University Advanced Quantum Mechanics · unit UAQM-08
Negative Energies, Hole Theory and Charge Conjugation
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.
Negative-energy components and Zitterbewegung: infer an input with uncertainty
Zitterbewegung is the oscillation of a free Dirac electron's position at angular frequency 2E/ħ, with amplitude of order ħ/2mc, produced by interference between the positive- and negative-energy components of its spinor. Use the supplied topic model Y_out = (1-2*w)*q/sqrt(1+q^2), where Y_out is DRIFT VELOCITY (c). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. q is Momentum p, measured in mc. Keep w (Negative-energy weight w₋) = 0.2 fixed. For this exercise use only 0.31875 mc ≤ q ≤ 2.68125 mc 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 = 0.52166489 c, with a bounded reading uncertainty ±0.0060763 c. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 mc, 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.
DRIFT VELOCITY against q
Predictions from the supplied Negative-energy components and Zitterbewegung 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): DRIFT VELOCITY / c
Horizontal axis (x): q: Momentum p / mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Momentum p / mc | DRIFT VELOCITY / c |
|---|---|
| 0.31875 | 0.18221711748636246 |
| 0.9093749999999999 | 0.40367271724603726 |
| 1.5 | 0.4992301766027062 |
| 2.0906249999999997 | 0.541266840295488 |
| 2.68125 | 0.5621735795698961 |
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(q) = 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(0.31875) = 0.182217 c and Y_out(2.68125) = 0.562174 c. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 0.31875 + (0.521665 − 0.182217)(2.68125 − 0.31875)/(0.562174 − 0.182217) = 2.42937 mc. 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 q = 1.75987 mc; substituting back gives Y_out = 0.52166489 c.
- The interpolation error in the input is |q_line − q| = 0.669499 mc. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.51558859 c and 0.52774119 c. Sort the roots even if the response decreases: 1.68016 mc ≤ q ≤ 1.84872 mc.
- The lower and upper input deviations are 0.0797148 mc and 0.0888439 mc. 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 q ≈ 1.75987 mc; compatible interval [1.68016, 1.84872] mc. Endpoint interpolation gives 2.42937 mc.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Klein step: transmission through a step higher than 2mc²: infer an input with uncertainty
The Klein step is a sharp electrostatic step met by a Dirac electron; once the step exceeds E + mc², the electron's energy lies in the far side's negative-energy continuum and a finite fraction of its flux is transmitted, however high the step. Use the supplied topic model Y_out = 4*sqrt(max((k-v)*(k+2)*(k+2-v)*k,0))/(sqrt(abs((k-v)*(k+2)))+sqrt(abs((k+2-v)*k)))^2, where Y_out is T, q BY GROUP VELOCITY (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|; max(a,b) selects the larger value. v is Step height V₀, measured in mc². Keep k (Kinetic energy E − mc²) = 1 mc² fixed. For this exercise use only 3.16 mc² ≤ v ≤ 7.15 mc² 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 = 0.84408556, with a bounded reading uncertainty ±0.0043123. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000399 mc², 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.
T, q BY GROUP VELOCITY against v
Predictions from the supplied The Klein step: transmission through a step higher than 2mc² 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): T, q BY GROUP VELOCITY
Horizontal axis (x): v: Step height V₀ / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v: Step height V₀ / mc² | T, q BY GROUP VELOCITY |
|---|---|
| 3.16 | 0.4694231892919099 |
| 4.1575 | 0.7677157339571586 |
| 5.154999999999999 | 0.8297318790176073 |
| 6.1525 | 0.8572776702469411 |
| 7.1499999999999995 | 0.8728340170656331 |
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(v) = 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(3.16) = 0.469423 and Y_out(7.15) = 0.872834. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v_line = 3.16 + (0.844086 − 0.469423)(7.15 − 3.16)/(0.872834 − 0.469423) = 6.86566 mc². 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 v = 5.5939 mc²; substituting back gives Y_out = 0.84408556.
- The interpolation error in the input is |v_line − v| = 1.27176 mc². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.83977326 and 0.84839786. Sort the roots even if the response decreases: 5.44719 mc² ≤ v ≤ 5.75634 mc².
- The lower and upper input deviations are 0.146714 mc² and 0.162438 mc². 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 v ≈ 5.5939 mc²; compatible interval [5.44719, 5.75634] mc². Endpoint interpolation gives 6.86566 mc².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Dirac's sea and the radiative catastrophe: infer an input with uncertainty
Dirac's sea is hole theory's vacuum: the state in which every negative-energy solution of the free Dirac equation, two spin states for each momentum, holds one electron, so that the Pauli principle forbids any further electron from radiating into the negative-energy continuum. Use the supplied topic model Y_out = wp+sqrt(1-1/137.036^2)-sqrt(wp^2-1), where Y_out is MOMENTUM TO NUCLEUS ≥ (mc). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. wp is Landing energy |E′| of the electron, measured in mc². Keep we (Sea empty down to |E| (filled below)) = 2 mc² fixed. For this exercise use only 1.53125 mc² ≤ wp ≤ 5.46875 mc² 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 = 1.1292224 mc, with a bounded reading uncertainty ±0.0055565 mc. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown wp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00039375 mc², 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.
MOMENTUM TO NUCLEUS ≥ against wp
Predictions from the supplied Dirac's sea and the radiative catastrophe 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): MOMENTUM TO NUCLEUS ≥ / mc
Horizontal axis (x): wp: Landing energy |E′| of the electron / mc²
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| wp: Landing energy |E′| of the electron / mc² | MOMENTUM TO NUCLEUS ≥ / mc |
|---|---|
| 1.53125 | 1.371599916763656 |
| 2.515625 | 1.207272326672661 |
| 3.5 | 1.1458714077189383 |
| 4.484375 | 1.112893336955775 |
| 5.46875 | 1.0921792643662478 |
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(wp) = 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(1.53125) = 1.3716 mc and Y_out(5.46875) = 1.09218 mc. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives wp_line = 1.53125 + (1.12922 − 1.3716)(5.46875 − 1.53125)/(1.09218 − 1.3716) = 4.94675 mc². 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 wp = 3.93313 mc²; substituting back gives Y_out = 1.1292224 mc.
- The interpolation error in the input is |wp_line − wp| = 1.01363 mc². This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.1236659 mc and 1.1347789 mc. Sort the roots even if the response decreases: 3.77645 mc² ≤ wp ≤ 4.10413 mc².
- The lower and upper input deviations are 0.156676 mc² and 0.171002 mc². 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 wp ≈ 3.93313 mc²; compatible interval [3.77645, 4.10413] mc². Endpoint interpolation gives 4.94675 mc².
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Holes as positrons: charge, energy, momentum and spin: infer an input with uncertainty
A hole is an empty negative-energy state in the filled Dirac sea. Measured from the vacuum it carries minus the charge, energy, momentum and angular momentum of the missing electron, which makes it a particle of charge +e, positive energy, mass m and spin ½: the positron. Use the supplied topic model Y_out = 0.511*(sqrt(1+p^2)+1), where Y_out is MIN PAIR COST |E| + mc² (MeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. p is Emptied state's momentum p_z, measured in mc. Keep sz (Emptied state's spin J_z) = 1 ħ/2 fixed. For this exercise use only -1.575 mc ≤ p ≤ -0.105 mc 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 = 1.1284629 MeV, with a bounded reading uncertainty ±0.015548 MeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000147 mc, 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.
MIN PAIR COST |E| + mc² against p
Predictions from the supplied Holes as positrons: charge, energy, momentum and spin 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): MIN PAIR COST |E| + mc² / MeV
Horizontal axis (x): p: Emptied state's momentum p_z / mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Emptied state's momentum p_z / mc | MIN PAIR COST |E| + mc² / MeV |
|---|---|
| -1.575 | 1.4643437368677679 |
| -1.2075 | 1.312155481823753 |
| -0.84 | 1.1783589570838169 |
| -0.4724999999999999 | 1.0761706779869686 |
| -0.10499999999999998 | 1.024809165960476 |
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(p) = 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(-1.575) = 1.46434 MeV and Y_out(-0.105) = 1.02481 MeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = -1.575 + (1.12846 − 1.46434)(-0.105 − -1.575)/(1.02481 − 1.46434) = -0.451664 mc. 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 p = -0.6783 mc; substituting back gives Y_out = 1.1284629 MeV.
- The interpolation error in the input is |p_line − p| = 0.226636 mc. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.1129149 MeV and 1.1440109 MeV. Sort the roots even if the response decreases: -0.731128 mc ≤ p ≤ -0.622483 mc.
- The lower and upper input deviations are 0.052828 mc and 0.0558169 mc. 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 p ≈ -0.6783 mc; compatible interval [-0.731128, -0.622483] mc. Endpoint interpolation gives -0.451664 mc.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.