University Quantum Mechanics II · unit UQMII-12
Time-Dependent Perturbation Theory
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Coefficient equations for H₀ + H′(t): infer an input with uncertainty
The coefficient equations are the Schrödinger equation for H₀ + H′(t) rewritten, with no approximation, as coupled first-order equations for the amplitudes cₙ(t) of the unperturbed stationary states, each state keeping its own phase e^(−iEₙt/ħ). Use the supplied topic model Y_out = (1-bz/sqrt(bz^2+bx^2))/2, where Y_out is UPPER STATE OF H₀ + H′ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. bz is Static field B₀ along z, measured in μT. Keep bx (Transverse field B₁ along x) = 30 μT; tau (Switch-off time τ) = 0.3 ms fixed. For this exercise use only 19.5625 μT ≤ bz ≤ 90.4375 μT on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.048840679, with a bounded reading uncertainty ±0.0035117. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown bz by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0070875 μT, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
UPPER STATE OF H₀ + H′ against bz
Predictions from the supplied Coefficient equations for H₀ + H′(t) model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): UPPER STATE OF H₀ + H′
Horizontal axis (x): bz: Static field B₀ along z / μT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| bz: Static field B₀ along z / μT | UPPER STATE OF H₀ + H′ |
|---|---|
| 19.5625 | 0.2268927264393928 |
| 37.28125 | 0.11045915002833195 |
| 55 | 0.06105221354280782 |
| 72.71875 | 0.037788577136887824 |
| 90.4375 | 0.025429272890197974 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(bz) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(19.5625) = 0.226893 and Y_out(90.4375) = 0.0254293. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives bz_line = 19.5625 + (0.0488407 − 0.226893)(90.4375 − 19.5625)/(0.0254293 − 0.226893) = 82.2013 μT. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives bz = 62.7962 μT; substituting back gives Y_out = 0.048840679.
- The interpolation error in the input is |bz_line − bz| = 19.4051 μT. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.045328979 and 0.052352379. Sort the roots even if the response decreases: 60.2929 μT ≤ bz ≤ 65.5698 μT.
- The lower and upper input deviations are 2.50339 μT and 2.77352 μT. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer bz ≈ 62.7962 μT; compatible interval [60.2929, 65.5698] μT. Endpoint interpolation gives 82.2013 μT.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The interaction picture and the Dyson series: infer an input with uncertainty
The interaction picture is the frame that turns with the solved Hamiltonian H₀, so a state changes only under the dressed perturbation H′_I(t); the Dyson series expands that change in the number of times H′_I acts, each later action written to the left. Use the supplied topic model Y_out = 1/(1+d^2), where Y_out is CEILING ω₁²/Ω² (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Detuning Δ/ω₁, dimensionless. Keep s (Elapsed time ω₁t) = 6.25 rad fixed. For this exercise use only 0.31875 ≤ d ≤ 2.68125 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.2440715, with a bounded reading uncertainty ±0.018294. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CEILING ω₁²/Ω² against d
Predictions from the supplied The interaction picture and the Dyson series model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CEILING ω₁²/Ω²
Horizontal axis (x): d: Detuning Δ/ω₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Detuning Δ/ω₁ | CEILING ω₁²/Ω² |
|---|---|
| 0.31875 | 0.9077692280415588 |
| 0.9093749999999999 | 0.5473564926422245 |
| 1.5 | 0.3076923076923077 |
| 2.0906249999999997 | 0.18619502110149633 |
| 2.68125 | 0.12211351787102714 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(d) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.31875) = 0.907769 and Y_out(2.68125) = 0.122114. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.31875 + (0.244071 − 0.907769)(2.68125 − 0.31875)/(0.122114 − 0.907769) = 2.31452. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives d = 1.75988; substituting back gives Y_out = 0.2440715.
- The interpolation error in the input is |d_line − d| = 0.554642. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.2257775 and 0.2623655. Sort the roots even if the response decreases: 1.67675 ≤ d ≤ 1.85179.
- The lower and upper input deviations are 0.0831291 and 0.0919185. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer d ≈ 1.75988; compatible interval [1.67675, 1.85179]. Endpoint interpolation gives 2.31452.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
First-order amplitudes and a perturbation turned on at t = 0: infer an input with uncertainty
The first-order amplitude cₘ(t) is the amplitude for reaching |m⟩ when the perturbation may act only once on the initial state |n⟩, at any moment t′ between 0 and t, each moment weighted by the phase e^(iωₘₙt′). Use the supplied topic model Y_out = 1000/(t-ramp/2), where Y_out is FIRST ZEROS ±1/(t − T/2) (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t is Time since switch-on t, measured in ns. Keep ramp (Linear ramp duration T) = 0 ns fixed. For this exercise use only 46.375 ns ≤ t ≤ 93.625 ns on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 13.298314 MHz, with a bounded reading uncertainty ±0.39261 MHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.004725 ns, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
FIRST ZEROS ±1/(t − T/2) against t
Predictions from the supplied First-order amplitudes and a perturbation turned on at t = 0 model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): FIRST ZEROS ±1/(t − T/2) / MHz
Horizontal axis (x): t: Time since switch-on t / ns
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since switch-on t / ns | FIRST ZEROS ±1/(t − T/2) / MHz |
|---|---|
| 46.375 | 21.5633423180593 |
| 58.1875 | 17.185821697099893 |
| 70 | 14.285714285714286 |
| 81.8125 | 12.223071046600458 |
| 93.625 | 10.68090787716956 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(t) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(46.375) = 21.5633 MHz and Y_out(93.625) = 10.6809 MHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 46.375 + (13.2983 − 21.5633)(93.625 − 46.375)/(10.6809 − 21.5633) = 82.2606 ns. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives t = 75.1975 ns; substituting back gives Y_out = 13.298314 MHz.
- The interpolation error in the input is |t_line − t| = 7.06309 ns. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 12.905704 MHz and 13.690924 MHz. Sort the roots even if the response decreases: 73.0411 ns ≤ t ≤ 77.4851 ns.
- The lower and upper input deviations are 2.15641 ns and 2.28762 ns. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer t ≈ 75.1975 ns; compatible interval [73.0411, 77.4851] ns. Endpoint interpolation gives 82.2606 ns.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Harmonic perturbations: absorption and stimulated emission: infer an input with uncertainty
A harmonic perturbation H′ = V cos ωt drives a pair of levels with a first-order transition probability that peaks where ħω matches their spacing, and is the same whether the system starts in the lower level (absorption) or the upper one (stimulated emission). Use the supplied topic model Y_out = (4*(sin((1+u)*pi*n)^2/(1+u)+sin((1-u+1e-7)*pi*n)^2/(1-u+1e-7))^2+(sin((1+u)*2*pi*n)/(1+u)+sin((1-u+1e-7)*2*pi*n)/(1-u+1e-7))^2)/(2*pi*n)^2, where Y_out is BOTH TERMS, a → b (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. u is Drive frequency ω/ω₀, dimensionless. Keep n (Drive duration ω₀t/2π) = 2 cycles fixed. For this exercise use only 1.105 ≤ u ≤ 1.4725 on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.15635231, with a bounded reading uncertainty ±0.02275. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown u by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00003675, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
BOTH TERMS, a → b against u
Predictions from the supplied Harmonic perturbations: absorption and stimulated emission model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BOTH TERMS, a → b
Horizontal axis (x): u: Drive frequency ω/ω₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| u: Drive frequency ω/ω₀ | BOTH TERMS, a → b |
|---|---|
| 1.105 | 0.8866404027578266 |
| 1.196875 | 0.5059442707267284 |
| 1.2887499999999998 | 0.2268950494988184 |
| 1.380625 | 0.08515864751433028 |
| 1.4725 | 0.004682347369039442 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(u) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.105) = 0.88664 and Y_out(1.4725) = 0.00468235. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives u_line = 1.105 + (0.156352 − 0.88664)(1.4725 − 1.105)/(0.00468235 − 0.88664) = 1.4093. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives u = 1.32918; substituting back gives Y_out = 0.15635231.
- The interpolation error in the input is |u_line − u| = 0.0801262. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.13360231 and 0.17910231. Sort the roots even if the response decreases: 1.3149 ≤ u ≤ 1.34455.
- The lower and upper input deviations are 0.0142714 and 0.0153767. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer u ≈ 1.32918; compatible interval [1.3149, 1.34455]. Endpoint interpolation gives 1.4093.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.