University Quantum Mechanics I · unit UQMI-11
Two-State Quantum Systems
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 two-state idealisation and when it is legitimate: infer an input with uncertainty
The two-state idealisation writes a system's state as c₁|1⟩ + c₂|2⟩ and discards every other level; it is legitimate when each discarded level lies further away in energy than both the couplings that reach it and ħ/τ for the timescale τ of the process. Use the supplied topic model Y_out = w/gap, where Y_out is COUPLING RATIO |W|/Δ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gap is Distance Δ to the discarded level, measured in meV. Keep w (Coupling |W| to that level) = 1.5 meV; tau (Timescale probed τ) = 150 fs fixed. For this exercise use only 21.9063 meV ≤ gap ≤ 73.0938 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 = 0.028232305, with a bounded reading uncertainty ±0.0011566. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gap by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00511875 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.
COUPLING RATIO |W|/Δ against gap
Predictions from the supplied The two-state idealisation and when it is legitimate 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): COUPLING RATIO |W|/Δ
Horizontal axis (x): gap: Distance Δ to the discarded level / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gap: Distance Δ to the discarded level / meV | COUPLING RATIO |W|/Δ |
|---|---|
| 21.90625 | 0.06847360912981455 |
| 34.703125 | 0.04322377307519135 |
| 47.5 | 0.031578947368421054 |
| 60.296875 | 0.024876911116869654 |
| 73.09375 | 0.020521590423257803 |
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(gap) = 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(21.9063) = 0.0684736 and Y_out(73.0938) = 0.0205216. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives gap_line = 21.9063 + (0.0282323 − 0.0684736)(73.0938 − 21.9063)/(0.0205216 − 0.0684736) = 64.8628 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 gap = 53.1306 meV; substituting back gives Y_out = 0.028232305.
- The interpolation error in the input is |gap_line − gap| = 11.7321 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.027075705 and 0.029388905. Sort the roots even if the response decreases: 51.0397 meV ≤ gap ≤ 55.4002 meV.
- The lower and upper input deviations are 2.09096 meV and 2.26959 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 gap ≈ 53.1306 meV; compatible interval [51.0397, 55.4002] meV. Endpoint interpolation gives 64.8628 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
State vectors: |ψ⟩ = a|1⟩ + b|2⟩, normalisation, and phase: infer an input with uncertainty
A two-state ket, written in an orthonormal basis (|1⟩, |2⟩), is the column (a, b) of complex coordinates a = ⟨1|ψ⟩ and b = ⟨2|ψ⟩, scaled so that |a|² + |b|² = 1 and defined only up to one overall phase factor. Use the supplied topic model Y_out = ((cosd(theta/2)*cosd(alpha) - sind(theta/2)*cosd(phi+alpha))^2 + (cosd(theta/2)*sind(alpha) - sind(theta/2)*sind(phi+alpha))^2)/2, where Y_out is P₋ = |a − b|²/2 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Polar angle θ, measured in °. Keep phi (Relative phase φ) = 45 °; alpha (Global phase α) = 0 ° fixed. For this exercise use only 94.725 ° ≤ theta ≤ 160.875 ° 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 = 0.25033402, with a bounded reading uncertainty ±0.015403. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.006615 °, 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₋ = |a − b|²/2 against theta
Predictions from the supplied State vectors: |ψ⟩ = a|1⟩ + b|2⟩, normalisation, and phase 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₋ = |a − b|²/2
Horizontal axis (x): theta: Polar angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Polar angle θ / ° | P₋ = |a − b|²/2 |
|---|---|
| 94.725 | 0.1476481466296353 |
| 111.26249999999999 | 0.17051342166960437 |
| 127.8 | 0.22063801628030438 |
| 144.3375 | 0.2938749866020792 |
| 160.875 | 0.3841652391552801 |
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(theta) = 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(94.725) = 0.147648 and Y_out(160.875) = 0.384165. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = 94.725 + (0.250334 − 0.147648)(160.875 − 94.725)/(0.384165 − 0.147648) = 123.445 °. 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 theta = 135.077 °; substituting back gives Y_out = 0.25033402.
- The interpolation error in the input is |theta_line − theta| = 11.6319 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.23493102 and 0.26573702. Sort the roots even if the response decreases: 131.433 ° ≤ theta ≤ 138.502 °.
- The lower and upper input deviations are 3.64335 ° and 3.42536 °. 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 theta ≈ 135.077 °; compatible interval [131.433, 138.502] °. Endpoint interpolation gives 123.445 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Operators as matrices: Aₘₙ = ⟨m|Â|n⟩ and Hermiticity: infer an input with uncertainty
In an orthonormal basis an operator becomes the table Aₘₙ = ⟨m|Â|n⟩, whose column n lists the components of Â|n⟩; it can represent an observable only if the table equals its own conjugate transpose, Aₘₙ = Aₙₘ*. Use the supplied topic model Y_out = 1+sqrt(d*d+v*v), where Y_out is LARGER EIGENVALUE a₀ + |a⃗| (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. d is Half-splitting a₃ = (A₁₁ − A₂₂)/2, measured in meV. Keep v (Coupling size |A₁₂|) = 2 meV; beta (Coupling phase β = arg A₁₂) = 60 ° fixed. For this exercise use only -2.3625 meV ≤ d ≤ -0.1575 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 = 3.2439261 meV, with a bounded reading uncertainty ±0.03566 meV. 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 0.0002205 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.
LARGER EIGENVALUE a₀ + |a⃗| against d
Predictions from the supplied Operators as matrices: Aₘₙ = ⟨m|Â|n⟩ and Hermiticity 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): LARGER EIGENVALUE a₀ + |a⃗| / meV
Horizontal axis (x): d: Half-splitting a₃ = (A₁₁ − A₂₂)/2 / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Half-splitting a₃ = (A₁₁ − A₂₂)/2 / meV | LARGER EIGENVALUE a₀ + |a⃗| / meV |
|---|---|
| -2.3625 | 4.09538466914857 |
| -1.8112499999999998 | 3.6982636199044747 |
| -1.2599999999999998 | 3.363810483097154 |
| -0.7087499999999998 | 3.1218686487386536 |
| -0.15749999999999975 | 3.0061919773541117 |
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(-2.3625) = 4.09538 meV and Y_out(-0.1575) = 3.00619 meV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = -2.3625 + (3.24393 − 4.09538)(-0.1575 − -2.3625)/(3.00619 − 4.09538) = -0.638777 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 d = -1.01745 meV; substituting back gives Y_out = 3.2439261 meV.
- The interpolation error in the input is |d_line − d| = 0.378673 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.2082661 meV and 3.2795861 meV. Sort the roots even if the response decreases: -1.09385 meV ≤ d ≤ -0.936183 meV.
- The lower and upper input deviations are 0.0764023 meV and 0.0812666 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 d ≈ -1.01745 meV; compatible interval [-1.09385, -0.936183] meV. Endpoint interpolation gives -0.638777 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The 2×2 eigenvalue problem and the mixing angle: infer an input with uncertainty
The mixing angle χ of a two-state Hamiltonian is the angle of the rotation that diagonalises it: with the coupling made real, the upper eigenstate is cos χ|1⟩ + sin χ|2⟩, where tan 2χ = 2|V|/(E₁ − E₂) and 0 ≤ χ ≤ 90°. Use the supplied topic model Y_out = sqrt(d^2/4+v^2)-abs(d)/2, where Y_out is PUSH E₊ − max(E₁, E₂) (meV). 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|. d is Uncoupled offset E₁ − E₂, measured in meV. Keep v (Coupling |V|) = 5 meV fixed. For this exercise use only 2.1 meV ≤ d ≤ 31.5 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 = 1.1785477 meV, with a bounded reading uncertainty ±0.060592 meV. 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 0.00294 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.
PUSH E₊ − max(E₁, E₂) against d
Predictions from the supplied The 2×2 eigenvalue problem and the mixing angle 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 E₊ − max(E₁, E₂) / meV
Horizontal axis (x): d: Uncoupled offset E₁ − E₂ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Uncoupled offset E₁ − E₂ / meV | PUSH E₊ − max(E₁, E₂) / meV |
|---|---|
| 2.1000000000000014 | 4.059060579010587 |
| 9.450000000000001 | 2.1543622524184602 |
| 16.8 | 1.3754795278799499 |
| 24.15 | 0.9942626035289379 |
| 31.5 | 0.7746028696607397 |
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(2.1) = 4.05906 meV and Y_out(31.5) = 0.774603 meV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 2.1 + (1.17855 − 4.05906)(31.5 − 2.1)/(0.774603 − 4.05906) = 27.8842 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 d = 20.034 meV; substituting back gives Y_out = 1.1785477 meV.
- The interpolation error in the input is |d_line − d| = 7.85019 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.1179557 meV and 1.2391397 meV. Sort the roots even if the response decreases: 18.9361 meV ≤ d ≤ 21.2443 meV.
- The lower and upper input deviations are 1.09785 meV and 1.21029 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 d ≈ 20.034 meV; compatible interval [18.9361, 21.2443] meV. Endpoint interpolation gives 27.8842 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.