University Quantum Mechanics II · topic UQMII-07.04
Exchange effects: two particles in a well, and their mean separation
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.
Exchange effects and mean separation in a well: predict a finite change
An exchange effect is a difference between identical and distinguishable particles that comes from the symmetry of the state alone, with no interaction in the Hamiltonian; for positions it appears as a correlation between where the two particles are found. Use the supplied topic model Y_out = sqrt((1/6-(1+1/m^2)/(2*pi^2))-64*m^2*(1+cos(pi*m))/(pi^4*(m^2-1)^4)), where Y_out is RMS x₁ − x₂, BOSONS (a). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. m is Second level m (first is n = 1), dimensionless. Keep p (Particle 1 found at x₁) = 0.3 a fixed. For this exercise use only 2 ≤ m ≤ 5, at the specified test settings separated by 1. 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 = 3 and 4. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; explain why a finite difference between these settings does not require interpolation between them.
RMS x₁ − x₂, BOSONS against m
Predictions from the supplied Exchange effects and mean separation in a well model, with all other inputs fixed as stated in the question. 4 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RMS x₁ − x₂, BOSONS / a
Horizontal axis (x): m: Second level m (first is n = 1)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Second level m (first is n = 1) | RMS x₁ − x₂, BOSONS / a |
|---|---|
| 2 | 0.19608615402130644 |
| 3 | 0.33223052267770525 |
| 4 | 0.335297605889589 |
| 5 | 0.33760872496523403 |
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 = 3, substitute into Y_out = sqrt((1/6-(1+1/m^2)/(2*pi^2))-64*m^2*(1+cos(pi*m))/(pi^4*(m^2-1)^4)): Y₀ = 0.332231 a.
- At m = 4, the same fixed-parameter model gives Y₁ = 0.335298 a.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.00306708 a; the input change is Δm = 1.
- Average sensitivity is ΔY_out/Δm = 0.00306708 a per unit input. Its positive sign means that this response increases over the specified interval.
- This finite difference compares two specified test settings without interpolating between them. A point derivative is not required. Any quantum numbers or particle counts must also obey their physical admissibility conditions.
Answer Y₀ = 0.332231 a, Y₁ = 0.335298 a; ΔY_out = 0.00306708 a; average sensitivity = 0.00306708 a 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.
Exchange effects and mean separation in a well: infer an input with uncertainty
An exchange effect is a difference between identical and distinguishable particles that comes from the symmetry of the state alone, with no interaction in the Hamiltonian; for positions it appears as a correlation between where the two particles are found. Use the supplied topic model Y_out = sqrt((1/6-(1+1/m^2)/(2*pi^2))-64*m^2*(1+cos(pi*m))/(pi^4*(m^2-1)^4)), where Y_out is RMS x₁ − x₂, BOSONS (a). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. m is Second level m (first is n = 1), dimensionless. Keep p (Particle 1 found at x₁) = 0.3 a fixed. For this exercise use only 2 ≤ m ≤ 5, at the specified test settings separated by 1. 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 = 0.33529761 a, with a bounded reading uncertainty ±0.011322 a. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of m consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
RMS x₁ − x₂, BOSONS against m
Predictions from the supplied Exchange effects and mean separation in a well model, with all other inputs fixed as stated in the question. 4 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): RMS x₁ − x₂, BOSONS / a
Horizontal axis (x): m: Second level m (first is n = 1)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Second level m (first is n = 1) | RMS x₁ − x₂, BOSONS / a |
|---|---|
| 2 | 0.19608615402130644 |
| 3 | 0.33223052267770525 |
| 4 | 0.335297605889589 |
| 5 | 0.33760872496523403 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(2) = 0.196086 a and Y_out(5) = 0.337609 a. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives m_line = 2 + (0.335298 − 0.196086)(5 − 2)/(0.337609 − 0.196086) = 4.95101. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to m = 4; only the allowed settings 3, 4, 5 lie inside the stated reading bound.
- The interpolation error in the input is |m_line − m| = 0.951009.
- The compatible allowed values span 3 to 5. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible m: 3, 4, 5. Endpoint interpolation predicts 4.95101; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.