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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.

Extra challenging · 01

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.

Explore the question data

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

SERIES TO ORDER K, E/Δ against lamPredictions 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. Horizontal axis: lam: Bookkeeping parameter λ. Vertical axis: SERIES TO ORDER K, E/Δ. Exact values are available in the data table below.-0.1328-0.09883-0.06481-0.030790.0032340.074750.28740.50.71260.9252xy

Horizontal axis (x): lam: Bookkeeping parameter λ

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1lam: Bookkeeping parameter λ: 0.1062SERIES TO ORDER K, E/Δ: -0.001806
Read the exact data as a table
SERIES TO ORDER K, E/Δ against lam — supplied values
lam: Bookkeeping parameter λSERIES 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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |lam_line − lam| = 0.147535. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -0.063048822 and -0.047072422. Sort the roots even if the response decreases: 0.542404 ≤ lam ≤ 0.627738.
  6. 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.

Extra challenging · 02

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.

Explore the question data

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

LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ against x0Predictions 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. Horizontal axis: x0: Bump centre x₀ / nm. Vertical axis: LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV. Exact values are available in the data table below.0.401950.40550.409060.412610.416160.50580.55540.6050.65460.7042xy

Horizontal axis (x): x0: Bump centre x₀ / nm

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1x0: Bump centre x₀ / nm: 0.5131LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV: 0.4156
Read the exact data as a table
LEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ against x0 — supplied values
x0: Bump centre x₀ / nmLEVEL Eₙ⁽⁰⁾ + E⁽¹⁾ / eV
0.5131250.4156357761964956
0.55906250.4143634926514627
0.6050.41157442946227474
0.65093750.40749933340217587
0.69687499999999990.40247534823692

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |x0_line − x0| = 0.0323109 nm. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 03

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.

Explore the question data

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

P(STAY IN |0⟩), EXACT against hwPredictions 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. Horizontal axis: hw: Level spacing ħω / meV. Vertical axis: P(STAY IN |0⟩), EXACT. Exact values are available in the data table below.0.77380.83060.88740.94421.0012.5984.29967.7019.402xy

Horizontal axis (x): hw: Level spacing ħω / meV

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1hw: Level spacing ħω / meV: 2.85P(STAY IN |0⟩), EXACT: 0.7822
Read the exact data as a table
P(STAY IN |0⟩), EXACT against hw — supplied values
hw: Level spacing ħω / meVP(STAY IN |0⟩), EXACT
2.850.782204913244314
4.4250.9364788747347376
60.974017953841428
7.5749999999999990.9870029291772693
9.1499999999999990.9926046802986603

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |hw_line − hw| = 2.1159 meV. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 04

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.

Explore the question data

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

PUSH FROM LEVEL n + 1 against dPredictions 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. Horizontal axis: d: Field as displacement d/x₀ = qF/(mω²x₀). Vertical axis: PUSH FROM LEVEL n + 1 / ħω. Exact values are available in the data table below.-0.8137-0.6053-0.3969-0.18860.019810.10460.40230.70.99771.295xy

Horizontal axis (x): d: Field as displacement d/x₀ = qF/(mω²x₀)

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1d: Field as displacement d/x₀ = qF/(mω²x₀): 0.1487PUSH FROM LEVEL n + 1 / ħω: -0.01106
Read the exact data as a table
PUSH FROM LEVEL n + 1 against d — supplied values
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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |d_line − d| = 0.206549. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

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