University Physics V · unit UPV-20
Introduction to Quantum Information
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 classical bit and the qubit: infer an input with uncertainty
A qubit is a unit vector α|0⟩ + β|1⟩ in the complex space C², where |0⟩ and |1⟩ are the σ_z eigenvectors, |α|² + |β|² = 1, and vectors that differ only by an overall phase e^(iγ) are the same state. Use the supplied topic model Y_out = sind(t)*cosd(ph), where Y_out is ⟨σ_x⟩ = sin θ cos φ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. t is Polar angle θ, measured in °. Keep ph (Relative phase φ) = 60 °; gm (Global phase γ) = 0 ° fixed. For this exercise use only 19.125 ° ≤ t ≤ 85.275 ° 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 = 0.43071046, with a bounded reading uncertainty ±0.010139. 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 0.006615 °, 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.
⟨σ_x⟩ = sin θ cos φ against t
Predictions from the supplied The classical bit and the qubit 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): ⟨σ_x⟩ = sin θ cos φ
Horizontal axis (x): t: Polar angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Polar angle θ / ° | ⟨σ_x⟩ = sin θ cos φ |
|---|---|
| 19.125 | 0.16381508978084677 |
| 35.6625 | 0.2915047894916755 |
| 52.2 | 0.3950775061878453 |
| 68.73750000000001 | 0.46596438769475074 |
| 85.275 | 0.498300769763666 |
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(19.125) = 0.163815 and Y_out(85.275) = 0.498301. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 19.125 + (0.43071 − 0.163815)(85.275 − 19.125)/(0.498301 − 0.163815) = 71.9079 °. 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 = 59.4765 °; substituting back gives Y_out = 0.43071046.
- The interpolation error in the input is |t_line − t| = 12.4314 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.42057146 and 0.44084946. Sort the roots even if the response decreases: 57.261 ° ≤ t ≤ 61.848 °.
- The lower and upper input deviations are 2.21549 ° and 2.37149 °. 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 ≈ 59.4765 °; compatible interval [57.261, 61.848] °. Endpoint interpolation gives 71.9079 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Bloch sphere, density operators and mixed states: infer an input with uncertainty
A density operator is a Hermitian, positive, unit-trace operator whose expectation values Tr(ρA) reproduce every measurement; for one qubit it is ρ = (I + r⋅σ)/2, and the Bloch vector r fills the unit ball with the pure states on its surface. Use the supplied topic model Y_out = -((1+m)/2*ln((1+m)/2)+max((1-m)/2,0.0000001)*ln(max((1-m)/2,0.0000001)))/ln(2), where Y_out is ENTROPY S (bits). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; max(a,b) selects the larger value. m is Bloch length |r|, dimensionless. Keep th (Polar angle θ of r from z) = 60 °; a (Measurement axis n from z) = 90 ° fixed. For this exercise use only 0.10625 ≤ m ≤ 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 = 0.73510312 bits, with a bounded reading uncertainty ±0.038511 bits. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m 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.
ENTROPY S against m
Predictions from the supplied Bloch sphere, density operators and mixed states 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): ENTROPY S / bits
Horizontal axis (x): m: Bloch length |r|
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Bloch length |r| | ENTROPY S / bits |
|---|---|
| 0.10625 | 0.9918412713748618 |
| 0.303125 | 0.9326648639363424 |
| 0.5 | 0.8112781244591328 |
| 0.6968749999999999 | 0.6137366882156129 |
| 0.8937499999999999 | 0.29952626584498787 |
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(m) = 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) = 0.991841 bits and Y_out(0.89375) = 0.299526 bits. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 0.10625 + (0.735103 − 0.991841)(0.89375 − 0.10625)/(0.299526 − 0.991841) = 0.398287. 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 m = 0.586625; substituting back gives Y_out = 0.73510312 bits.
- The interpolation error in the input is |m_line − m| = 0.188338. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.69659212 bits and 0.77361412 bits. Sort the roots even if the response decreases: 0.545015 ≤ m ≤ 0.624641.
- The lower and upper input deviations are 0.0416096 and 0.0380157. 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 m ≈ 0.586625; compatible interval [0.545015, 0.624641]. Endpoint interpolation gives 0.398287.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Measurement, the Born rule and POVMs: infer an input with uncertainty
A measurement is a set of positive operators E_k on the state space that sum to the identity; outcome k occurs with probability Tr(ρ E_k), and when the E_k are orthogonal projectors P_k the state left behind is P_k ρ P_k divided by that probability. Use the supplied topic model Y_out = 1/(1+cosd(alpha/2)), where Y_out is LARGEST VALID a = 1/(1 + s) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. alpha is Bloch angle α between ψ_1 and ψ_2, measured in °. Keep a (POVM scale a) = 0.5 fixed. For this exercise use only 35.9375 ° ≤ alpha ≤ 154.063 ° 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 = 0.62979059, with a bounded reading uncertainty ±0.017594. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown alpha by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0118125 °, 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.
LARGEST VALID a = 1/(1 + s) against alpha
Predictions from the supplied Measurement, the Born rule and POVMs 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): LARGEST VALID a = 1/(1 + s)
Horizontal axis (x): alpha: Bloch angle α between ψ_1 and ψ_2 / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| alpha: Bloch angle α between ψ_1 and ψ_2 / ° | LARGEST VALID a = 1/(1 + s) |
|---|---|
| 35.9375 | 0.5124985802792946 |
| 65.46875 | 0.5431280290135092 |
| 95 | 0.5968046336478565 |
| 124.53125 | 0.6824200328254717 |
| 154.0625 | 0.8167134964798659 |
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(alpha) = 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(35.9375) = 0.512499 and Y_out(154.063) = 0.816713. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives alpha_line = 35.9375 + (0.629791 − 0.512499)(154.063 − 35.9375)/(0.816713 − 0.512499) = 81.4814 °. 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 alpha = 107.994 °; substituting back gives Y_out = 0.62979059.
- The interpolation error in the input is |alpha_line − alpha| = 26.5124 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.61219659 and 0.64738459. Sort the roots even if the response decreases: 101.388 ° ≤ alpha ≤ 113.995 °.
- The lower and upper input deviations are 6.6058 ° and 6.00108 °. 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 alpha ≈ 107.994 °; compatible interval [101.388, 113.995] °. Endpoint interpolation gives 81.4814 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Tensor products, entanglement and Schmidt form: infer an input with uncertainty
A pure state of two systems is entangled when it cannot be written as one vector tensor another; equivalently its coefficient matrix has rank above one, its Schmidt decomposition has more than one term, and the reduced state of either half is mixed. Use the supplied topic model Y_out = abs(sind(2*t)), where Y_out is CONCURRENCE 2|det C| (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. t is Schmidt angle θ, measured in °. Keep a (Local rotation α on qubit A) = 0 °; b (Local rotation β on qubit B) = 0 ° fixed. For this exercise use only 47.3625 ° ≤ t ≤ 80.4375 ° 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 = 0.70616204, with a bounded reading uncertainty ±0.043566. 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 0.0033075 °, 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.
CONCURRENCE 2|det C| against t
Predictions from the supplied Tensor products, entanglement and Schmidt 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): CONCURRENCE 2|det C|
Horizontal axis (x): t: Schmidt angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Schmidt angle θ / ° | CONCURRENCE 2|det C| |
|---|---|
| 47.3625 | 0.9966015395273318 |
| 55.631249999999994 | 0.9319287753895013 |
| 63.9 | 0.7901550123756905 |
| 72.16875 | 0.5830095789833507 |
| 80.4375 | 0.3276301795616937 |
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(47.3625) = 0.996602 and Y_out(80.4375) = 0.32763. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 47.3625 + (0.706162 − 0.996602)(80.4375 − 47.3625)/(0.32763 − 0.996602) = 61.7223 °. 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 = 67.5382 °; substituting back gives Y_out = 0.70616204.
- The interpolation error in the input is |t_line − t| = 5.81596 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.66259604 and 0.74972804. Sort the roots even if the response decreases: 65.7166 ° ≤ t ≤ 69.2509 °.
- The lower and upper input deviations are 1.82166 ° and 1.71267 °. 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 ≈ 67.5382 °; compatible interval [65.7166, 69.2509] °. Endpoint interpolation gives 61.7223 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.