University Quantum Mechanics I · unit UQMI-10
Dirac Notation and Quantum State Space
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.
What the solved systems were already telling us: infer an input with uncertainty
Reading a solved system as a vector space means treating its orthonormal eigenfunctions ψₙ as perpendicular unit axes, the overlap integral ∫φ*ψ dx as the dot product, and the coefficients cₙ = ∫ψₙ*ψ dx as the components of a state along those axes. Use the supplied topic model Y_out = (100*r*cosd(theta-beta))/100, where Y_out is OVERLAP ∫ΦΨ dx = b₁c₁ + b₂c₂ (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. theta is Direction of Ψ, θ, measured in °. Keep r (Length of Ψ, r) = 1; beta (Direction of Φ, β) = 0 ° fixed. For this exercise use only -70.875 ° ≤ theta ≤ -4.725 ° 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 = 0.86142092, with a bounded reading uncertainty ±0.020277. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta 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.
OVERLAP ∫ΦΨ dx = b₁c₁ + b₂c₂ (before display rounding) against theta
Predictions from the supplied What the solved systems were already telling us 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): OVERLAP ∫ΦΨ dx = b₁c₁ + b₂c₂ (before display rounding)
Horizontal axis (x): theta: Direction of Ψ, θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Direction of Ψ, θ / ° | OVERLAP ∫ΦΨ dx = b₁c₁ + b₂c₂ (before display rounding) |
|---|---|
| -70.875 | 0.32763017956169366 |
| -54.3375 | 0.5830095789833509 |
| -37.8 | 0.7901550123756904 |
| -21.262499999999996 | 0.9319287753895013 |
| -4.724999999999994 | 0.9966015395273318 |
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(theta) = 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(-70.875) = 0.32763 and Y_out(-4.725) = 0.996602. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = -70.875 + (0.861421 − 0.32763)(-4.725 − -70.875)/(0.996602 − 0.32763) = -18.0921 °. 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 theta = -30.5235 °; substituting back gives Y_out = 0.86142092.
- The interpolation error in the input is |theta_line − theta| = 12.4314 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.84114392 and 0.88169792. Sort the roots even if the response decreases: -32.7389 ° ≤ theta ≤ -28.1521 °.
- The lower and upper input deviations are 2.21539 ° and 2.37137 °. 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 theta ≈ -30.5235 °; compatible interval [-32.7389, -28.1521] °. Endpoint interpolation gives -18.0921 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Kets: a state as a vector over the complex numbers: infer an input with uncertainty
A ket |ψ⟩ is a member of a complex vector space: kets can be added and multiplied by complex numbers, any such combination is again a ket, and two kets that differ only by a nonzero complex factor describe the same physical state. Use the supplied topic model Y_out = cosd(th)*cosd(al)*sin(pi/4)+sind(th)*cosd(al+ph), where Y_out is Re ψ(L/4) ÷ √(2/L) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; unmarked trigonometric functions use radians. th is Mixing angle θ, measured in °. Keep ph (Relative phase φ) = 60 °; al (Global phase α) = 0 ° fixed. For this exercise use only 37.9125 ° ≤ th ≤ 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 = 0.76043976, with a bounded reading uncertainty ±0.015699. 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 0.0042525 °, 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.
Re ψ(L/4) ÷ √(2/L) against th
Predictions from the supplied Kets: a state as a vector over the complex numbers 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): Re ψ(L/4) ÷ √(2/L)
Horizontal axis (x): th: Mixing angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Mixing angle θ / ° | Re ψ(L/4) ÷ √(2/L) |
|---|---|
| 37.9125 | 0.8651005987025565 |
| 48.54375 | 0.8428693398828988 |
| 59.175 | 0.7917021626304335 |
| 69.80625 | 0.7133556491309824 |
| 80.4375 | 0.6105194550842424 |
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(37.9125) = 0.865101 and Y_out(80.4375) = 0.610519. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 37.9125 + (0.76044 − 0.865101)(80.4375 − 37.9125)/(0.610519 − 0.865101) = 55.395 °. 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 = 63.8528 °; substituting back gives Y_out = 0.76043976.
- The interpolation error in the input is |th_line − th| = 8.4578 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.74474076 and 0.77613876. Sort the roots even if the response decreases: 61.6004 ° ≤ th ≤ 65.9531 °.
- The lower and upper input deviations are 2.25236 ° and 2.10038 °. 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 ≈ 63.8528 °; compatible interval [61.6004, 65.9531] °. Endpoint interpolation gives 55.395 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Bras, the dual space, and the inner product: infer an input with uncertainty
A bra ⟨φ| is a linear rule that turns any ket |ψ⟩ into a complex number ⟨φ|ψ⟩; the bra paired with a state φ does it by conjugating φ, multiplying by ψ, and summing over components or integrating over x. Use the supplied topic model Y_out = exp(-2*k^2)*cosd(deg(2*k*d)+2*alpha), where Y_out is RE ∫ψψ dx, STAR OMITTED (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; sind/cosd/tand use degrees; deg converts radians to degrees. k is Wavenumber k of ψ, measured in σ⁻¹. Keep d (Separation d of ψ from φ) = 1.5 σ; alpha (Global phase α of ψ) = 0 ° fixed. For this exercise use only 0.84 σ⁻¹ ≤ k ≤ 1.575 σ⁻¹ 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 = -0.027105688, with a bounded reading uncertainty ±0.0040791. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown k by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000735 σ⁻¹, 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.
RE ∫ψψ dx, STAR OMITTED against k
Predictions from the supplied Bras, the dual space, and the inner product 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): RE ∫ψψ dx, STAR OMITTED
Horizontal axis (x): k: Wavenumber k of ψ / σ⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: Wavenumber k of ψ / σ⁻¹ | RE ∫ψψ dx, STAR OMITTED |
|---|---|
| 0.8399999999999999 | -0.19823875009471828 |
| 1.0237499999999997 | -0.12262747075244995 |
| 1.2074999999999998 | -0.048002536847614456 |
| 1.3912499999999999 | -0.010687348375457015 |
| 1.5749999999999997 | 0.0000883273480346603 |
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(k) = 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(0.84) = -0.198239 and Y_out(1.575) = 0.0000883273. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives k_line = 0.84 + (-0.0271057 − -0.198239)(1.575 − 0.84)/(0.0000883273 − -0.198239) = 1.47422 σ⁻¹. 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 k = 1.28835 σ⁻¹; substituting back gives Y_out = -0.027105688.
- The interpolation error in the input is |k_line − k| = 0.185869 σ⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.031184788 and -0.023026588. Sort the roots even if the response decreases: 1.26996 σ⁻¹ ≤ k ≤ 1.3087 σ⁻¹.
- The lower and upper input deviations are 0.0183888 σ⁻¹ and 0.0203537 σ⁻¹. 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 k ≈ 1.28835 σ⁻¹; compatible interval [1.26996, 1.3087] σ⁻¹. Endpoint interpolation gives 1.47422 σ⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Orthogonality, normalisation, and ⟨m|n⟩ = δₘₙ: infer an input with uncertainty
Kets |m⟩ and |n⟩ are orthogonal when ⟨m|n⟩ = 0, and a ket is normalised when ⟨n|n⟩ = 1; a set obeying ⟨m|n⟩ = δₘₙ is orthonormal, and the eigenkets of a Hermitian operator with a discrete spectrum can always be chosen that way. Use the supplied topic model Y_out = c*sind(theta), where Y_out is REMAINDER LENGTH √⟨w|w⟩ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Angle from |e₁⟩ to |b⟩, measured in °. Keep phi (Direction of the first pick |e₁⟩) = 20 °; c (Length of |b⟩, √⟨b|b⟩) = 1.2 fixed. For this exercise use only 34.875 ° ≤ theta ≤ 86.325 ° 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 = 1.0984539, with a bounded reading uncertainty ±0.014862. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.005145 °, 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.
REMAINDER LENGTH √⟨w|w⟩ against theta
Predictions from the supplied Orthogonality, normalisation, and ⟨m|n⟩ = δₘₙ 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): REMAINDER LENGTH √⟨w|w⟩
Horizontal axis (x): theta: Angle from |e₁⟩ to |b⟩ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Angle from |e₁⟩ to |b⟩ / ° | REMAINDER LENGTH √⟨w|w⟩ |
|---|---|
| 34.875 | 0.6861455522731347 |
| 47.7375 | 0.8880857066279694 |
| 60.6 | 1.0454565733442271 |
| 73.4625 | 1.1503603702909018 |
| 86.325 | 1.197532416969702 |
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(theta) = 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(34.875) = 0.686146 and Y_out(86.325) = 1.19753. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = 34.875 + (1.09845 − 0.686146)(86.325 − 34.875)/(1.19753 − 0.686146) = 76.3568 °. 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 theta = 66.2595 °; substituting back gives Y_out = 1.0984539.
- The interpolation error in the input is |theta_line − theta| = 10.0973 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.0835919 and 1.1133159. Sort the roots even if the response decreases: 64.5543 ° ≤ theta ≤ 68.0888 °.
- The lower and upper input deviations are 1.70515 ° and 1.82929 °. 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 theta ≈ 66.2595 °; compatible interval [64.5543, 68.0888] °. Endpoint interpolation gives 76.3568 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.