 * 1 2 dx  C 2 1 2 dx 2 2 dx 2 * 1 dx 1 * 2 dx . The first two integrals on the right are unity, while the last two are, in fact, the same integral since 1 and 2 are both real. Using the waveforms for the infinite square well, we find 2 1 dx 2 L sin x L sin 2 x L dx 0 L 1 L cos x L   cos 3 x L dx 0 L where, in writing the last line, we have used the trigonometric exponential identities of sine and cosine. Both of the integrals remaining are readily evaluated, and are zero. Thus, 1 C 2 1 0 0 0 2 C 2 , or C 1 2 . Since 1,2 are stationary states, they develop in time according to their respective energies E as e iEt . Then x , t C 1 e iE 1 t 2 e iE 2 t .
Background image of page 1

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

View Full DocumentRight Arrow Icon
(c) x , t  is a stationary state only if it is an eigenfunction of the energy operator E  i t . Applying E to gives E  Ci iE 1 1 e iE 1 t i iE 2 2 e iE 2 t CE 1 1 e iE 1 t E 2 2 e iE 2 t . Since E 1 E 2 , the operations E   does not return a multiple of the wavefunction, and so is not a stationary state. Nonetheless, we may calculate the average energy for this state as E  * E dx C 2 1 * e iE 1 t 2 * e iE 2 t E 1 1 e iE 1 t E 2 2 e iE 2 t dx C 2 E 1 1 2 dx E 2 2 2 dx  with the cross terms vanishing as in part (a). Since 1,2 are normalized and C 2 1 2 we get finally E E 1 E 2 2 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

hw8 - 6-35 d Applying the momentum operator px to each of...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online