University Quantum Mechanics II · unit UQMII-06
Time-Independent 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.
Expanding the eigenvalue problem in powers of λ: infer an input with uncertainty
Rayleigh–Schrödinger perturbation theory writes H = H₀ + λH′, expands an eigenvalue and its eigenstate in powers of λ, and solves the eigenvalue equation one power of λ at a time, starting from the known eigenstates of H₀. Use the supplied topic model Y_out = -(lam*v)^2 + clamp((kmax-2)/2,0,1)*(lam*v)^4 - 2*clamp((kmax-4)/2,0,1)*(lam*v)^6 + 5*clamp((kmax-6)/2,0,1)*(lam*v)^8, where Y_out is SERIES TO ORDER K, E/Δ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; clamp(x,a,b) confines x to [a,b]. lam is Bookkeeping parameter λ, dimensionless. Keep v (Coupling V/Δ) = 0.4; kmax (Series truncated at order K) = 2 fixed. For this exercise use only 0.10625 ≤ lam ≤ 0.89375 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 = -0.055060622, with a bounded reading uncertainty ±0.0079882. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
SERIES TO ORDER K, E/Δ against lam
Predictions from the supplied Expanding the eigenvalue problem in powers of λ 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): SERIES TO ORDER K, E/Δ
Horizontal axis (x): lam: Bookkeeping parameter λ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Bookkeeping parameter λ | SERIES TO ORDER K, E/Δ |
|---|---|
| 0.10625 | -0.0018062500000000003 |
| 0.303125 | -0.0147015625 |
| 0.5 | -0.04000000000000001 |
| 0.6968749999999999 | -0.0777015625 |
| 0.8937499999999999 | -0.12780624999999998 |
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(lam) = 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(0.10625) = -0.00180625 and Y_out(0.89375) = -0.127806. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lam_line = 0.10625 + (-0.0550606 − -0.00180625)(0.89375 − 0.10625)/(-0.127806 − -0.00180625) = 0.43909. 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 lam = 0.586625; substituting back gives Y_out = -0.055060622.
- The interpolation error in the input is |lam_line − lam| = 0.147535. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.063048822 and -0.047072422. Sort the roots even if the response decreases: 0.542404 ≤ lam ≤ 0.627738.
- The lower and upper input deviations are 0.0442205 and 0.0411131. 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 lam ≈ 0.586625; compatible interval [0.542404, 0.627738]. Endpoint interpolation gives 0.43909.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The first-order energy shift E⁽¹⁾ = ⟨n|H′|n⟩: infer an input with uncertainty
The first-order shift of a non-degenerate level is the expectation value of the perturbation in the unperturbed state, Eₙ⁽¹⁾ = ⟨n⁽⁰⁾|H′|n⁽⁰⁾⟩; for a potential it is H′ averaged over the unperturbed probability density. Use the supplied topic model Y_out = 0.37603*n^2+0.001*20/max(w,0.05)*(w-cos(2*pi*n*x0)*sin(pi*n*w)/(pi*n)), where Y_out is LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; unmarked trigonometric functions use radians. x0 is Bump centre x₀, measured in nm. Keep n (Level n) = 1; w (Bump width w) = 0.1 nm fixed. For this exercise use only 0.513125 nm ≤ x0 ≤ 0.696875 nm 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 = 0.40992207 eV, with a bounded reading uncertainty ±0.00085706 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown x0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000018375 nm, 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.
LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ against x0
Predictions from the supplied The first-order energy shift E⁽¹⁾ = ⟨n|H′|n⟩ 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): LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV
Horizontal axis (x): x0: Bump centre x₀ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| x0: Bump centre x₀ / nm | LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV |
|---|---|
| 0.513125 | 0.4156357761964956 |
| 0.5590625 | 0.4143634926514627 |
| 0.605 | 0.41157442946227474 |
| 0.6509375 | 0.40749933340217587 |
| 0.6968749999999999 | 0.40247534823692 |
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(x0) = 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(0.513125) = 0.415636 eV and Y_out(0.696875) = 0.402475 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives x0_line = 0.513125 + (0.409922 − 0.415636)(0.696875 − 0.513125)/(0.402475 − 0.415636) = 0.592902 nm. 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 x0 = 0.625212 nm; substituting back gives Y_out = 0.40992207 eV.
- The interpolation error in the input is |x0_line − x0| = 0.0323109 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.40906501 eV and 0.41077913 eV. Sort the roots even if the response decreases: 0.615092 nm ≤ x0 ≤ 0.634727 nm.
- The lower and upper input deviations are 0.0101204 nm and 0.00951485 nm. 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 x0 ≈ 0.625212 nm; compatible interval [0.615092, 0.634727] nm. Endpoint interpolation gives 0.592902 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The first-order state and the energy-denominator condition: infer an input with uncertainty
The first-order state is the part of the corrected eigenstate linear in H′: every other unperturbed state mixed in with amplitude ⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩/(Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾), trustworthy only while each amplitude is much smaller than 1. Use the supplied topic model Y_out = exp(-(0.023846*f/hw^1.5)^2), where Y_out is P(STAY IN |0⟩), EXACT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. hw is Level spacing ħω, measured in meV. Keep f (Field F) = 100 kV m⁻¹ fixed. For this exercise use only 2.85 meV ≤ hw ≤ 9.15 meV 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 = 0.98121303, with a bounded reading uncertainty ±0.0017087. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown hw by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 meV, 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.
P(STAY IN |0⟩), EXACT against hw
Predictions from the supplied The first-order state and the energy-denominator condition 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): P(STAY IN |0⟩), EXACT
Horizontal axis (x): hw: Level spacing ħω / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| hw: Level spacing ħω / meV | P(STAY IN |0⟩), EXACT |
|---|---|
| 2.85 | 0.782204913244314 |
| 4.425 | 0.9364788747347376 |
| 6 | 0.974017953841428 |
| 7.574999999999999 | 0.9870029291772693 |
| 9.149999999999999 | 0.9926046802986603 |
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(hw) = 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(2.85) = 0.782205 and Y_out(9.15) = 0.992605. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives hw_line = 2.85 + (0.981213 − 0.782205)(9.15 − 2.85)/(0.992605 − 0.782205) = 8.8089 meV. 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 hw = 6.693 meV; substituting back gives Y_out = 0.98121303.
- The interpolation error in the input is |hw_line − hw| = 2.1159 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.97950433 and 0.98292173. Sort the roots even if the response decreases: 6.4997 meV ≤ hw ≤ 6.91116 meV.
- The lower and upper input deviations are 0.1933 meV and 0.218156 meV. 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 hw ≈ 6.693 meV; compatible interval [6.4997, 6.91116] meV. Endpoint interpolation gives 8.8089 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Second-order energy, and why a ground state always goes down: infer an input with uncertainty
The second-order energy shift Eₙ⁽²⁾ is the part of a level's energy quadratic in the perturbation: the sum, over every other unperturbed state m, of |⟨m⁽⁰⁾|H′|n⁽⁰⁾⟩|² divided by the energy gap Eₙ⁽⁰⁾ − Eₘ⁽⁰⁾. Use the supplied topic model Y_out = -(n+1)*d^2/2, where Y_out is PUSH FROM LEVEL n + 1 (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Field as displacement d/x₀ = qF/(mω²x₀), dimensionless. Keep n (Level n examined) = 0 fixed. For this exercise use only 0.14875 ≤ d ≤ 1.25125 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 = -0.33724631 ħω, with a bounded reading uncertainty ±0.048927 ħω. 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 0.00011025, 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.
PUSH FROM LEVEL n + 1 against d
Predictions from the supplied Second-order energy, and why a ground state always goes down 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): PUSH FROM LEVEL n + 1 / ħω
Horizontal axis (x): d: Field as displacement d/x₀ = qF/(mω²x₀)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Field as displacement d/x₀ = qF/(mω²x₀) | PUSH FROM LEVEL n + 1 / ħω |
|---|---|
| 0.14875 | -0.01106328125 |
| 0.42437499999999995 | -0.09004707031249998 |
| 0.7 | -0.24499999999999997 |
| 0.9756249999999997 | -0.4759220703124997 |
| 1.2512499999999998 | -0.7828132812499997 |
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(0.14875) = -0.0110633 ħω and Y_out(1.25125) = -0.782813 ħω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.14875 + (-0.337246 − -0.0110633)(1.25125 − 0.14875)/(-0.782813 − -0.0110633) = 0.614726. 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 = 0.821275; substituting back gives Y_out = -0.33724631 ħω.
- The interpolation error in the input is |d_line − d| = 0.206549. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.38617331 ħω and -0.28831931 ħω. Sort the roots even if the response decreases: 0.759367 ≤ d ≤ 0.878833.
- The lower and upper input deviations are 0.0619077 and 0.0575575. 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 ≈ 0.821275; compatible interval [0.759367, 0.878833]. Endpoint interpolation gives 0.614726.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.