{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Physics512

# Physics512 - Physics 512 Winter 2003 Final 4:007:00(3 hours...

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

Physics 512 Winter 2003 Final – 4:00–7:00 (3 hours), April 23, 2003 – Solutions [30 pts] 1. A particle of mass m and charge q sits in an isotropic three-dimensional harmonic oscillator potential, and is described by the Hamiltonian H 0 = p 2 2 m + 1 2 2 r 2 Note that eigenstates of H 0 , L 2 and L z may be labeled by | n, , m where n = 0 , 1 , 2 , . . . and E n = ( n + 3 2 . At time t = −∞ the oscillator is in its ground state, | 0 , 0 , 0 . It is then acted upon by a spatially uniform but time dependent electric field E ( t ) = E 0 e ( t/τ ) 2 ˆ z (where E 0 and τ are constant). [5] a ) Show that, to first order in the perturbation, the only possible excited state the oscillator could end up in is the | 1 , 1 , 0 state. The transition amplitude is given by first order time dependent perturbation the- ory: A k s = i ¯ h −∞ e ks t k | V ( t ) | s dt Since, for a constant electric field in the ˆ z direction, the perturbation interaction is given by V ( t ) = q E · d = q E 0 ze ( t/τ ) 2 we find A k s = i ¯ h q E 0 −∞ e ks t t 2 2 k | z | s dt (1) The possible excited states that the oscillator could end up in (to first order) are determined by selection rules on the matrix element k | z | s . First of all, it is easy to see that O = z (which is a dipole operator) is the q = 0 component of a spherical vector T k =1 q =0 . The Wigner-Eckart theorem immediately yields the selection rules δ = ± 1 , δm = 0 . As a result, from the ground state | 0 , 0 , 0 we can only end up in an angular momentum one state | n, 1 , 0 . Note, however, that the angular momentum selection rule does not provide a constraint on the principle quantum number n . To show that the only possible final state has n = 1 , we recall that the eigen- states of the three-dimensional harmonic oscillator can also be given in a basis of individual x , y and z oscillators. Then, since z = ¯ h 2 ( a z + a z ) (2) 1

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

View Full Document
the action of z can only increase the z -oscillator level, n z , by one. Since n = n x + n y + n z , this indicates that we can only end up in the n = 1 state | 1 , 1 , 0 . [15] b ) What is the probability for the oscillator to be found in this excited state at time t = ? [Note that −∞ e x 2 dx = π ] We may perform the time integral in (1) by completing the square A k s = i ¯ h q E 0 1 , 1 , 0 | z | 0 , 0 , 0 −∞ e ( t ks τ 2 / 2) 2 2 ω 2 ks τ 2 / 4 dt = iq E 0 ¯ h τ πe τ 2 ω 2 ks / 4 1 , 1 , 0 | z | 0 , 0 , 0 Note that we have assumed it to be valid to shift the contour of integration away from the real axis. Now, the matrix element may be evaluated using (2) and noting that the properly normalized | 1 , 1 , 0 state is identical to the state with a single excitation in the z oscillator, n z = 1 . Since 1 , 1 , 0 | z | 0 , 0 , 0 = ¯ h 2 and since ω ks = ω , we obtain A k s = iq E 0 2 m ¯ τ πe τ 2 ω 2 / 4 The probability is P = |A k s | 2 = q 2 E 2 0 π 2 m ¯
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}