University Quantum Mechanics II · unit UQMII-11
Semiclassical Methods: WKB and the Path Integral
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.
The propagator, and the free particle's in closed form: infer an input with uncertainty
The propagator K(x, t; x′, 0) = ⟨x|e^(−iHt/ħ)|x′⟩ is the amplitude for a particle released exactly at x′ to be found at x a time t later; integrating it against any initial wavefunction gives the wavefunction at t. Use the supplied topic model Y_out = 1.374779/t, where Y_out is |K₀|² AT EVERY x (nm⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t is Time since release t, measured in fs. Keep d (Marked point x = d) = 2 nm; s (Width of a launched packet σ₀) = 0.5 nm fixed. For this exercise use only 1.95625 fs ≤ t ≤ 9.04375 fs 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 = 0.21892693 nm⁻², with a bounded reading uncertainty ±0.010037 nm⁻². 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 0.00070875 fs, 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.
|K₀|² AT EVERY x against t
Predictions from the supplied The propagator, and the free particle's in closed form 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): |K₀|² AT EVERY x / nm⁻²
Horizontal axis (x): t: Time since release t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since release t / fs | |K₀|² AT EVERY x / nm⁻² |
|---|---|
| 1.9562499999999998 | 0.702762428115016 |
| 3.7281249999999995 | 0.3687588264878458 |
| 5.5 | 0.24995981818181817 |
| 7.271875 | 0.18905426729694888 |
| 9.04375 | 0.15201426399447132 |
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(1.95625) = 0.702762 nm⁻² and Y_out(9.04375) = 0.152014 nm⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 1.95625 + (0.218927 − 0.702762)(9.04375 − 1.95625)/(0.152014 − 0.702762) = 8.18266 fs. 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 = 6.27962 fs; substituting back gives Y_out = 0.21892693 nm⁻².
- The interpolation error in the input is |t_line − t| = 1.90304 fs. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.20888993 nm⁻² and 0.22896393 nm⁻². Sort the roots even if the response decreases: 6.00435 fs ≤ t ≤ 6.58136 fs.
- The lower and upper input deviations are 0.275277 fs and 0.301731 fs. 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 ≈ 6.27962 fs; compatible interval [6.00435, 6.58136] fs. Endpoint interpolation gives 8.18266 fs.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Time slicing and Feynman's path integral: infer an input with uncertainty
Feynman's path integral is the propagator written as the N → ∞ limit of (m/(2πiħε))^(N/2) times an integral over the N − 1 intermediate positions of a path cut into slices of time ε = t/N, each path weighted by exp(iS_N/ħ). Use the supplied topic model Y_out = c^2*(2*n-3), where Y_out is ZIGZAG EXTRA PHASE (rad). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. c is Zigzag tooth c, in Fresnel lengths, measured in √(ħε/m). Keep n (Number of slices N) = 6 fixed. For this exercise use only 0.2125 √(ħε/m) ≤ c ≤ 1.7875 √(ħε/m) 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 = 12.38864 rad, with a bounded reading uncertainty ±1.7973 rad. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown c by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575 √(ħε/m), 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.
ZIGZAG EXTRA PHASE against c
Predictions from the supplied Time slicing and Feynman's path integral 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): ZIGZAG EXTRA PHASE / rad
Horizontal axis (x): c: Zigzag tooth c, in Fresnel lengths / √(ħε/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Zigzag tooth c, in Fresnel lengths / √(ħε/m) | ZIGZAG EXTRA PHASE / rad |
|---|---|
| 0.2125 | 0.40640624999999997 |
| 0.60625 | 3.3078515625 |
| 1 | 9 |
| 1.3937499999999998 | 17.482851562499995 |
| 1.7874999999999999 | 28.756406249999998 |
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(c) = 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(0.2125) = 0.406406 rad and Y_out(1.7875) = 28.7564 rad. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives c_line = 0.2125 + (12.3886 − 0.406406)(1.7875 − 0.2125)/(28.7564 − 0.406406) = 0.87818 √(ħε/m). 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 c = 1.17325 √(ħε/m); substituting back gives Y_out = 12.38864 rad.
- The interpolation error in the input is |c_line − c| = 0.29507 √(ħε/m). This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.59134 rad and 14.18594 rad. Sort the roots even if the response decreases: 1.08481 √(ħε/m) ≤ c ≤ 1.25547 √(ħε/m).
- The lower and upper input deviations are 0.0884387 √(ħε/m) and 0.0822242 √(ħε/m). 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 c ≈ 1.17325 √(ħε/m); compatible interval [1.08481, 1.25547] √(ħε/m). Endpoint interpolation gives 0.87818 √(ħε/m).
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Stationary phase and the classical limit: infer an input with uncertainty
Stationary phase approximates an integral of e^(iφ/ħ) with φ ≫ ħ by keeping only the neighbourhood of the points where φ is stationary; applied to the path integral, it keeps the classical path and the Gaussian fluctuations around it. Use the supplied topic model Y_out = sqrt(pi*wt/180/max(abs(sind(wt)),0.001)), where Y_out is PREFACTOR |K_osc|/|K_free| (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|; max(a,b) selects the larger value; sind/cosd/tand use degrees. wt is End time ωt, measured in °. Keep act (Action scale mωL²/ħ) = 10 fixed. For this exercise use only 82.05 ° ≤ wt ≤ 172.14 ° 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 = 1.8725567, with a bounded reading uncertainty ±0.10051. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown wt by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.009009 °, 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.
PREFACTOR |K_osc|/|K_free| against wt
Predictions from the supplied Stationary phase and the classical limit 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): PREFACTOR |K_osc|/|K_free|
Horizontal axis (x): wt: End time ωt / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| wt: End time ωt / ° | PREFACTOR |K_osc|/|K_free| |
|---|---|
| 82.05 | 1.202472189951345 |
| 104.57249999999999 | 1.3732445352493772 |
| 127.095 | 1.6676326486432869 |
| 149.6175 | 2.272234862499181 |
| 172.14 | 4.687174707745816 |
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(wt) = 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(82.05) = 1.20247 and Y_out(172.14) = 4.68717. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives wt_line = 82.05 + (1.87256 − 1.20247)(172.14 − 82.05)/(4.68717 − 1.20247) = 99.3737 °. 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 wt = 137.005 °; substituting back gives Y_out = 1.8725567.
- The interpolation error in the input is |wt_line − wt| = 37.6312 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.7720467 and 1.9730667. Sort the roots even if the response decreases: 132.548 ° ≤ wt ≤ 140.844 °.
- The lower and upper input deviations are 4.45719 ° and 3.83876 °. 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 wt ≈ 137.005 °; compatible interval [132.548, 140.844] °. Endpoint interpolation gives 99.3737 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The WKB wavefunction from an ħ expansion, and its validity: infer an input with uncertainty
The WKB wavefunction is the solution of the Schrödinger equation correct through first order in ħ, ψ ≈ (C/√p) exp(±(i/ħ)∫p dx), trustworthy wherever the reduced wavelength ħ/p changes by a small fraction of itself over a distance ħ/p. Use the supplied topic model Y_out = up/((2*n+1)*(1-up^2)^1.5), where Y_out is δ AT THE PROBE (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. up is Probe position u = x/x_t, dimensionless. Keep n (Quantum number n) = 10 fixed. For this exercise use only 0.104125 ≤ up ≤ 0.875875 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 = 0.049973852, with a bounded reading uncertainty ±0.0067401. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown up by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000077175, 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.
δ AT THE PROBE against up
Predictions from the supplied The WKB wavefunction from an ħ expansion, and its validity 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): δ AT THE PROBE
Horizontal axis (x): up: Probe position u = x/x_t
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| up: Probe position u = x/x_t | δ AT THE PROBE |
|---|---|
| 0.104125 | 0.005040077659728858 |
| 0.2970625 | 0.01624845653703022 |
| 0.49 | 0.035224279499582435 |
| 0.6829375 | 0.08343386544295467 |
| 0.875875 | 0.3712172081652925 |
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(up) = 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(0.104125) = 0.00504008 and Y_out(0.875875) = 0.371217. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives up_line = 0.104125 + (0.0499739 − 0.00504008)(0.875875 − 0.104125)/(0.371217 − 0.00504008) = 0.198827. 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 up = 0.574893; substituting back gives Y_out = 0.049973852.
- The interpolation error in the input is |up_line − up| = 0.376066. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.043233752 and 0.056713952. Sort the roots even if the response decreases: 0.540593 ≤ up ≤ 0.603602.
- The lower and upper input deviations are 0.0342994 and 0.0287098. 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 up ≈ 0.574893; compatible interval [0.540593, 0.603602]. Endpoint interpolation gives 0.198827.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.