Transition probability when perturbation is sinusoidal

Quantum Mechanics

Problem 9.15d Griffiths Quantum Mechanics 1st edition
Given H'=Vcos(\omega t) . Show that transitions only occur to states with energy E_m=E_N\pm \hbar\omega , the transition probability is P_{N\rightarrow m}=|V_{mN}|^2\,\, \frac{sin^2[(E_N-E_m\pm\hbar\omega)t/2\hbar]}{(E_N-Em\pm\hbar\omega)^2}

SOLUTION:
We start from the wavefunction C_m (t) then substitute the perturbation H', integrate then determine the transition probability |C_m (t)|^2. We’ll try to solve this step-by-step and down to algebra level (hopefully).

The transition probability is given by
P_{N\rightarrow m} = |C_m (t)|^2

where C_m (t) can be determined from the equation
\frac{\partial C_m}{\partial t} = \frac{-i}{\hbar} \sum_N C_N (t) e^{i(E_m - E_N)t/\hbar} \langle{\psi_m|H'(t)|\psi_N}\rangle

when m\ne N, C_N (t)=1
\frac{\partial C_m}{\partial t} = \frac{-i}{\hbar} \sum_N e^{i(E_m - E_N)t/\hbar} \langle{\psi_m|H'(t)|\psi_N}\rangle

Multiply \partial t both sides then integrate both sides, change dummy variable t \rightarrow t'
C_m (t) = \frac{-i}{\hbar} \int_0^t e^{i(E_m - E_N)t'/\hbar} \langle{\psi_m|H'(t')|\psi_N}\rangle\, dt'

Substitute the perturbation H' , that is
\langle{\psi_m|H'(t')|\psi_N}\rangle = H'_{mN} = V_{mN} cos(\omega t')

Thus
C_m (t) = \frac{-i}{\hbar} V_{mN} \int_0^t cos(\omega t') e^{i(E_m - E_N)t'/\hbar} \, dt'

Rewrite cos(\omega t') in exponent form
C_m (t) = \frac{-i}{\hbar} V_{mN} \int_0^t (\frac{e^{\omega t'} + e^{-\omega t'}}{2i}) e^{i(E_m - E_N)t'/\hbar} \, dt'

Simplify
C_m (t) = \frac{-1}{2\hbar} V_{mN} \int_0^t (e^{\omega t'} + e^{-\omega t'}) e^{i(E_m - E_N)t'/\hbar} \, dt'

Separate the integrals
C_m (t) = \frac{-1}{2\hbar} V_{mN} [\int_0^t e^{\omega t'} e^{i(E_m - E_N)t'/\hbar} \, dt' + \int_0^t e^{-\omega t'} e^{i(E_m - E_N)t'/\hbar} \, dt']

Simplify the integrals
C_m (t) = \frac{-1}{2\hbar} V_{mN} [\int_0^t e^{\omega t'+i(E_m - E_N)t'/\hbar} \, dt' + \int_0^t e^{-\omega t' + i(E_m - E_N)t'/\hbar} \, dt']

Factor \frac{it'}{\hbar} in the exponents
C_m (t) = \frac{-1}{2\hbar} V_{mN} [\int_0^t e^{i(\hbar\omega + E_m - E_N)t'/\hbar} \, dt' + \int_0^t e^{i(-\hbar\omega + E_m - E_N)t'/\hbar} \, dt']

We let u=i(\hbar\omega + E_m - E_N)t'/\hbar; du=i(\hbar\omega + E_m - E_N)t'/\hbar and u'=i(-\hbar\omega + E_m - E_N)t'/\hbar; du'=i(-\hbar\omega + E_m - E_N)t'/\hbar

C_m (t) = \frac{-1}{2\hbar} V_{mN} [\int \frac{e^u \, du}{i(\hbar\omega + E_m - E_N)/\hbar} + \int \frac{e^{u'} \, du'}{i(-\hbar\omega + E_m - E_N)/\hbar}]

Integrate \int e^u\,du
C_m (t) = \frac{-1}{2\hbar} V_{mN} [ \frac{e^u}{i(\hbar\omega + E_m - E_N)/\hbar} + \frac{e^{u'}}{i(-\hbar\omega + E_m - E_N)/\hbar}]

Return to variable t’
C_m (t) = \frac{-1}{2\hbar} V_{mN} [ \frac{e^{i(\hbar\omega + E_m - E_N)t'/\hbar}}{i(\hbar\omega + E_m - E_N)/\hbar}|_0^t + \frac{e^{i(-\hbar\omega + E_m - E_N)t'/\hbar}}{i(-\hbar\omega + E_m - E_N)/\hbar}|_0^t]

Evaluating we get
C_m (t) = \frac{-1}{2\hbar} V_{mN} [ \frac{e^{i(\hbar\omega + E_m - E_N)t/\hbar}-1}{i(\hbar\omega + E_m - E_N)/\hbar} + \frac{e^{i(-\hbar\omega + E_m - E_N)t/\hbar}-1}{i(-\hbar\omega + E_m - E_N)/\hbar}]

Factor out \frac{\hbar}{i}
C_m (t) = \frac{-1}{2i} V_{mN} [ \frac{e^{i(\hbar\omega + E_m - E_N)t/\hbar}-1}{\hbar\omega + E_m - E_N} + \frac{e^{i(-\hbar\omega + E_m - E_N)t/\hbar}-1}{-\hbar\omega + E_m - E_N}]

Since the transitions only occur to states with E_m=E_N\pm \hbar\omega , Either of the two terms vanish
C_m (t) = \frac{-1}{2i} V_{mN} [ \frac{e^{i(\pm\hbar\omega + E_m - E_N)t/\hbar}-1}{\pm\hbar\omega + E_m - E_N}]

Factor out e^{i(\pm \hbar\omega +E_m - E_N)t/2\hbar}
C_m (t) = \frac{-1}{2i} V_{mN} \frac{e^{i(\pm\hbar\omega + E_m - E_N)t/2\hbar}}{\pm\hbar\omega + E_m - E_N} [e^{i(\pm\hbar\omega + E_m - E_N)t/2\hbar} - e^{-i(\pm\hbar\omega + E_m - E_N)t/2\hbar}]

Then use sin x= \frac{e^{ix} - e^{-ix}}{2i}
C_m (t) = -V_{mN} \frac{e^{i(\pm\hbar\omega + E_m - E_N)t/2\hbar}}{\pm\hbar\omega + E_m - E_N} sin[(\frac{\pm\hbar\omega + E_m - E_N)t}{2\hbar}]

Use the fact that the sine funciton is an odd function, so sin(-x)=-sin(x) and factor out (-1) in the denominator.
C_m (t) = -V_{mN} \frac{e^{i(\pm\hbar\omega + E_m - E_N)t/2\hbar}}{ E_N - E_m\pm\hbar\omega} sin[(\frac{E_N - E_m\pm\hbar\omega)t}{2\hbar}]

Get the transition probability
|C_m (t)|^2 = C_m*(t)C_m (t) = (-V_{mN} \frac{e^{-i(\pm\hbar\omega + E_m - E_N)t/2\hbar}}{ E_N - E_m\pm\hbar\omega} sin[(\frac{E_N - E_m\pm\hbar\omega)t}{2\hbar}]) (-V_{mN} \frac{e^{i(\pm\hbar\omega + E_m - E_N)t/2\hbar}}{ E_N - E_m\pm\hbar\omega} sin[(\frac{E_N - E_m\pm\hbar\omega)t}{2\hbar}])

The exponential factor vanishes since e^{ix}e^{-ix} = e^{ix-ix} = e^0 = 1
|C_m (t)|^2 = P_{N\rightarrow m} = V_{mN}^2 \frac{sin^2[(E_N - E_m\pm\hbar\omega)t/2\hbar]}{(E_N - E_m\pm\hbar\omega)^2}

Comment for corrections and further questions

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