University Physics V · topic UPV-08.02
Deriving [x-hat, p-hat] = i h-bar
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.
The canonical commutation relation: predict a finite change
The canonical commutation relation says position and momentum fail to commute by a fixed amount: x̂p̂ − p̂x̂ is iħ times the identity operator — the same number in every state, in every representation, and at every time. Use the supplied topic model Y_out = (b*b/(a*a))*exp(-b*b/(2*a*a)) - (b*b/(a*a) - 1)*exp(-b*b/(2*a*a)), where Y_out is DIFFERENCE ÷ iħ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. a is Gaussian width a, measured in nm. Keep b (Probe position x) = 1 nm fixed. For this exercise use only 0.79125 nm ≤ a ≤ 2.20875 nm 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 a = 1.1031 nm and 1.30155 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DIFFERENCE ÷ iħ against a
Predictions from the supplied The canonical commutation relation 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): DIFFERENCE ÷ iħ
Horizontal axis (x): a: Gaussian width a / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Gaussian width a / nm | DIFFERENCE ÷ iħ |
|---|---|
| 0.79125 | 0.44994750007515355 |
| 1.145625 | 0.6832018242185902 |
| 1.5 | 0.8007374029168081 |
| 1.8543749999999999 | 0.864673322251564 |
| 2.2087499999999998 | 0.9025881583348507 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 1.1031 nm, substitute into Y_out = (b*b/(a*a))*exp(-b*b/(2*a*a)) - (b*b/(a*a) - 1)*exp(-b*b/(2*a*a)): Y₀ = 0.663051.
- At a = 1.30155 nm, the same fixed-parameter model gives Y₁ = 0.744417.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0813665; the input change is Δa = 0.19845 nm.
- Average sensitivity is ΔY_out/Δa = 0.41001 1 per nm. Its positive sign means that this response increases 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 = 0.663051, Y₁ = 0.744417; ΔY_out = 0.0813665; average sensitivity = 0.41001 1 per nm.
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.
The canonical commutation relation: infer an input with uncertainty
The canonical commutation relation says position and momentum fail to commute by a fixed amount: x̂p̂ − p̂x̂ is iħ times the identity operator — the same number in every state, in every representation, and at every time. Use the supplied topic model Y_out = (b*b/(a*a))*exp(-b*b/(2*a*a)) - (b*b/(a*a) - 1)*exp(-b*b/(2*a*a)), where Y_out is DIFFERENCE ÷ iħ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. a is Gaussian width a, measured in nm. Keep b (Probe position x) = 1 nm fixed. For this exercise use only 0.79125 nm ≤ a ≤ 2.20875 nm 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 = 0.83331561, with a bounded reading uncertainty ±0.010391. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00014175 nm, 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.
DIFFERENCE ÷ iħ against a
Predictions from the supplied The canonical commutation relation 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): DIFFERENCE ÷ iħ
Horizontal axis (x): a: Gaussian width a / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Gaussian width a / nm | DIFFERENCE ÷ iħ |
|---|---|
| 0.79125 | 0.44994750007515355 |
| 1.145625 | 0.6832018242185902 |
| 1.5 | 0.8007374029168081 |
| 1.8543749999999999 | 0.864673322251564 |
| 2.2087499999999998 | 0.9025881583348507 |
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(a) = 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(0.79125) = 0.449948 and Y_out(2.20875) = 0.902588. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.79125 + (0.833316 − 0.449948)(2.20875 − 0.79125)/(0.902588 − 0.449948) = 1.99181 nm. 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 a = 1.65592 nm; substituting back gives Y_out = 0.83331561.
- The interpolation error in the input is |a_line − a| = 0.335889 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.82292461 and 0.84370661. Sort the roots even if the response decreases: 1.60173 nm ≤ a ≤ 1.71524 nm.
- The lower and upper input deviations are 0.0541944 nm and 0.0593108 nm. 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 a ≈ 1.65592 nm; compatible interval [1.60173, 1.71524] nm. Endpoint interpolation gives 1.99181 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.