University Physics IV · unit UPIV-08
The Quantum Harmonic Oscillator
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.
The classical oscillator and the harmonic approximation: infer an input with uncertainty
The harmonic approximation replaces a potential near a stable minimum by the leading quadratic term of its Taylor series, ½U″(x₀)ξ², so the system becomes a mass on a spring whose stiffness is the curvature of the curve. Use the supplied topic model Y_out = 1/(pi*sqrt(2*en/k)*sqrt(1-s*s)), where Y_out is DENSITY AT PROBE (/m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. s is Probe position x/A, measured in of A. Keep en (Total energy E) = 0.6 J; k (Curvature k = U″₀) = 3 N/m fixed. For this exercise use only -0.748125 of A ≤ s ≤ -0.049875 of A 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 = 0.53164239 /m, with a bounded reading uncertainty ±0.0041585 /m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000069825 of A, 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.
DENSITY AT PROBE against s
Predictions from the supplied The classical oscillator and the harmonic approximation 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): DENSITY AT PROBE / /m
Horizontal axis (x): s: Probe position x/A / of A
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Probe position x/A / of A | DENSITY AT PROBE / /m |
|---|---|
| -0.7481249999999999 | 0.7584751526560197 |
| -0.5735625 | 0.6143989130588873 |
| -0.39899999999999997 | 0.5488757434899616 |
| -0.22443749999999996 | 0.516468001905529 |
| -0.049875 | 0.5039192648176797 |
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(s) = 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(-0.748125) = 0.758475 /m and Y_out(-0.049875) = 0.503919 /m. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = -0.748125 + (0.531642 − 0.758475)(-0.049875 − -0.748125)/(0.503919 − 0.758475) = -0.12592 of A. 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 s = -0.322192 of A; substituting back gives Y_out = 0.53164239 /m.
- The interpolation error in the input is |s_line − s| = 0.196273 of A. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.52748389 /m and 0.53580089 /m. Sort the roots even if the response decreases: -0.343024 of A ≤ s ≤ -0.299369 of A.
- The lower and upper input deviations are 0.0208311 of A and 0.0228235 of A. 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 s ≈ -0.322192 of A; compatible interval [-0.343024, -0.299369] of A. Endpoint interpolation gives -0.12592 of A.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The oscillator equation and its scales: infer an input with uncertainty
Non-dimensionalising the oscillator means measuring x in units of the natural length x₀ = √(ħ/mω) and E in units of ħω, which reduces the Schrödinger equation to d²ψ/dξ² = (ξ² − ε)ψ with the single parameter ε = 2E/ħω. Use the supplied topic model Y_out = 25.201/sqrt(max(m*w,0.01)), where Y_out is NATURAL LENGTH x₀ (pm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. m is Particle mass m, measured in u. Keep w (Angular frequency ω) = 4 ×10¹⁴ rad/s; n (Level n) = 0 fixed. For this exercise use only 3.0625 u ≤ m ≤ 10.9375 u 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 = 4.4926637 pm, with a bounded reading uncertainty ±0.10239 pm. 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 0.0007875 u, 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.
NATURAL LENGTH x₀ against m
Predictions from the supplied The oscillator equation and its scales 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): NATURAL LENGTH x₀ / pm
Horizontal axis (x): m: Particle mass m / u
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Particle mass m / u | NATURAL LENGTH x₀ / pm |
|---|---|
| 3.0625 | 7.200285714285714 |
| 5.03125 | 5.617587294044586 |
| 7 | 4.762541342152767 |
| 8.96875 | 4.207477667267825 |
| 10.9375 | 3.8100330737222143 |
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(3.0625) = 7.20029 pm and Y_out(10.9375) = 3.81003 pm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m_line = 3.0625 + (4.49266 − 7.20029)(10.9375 − 3.0625)/(3.81003 − 7.20029) = 9.35186 u. 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 = 7.86625 u; substituting back gives Y_out = 4.4926637 pm.
- The interpolation error in the input is |m_line − m| = 1.48561 u. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.3902737 pm and 4.5950537 pm. Sort the roots even if the response decreases: 7.51959 u ≤ m ≤ 8.23744 u.
- The lower and upper input deviations are 0.346656 u and 0.371192 u. 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 ≈ 7.86625 u; compatible interval [7.51959, 8.23744] u. Endpoint interpolation gives 9.35186 u.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Asymptotic form, Hermite polynomials and quantisation: infer an input with uncertainty
The oscillator's bound states are Hermite polynomials times a Gaussian, and they exist only at E_n = (n + ½)ħω because only there does the power series multiplying the Gaussian terminate rather than outgrow it. Use the supplied topic model Y_out = ((1-p)*(1+(1-k)/2*16*(1+(5-k)/12*16*(1+(9-k)/30*16*(1+(13-k)/56*16*(1+(17-k)/90*16*(1+(21-k)/132*16*(1+(25-k)/182*16*(1+(29-k)/240*16*(1+(33-k)/306*16*(1+(37-k)/380*16*(1+(41-k)/462*16*(1+(45-k)/552*16*(1+(49-k)/650*16*(1+(53-k)/756*16*(1+(57-k)/870*16*(1+(61-k)/992*16*(1+(65-k)/1122*16*(1+(69-k)/1260*16*(1+(73-k)/1406*16*(1+(77-k)/1560*16*(1+(81-k)/1722*16*(1+(85-k)/1892*16*(1+(89-k)/2070*16*(1+(93-k)/2256*16*(1+(97-k)/2450*16*(1))))))))))))))))))))))))))+p*4*(1+(3-k)/6*16*(1+(7-k)/20*16*(1+(11-k)/42*16*(1+(15-k)/72*16*(1+(19-k)/110*16*(1+(23-k)/156*16*(1+(27-k)/210*16*(1+(31-k)/272*16*(1+(35-k)/342*16*(1+(39-k)/420*16*(1+(43-k)/506*16*(1+(47-k)/600*16*(1+(51-k)/702*16*(1+(55-k)/812*16*(1+(59-k)/930*16*(1+(63-k)/1056*16*(1+(67-k)/1190*16*(1+(71-k)/1332*16*(1+(75-k)/1482*16*(1+(79-k)/1640*16*(1+(83-k)/1806*16*(1+(87-k)/1980*16*(1+(91-k)/2162*16*(1+(95-k)/2352*16*(1))))))))))))))))))))))))))*exp(-8), where Y_out is TAIL ψ(ξ = 4) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. k is K = 2E/ħω, dimensionless. Keep p (Parity (0 even, 1 odd)) = 0 fixed. For this exercise use only 1.165 ≤ k ≤ 1.921 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 = -125.09425, with a bounded reading uncertainty ±2.4312. 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 0.0000756, 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.
TAIL ψ(ξ = 4) against k
Predictions from the supplied Asymptotic form, Hermite polynomials and quantisation 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): TAIL ψ(ξ = 4)
Horizontal axis (x): k: K = 2E/ħω
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| k: K = 2E/ħω | TAIL ψ(ξ = 4) |
|---|---|
| 1.165 | -48.78609064455918 |
| 1.354 | -89.3845032263566 |
| 1.543 | -116.633943246202 |
| 1.7319999999999998 | -133.190028587299 |
| 1.9209999999999998 | -141.30247000338258 |
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(1.165) = -48.7861 and Y_out(1.921) = -141.302. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives k_line = 1.165 + (-125.094 − -48.7861)(1.921 − 1.165)/(-141.302 − -48.7861) = 1.78855. 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 = 1.62616; substituting back gives Y_out = -125.09425.
- The interpolation error in the input is |k_line − k| = 0.162394. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -127.52545 and -122.66305. Sort the roots even if the response decreases: 1.60021 ≤ k ≤ 1.6543.
- The lower and upper input deviations are 0.0259543 and 0.0281421. 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 ≈ 1.62616; compatible interval [1.60021, 1.6543]. Endpoint interpolation gives 1.78855.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The ladder and zero-point energy: infer an input with uncertainty
The harmonic ladder is the spectrum Eₙ = (n + ½)ħω for n = 0, 1, 2, …: infinitely many levels separated by the same ħω, the lowest of them — the zero-point energy — sitting half a quantum above the bottom of the well rather than on it. Use the supplied topic model Y_out = (w-2*a*(n+1))/max(w,0.01), where Y_out is GAP AS A FRACTION OF ħω (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. a is Anharmonicity a, measured in eV. Keep w (Vibrational quantum ħω) = 0.27 eV; n (Marked level n) = 0 fixed. For this exercise use only 0.0017 eV ≤ a ≤ 0.0143 eV 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 = 0.93047407, with a bounded reading uncertainty ±0.00546. 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 0.00000126 eV, 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.
GAP AS A FRACTION OF ħω against a
Predictions from the supplied The ladder and zero-point energy 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): GAP AS A FRACTION OF ħω
Horizontal axis (x): a: Anharmonicity a / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Anharmonicity a / eV | GAP AS A FRACTION OF ħω |
|---|---|
| 0.0017 | 0.9874074074074074 |
| 0.004849999999999999 | 0.9640740740740741 |
| 0.007999999999999998 | 0.9407407407407407 |
| 0.011149999999999998 | 0.9174074074074074 |
| 0.014299999999999998 | 0.8940740740740741 |
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(0.0017) = 0.987407 and Y_out(0.0143) = 0.894074. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 0.0017 + (0.930474 − 0.987407)(0.0143 − 0.0017)/(0.894074 − 0.987407) = 0.009386 eV. 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 = 0.009386 eV; substituting back gives Y_out = 0.93047407.
- The interpolation error in the input is |a_line − a| = 1.21431e-17 eV. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.92501407 and 0.93593407. Sort the roots even if the response decreases: 0.0086489 eV ≤ a ≤ 0.0101231 eV.
- The lower and upper input deviations are 0.0007371 eV and 0.0007371 eV. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer a ≈ 0.009386 eV; compatible interval [0.0086489, 0.0101231] eV. Endpoint interpolation gives 0.009386 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.