University Quantum Mechanics II · unit UQMII-04
Composite Systems, Entanglement and Density Operators
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.
Tensor-product spaces for two subsystems: infer an input with uncertainty
The tensor product H_A ⊗ H_B is the state space of systems A and B taken together: it is spanned by the product kets |i⟩|j⟩ of basis kets of each part, so it has d_A d_B dimensions, and any superposition Σ cᵢⱼ|i⟩|j⟩ of them is a state. Use the supplied topic model Y_out = (200*(cosd(th1/2)*cosd(th2/2)*sind(th1/2)*sind(th2/2)+cosd(th1/2)*sind(th2/2)*sind(th1/2)*cosd(th2/2)))/100, where Y_out is ⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th1 is Spin 1 polar angle θ₁, measured in °. Keep th2 (Spin 2 polar angle θ₂) = 150 ° fixed. For this exercise use only 283.95 ° ≤ th1 ≤ 321.75 ° 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 = -0.39927574, with a bounded reading uncertainty ±0.012897. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00378 °, 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.
⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (before display rounding) against th1
Predictions from the supplied Tensor-product spaces for two subsystems 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): ⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (before display rounding)
Horizontal axis (x): th1: Spin 1 polar angle θ₁ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th1: Spin 1 polar angle θ₁ / ° | ⟨σₓ ⊗ σₓ⟩ FROM THE COLUMN (before display rounding) |
|---|---|
| 283.95 | -0.4852532367342713 |
| 293.4 | -0.4588773128419906 |
| 302.85 | -0.4200467769494708 |
| 312.3 | -0.36981554748930484 |
| 321.75 | -0.30954697465491715 |
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(th1) = 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(283.95) = -0.485253 and Y_out(321.75) = -0.309547. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th1_line = 283.95 + (-0.399276 − -0.485253)(321.75 − 283.95)/(-0.309547 − -0.485253) = 302.446 °. 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 th1 = 307.008 °; substituting back gives Y_out = -0.39927574.
- The interpolation error in the input is |th1_line − th1| = 4.56151 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.41217274 and -0.38637874. Sort the roots even if the response decreases: 304.478 ° ≤ th1 ≤ 309.398 °.
- The lower and upper input deviations are 2.53019 ° and 2.38984 °. 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 th1 ≈ 307.008 °; compatible interval [304.478, 309.398] °. Endpoint interpolation gives 302.446 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Product and entangled states: the Schmidt decomposition: infer an input with uncertainty
The Schmidt decomposition writes a pure state of two systems as |ψ⟩ = Σₖ √pₖ |uₖ⟩|vₖ⟩, pairing orthonormal kets of A with orthonormal kets of B, with weights pₖ ≥ 0 that sum to 1; the state is a product exactly when a single weight equals 1. Use the supplied topic model Y_out = abs(sind(2*th)*sind(ph))/2, where Y_out is |det C| = AREA (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; sind/cosd/tand use degrees. th is Weight angle θ, measured in °. Keep ph (Angle φ from |b₁⟩ to |b₂⟩) = 60 °; rot (Turn β of spin B's kets) = 20 ° fixed. For this exercise use only 13.5 ° ≤ th ≤ 42.9 ° 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 = 0.38536322, with a bounded reading uncertainty ±0.006973. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th 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 °, 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.
|det C| = AREA against th
Predictions from the supplied Product and entangled states: the Schmidt decomposition 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): |det C| = AREA
Horizontal axis (x): th: Weight angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Weight angle θ / ° | |det C| = AREA |
|---|---|
| 13.5 | 0.19658365292562002 |
| 20.85 | 0.28805319324960416 |
| 28.2 | 0.36066547690329975 |
| 35.55 | 0.4096669774465852 |
| 42.9 | 0.4318498364191888 |
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(th) = 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(13.5) = 0.196584 and Y_out(42.9) = 0.43185. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 13.5 + (0.385363 − 0.196584)(42.9 − 13.5)/(0.43185 − 0.196584) = 37.0908 °. 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 th = 31.434 °; substituting back gives Y_out = 0.38536322.
- The interpolation error in the input is |th_line − th| = 5.65681 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.37839022 and 0.39233622. Sort the roots even if the response decreases: 30.4549 ° ≤ th ≤ 32.4833 °.
- The lower and upper input deviations are 0.979136 ° and 1.04933 °. 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 th ≈ 31.434 °; compatible interval [30.4549, 32.4833] °. Endpoint interpolation gives 37.0908 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The density operator and ⟨A⟩ = Tr(ρA): infer an input with uncertainty
The density operator ρ = Σₖ pₖ|ψₖ⟩⟨ψₖ| of an ensemble is the probability-weighted sum of the projectors onto its members' states, the one object from which every mean, Tr(ρA), and every outcome probability, Tr(ρPₙ), follows. Use the supplied topic model Y_out = (1-2*p*(1-p)*sind(a)^2), where Y_out is PURITY Tr(ρ²) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. p is Weight p of the 0° member, dimensionless. Keep a (Polarisation α of the other member) = 60 °; chi (Polariser angle χ) = 30 ° fixed. For this exercise use only 0.52625 ≤ p ≤ 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 = 0.71906902, with a bounded reading uncertainty ±0.013955. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p 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.
PURITY Tr(ρ²) against p
Predictions from the supplied The density operator and ⟨A⟩ = Tr(ρA) 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): PURITY Tr(ρ²)
Horizontal axis (x): p: Weight p of the 0° member
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Weight p of the 0° member | PURITY Tr(ρ²) |
|---|---|
| 0.52625 | 0.6260335937500001 |
| 0.618125 | 0.6459302734375001 |
| 0.71 | 0.6911499999999999 |
| 0.8018749999999999 | 0.7616927734375 |
| 0.8937499999999999 | 0.85755859375 |
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(p) = 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(0.52625) = 0.626034 and Y_out(0.89375) = 0.857559. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p_line = 0.52625 + (0.719069 − 0.626034)(0.89375 − 0.52625)/(0.857559 − 0.626034) = 0.673925. 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 p = 0.750425; substituting back gives Y_out = 0.71906902.
- The interpolation error in the input is |p_line − p| = 0.0764997. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.70511402 and 0.73302402. Sort the roots even if the response decreases: 0.731105 ≤ p ≤ 0.768358.
- The lower and upper input deviations are 0.0193204 and 0.017933. 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 p ≈ 0.750425; compatible interval [0.731105, 0.768358]. Endpoint interpolation gives 0.673925.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Different ensembles, one density operator: infer an input with uncertainty
Two ensembles of pure states that produce the same density operator make the same predictions for every measurement, so they are one physical state, however differently they were prepared. Use the supplied topic model Y_out = (1-r^2)/(2-2*r*cosd(theta)), where Y_out is WEIGHT p₁ OF STATE 1 (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. r is Beam polarisation r, dimensionless. Keep theta (Axis of state 1, θ₁ from z) = 90 °; alpha (Analyser axis α from z) = 60 ° fixed. For this exercise use only 0.100937 ≤ r ≤ 0.849062 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 = 0.34471184, with a bounded reading uncertainty ±0.022529. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000748125, 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.
WEIGHT p₁ OF STATE 1 against r
Predictions from the supplied Different ensembles, one density operator 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): WEIGHT p₁ OF STATE 1
Horizontal axis (x): r: Beam polarisation r
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Beam polarisation r | WEIGHT p₁ OF STATE 1 |
|---|---|
| 0.1009375 | 0.494905810546875 |
| 0.28796875 | 0.4585369995117188 |
| 0.475 | 0.3871875 |
| 0.66203125 | 0.28085731201171876 |
| 0.8490624999999999 | 0.13954643554687507 |
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(r) = 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(0.100937) = 0.494906 and Y_out(0.849062) = 0.139546. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives r_line = 0.100937 + (0.344712 − 0.494906)(0.849062 − 0.100937)/(0.139546 − 0.494906) = 0.417135. 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 r = 0.557294; substituting back gives Y_out = 0.34471184.
- The interpolation error in the input is |r_line − r| = 0.140158. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.32218284 and 0.36724084. Sort the roots even if the response decreases: 0.515285 ≤ r ≤ 0.596351.
- The lower and upper input deviations are 0.042009 and 0.0390571. 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 r ≈ 0.557294; compatible interval [0.515285, 0.596351]. Endpoint interpolation gives 0.417135.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.