University Physics V · topic UPV-20.02
The Bloch sphere, density operators, and mixed states
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.
Further practice
Work through a fresh case before tackling the extension.
Bloch sphere, density operators and mixed states: predict a finite change
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. Calculate Y_out at m = 0.2795 and 0.38975. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 0.2795, substitute into Y_out = -((1+m)/2*ln((1+m)/2)+max((1-m)/2,0.0000001)*ln(max((1-m)/2,0.0000001)))/ln(2): Y₀ = 0.942891 bits.
- At m = 0.38975, the same fixed-parameter model gives Y₁ = 0.887466 bits.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0554248 bits; the input change is Δm = 0.11025.
- Average sensitivity is ΔY_out/Δm = -0.502719 bits per unit input. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.942891 bits, Y₁ = 0.887466 bits; ΔY_out = -0.0554248 bits; average sensitivity = -0.502719 bits per unit input.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
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.