University Physics V · Midterm Examination

|a⟩ and |b⟩ are orthonormal kets, and a system is prepared in |ψ⟩ = (|a⟩ + i|b⟩)/√2. Which of the following kets describes the same physical state as |ψ⟩?

1. A(|a⟩ − i|b⟩)/√2
2. B(i|a⟩ + |b⟩)/√2
3. C(i|a⟩ − |b⟩)/√2
4. D(|a⟩ + |b⟩)/√2

In a semiclassical model of the interaction of light with matter, the electromagnetic field is a classical wave and only the electrons in the detector are treated quantum mechanically. Which observation can this model NOT account for?

1. ALight from a single trapped atom, re-excited by one short laser pulse at a time, falls on a 50:50 beam splitter with a detector at each output, and coincident clicks at the two detectors within one pulse are far rarer than any classical wave of the same mean intensity allows.
2. BThe maximum kinetic energy of photoelectrons increases linearly with the frequency of the light, with gradient h.
3. CNo photoelectrons are emitted below a threshold frequency, however intense the light.
4. DPhotoelectrons are emitted with no measurable delay after very weak light is switched on.

A particle is in the second excited state, n = 3, of an infinitely deep one-dimensional well occupying 0 < x < L. What is the probability of finding it in the first quarter of the well, 0 < x < L/4?

1. A0.15
2. B0.20
3. C0.25
4. D0.30

A steady beam of non-relativistic particles, each with kinetic energy E, travels in the +x direction and meets a potential step of height V₀ = 3E/4 at x = 0. What fraction of the incident probability current is transmitted past the step?

1. A1/9
2. B8/9
3. C16/9
4. D16/25

For a one-dimensional harmonic oscillator of mass m and angular frequency ω, the position operator is x = √(ℏ/2mω)(a + a†), with a|n⟩ = √n |n − 1⟩ and a†|n⟩ = √(n + 1) |n + 1⟩. What is the matrix element ⟨3|x²|1⟩?

1. A0
2. B√2 ℏ/(2mω)
3. C√6 ℏ/(2mω)
4. D3ℏ/(2mω)

Two Hermitian operators A and B on a finite-dimensional state space satisfy [A, B] = 0. Which statement must be true?

1. AEvery eigenvector of A is also an eigenvector of B.
2. BΔA ΔB = 0 in every state.
3. CA measurement of B made straight after a measurement of A leaves the state produced by the A measurement unchanged.
4. DThere is an orthonormal basis made entirely of simultaneous eigenvectors of A and B.

X-ray photons of wavelength 60.0 pm scatter from free electrons initially at rest, and photons scattered through 120° are detected. Take h/(mₑc) = 2.426 pm and hc = 1240 eV nm. What is the kinetic energy of an electron that has scattered one of these detected photons?

1. A0.41 keV
2. B1.18 keV
3. C19.5 keV
4. D20.7 keV

Beam P consists of spin-½ atoms each prepared in (|↑⟩ + |↓⟩)/√2, where |↑⟩ and |↓⟩ are the eigenstates of S_z. Beam Q is an equal statistical mixture of atoms in |↑⟩ and atoms in |↓⟩. Which measurement, repeated on many atoms from each beam, distinguishes the two beams?

1. AS_x: every atom from P gives +ℏ/2, while atoms from Q give +ℏ/2 and −ℏ/2 equally often.
2. BS_z: each atom from P gives either result with probability ½, whereas each atom in Q already has a definite value.
3. CS_y: the relative phase between |↑⟩ and |↓⟩ in P can only be detected by a measurement along y.
4. DNone: in both beams each atom is found up along z with probability ½, so no measurement can separate them.

An electron shared between two coupled quantum dots is described in the orthonormal basis {|1⟩, |2⟩} of the two dot states by a Hamiltonian with matrix elements H₁₁ = 4.0 meV, H₂₂ = −4.0 meV, H₁₂ = 3.0i meV and H₂₁ = −3.0i meV. At t = 0 the electron is in |1⟩. Take ℏ = 6.582 × 10⁻¹⁶ eV s.

1. (a)

   Show that Show that the energy eigenvalues are E₊ = +5.0 meV and E₋ = −5.0 meV.

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   1. 1

      Sets det(H − EI) = 0 with the off-diagonal product written out: (4.0 − E)(−4.0 − E) − (3.0i)(−3.0i) = 0, with E in meV

   2. 2

      (3.0i)(−3.0i) = +9.0, so E² − 16 − 9 = 0, giving E² = 25 and E± = ±5.0 meV

2. (b)

   Determine Determine the normalised eigenvector |E₊⟩ as a combination of |1⟩ and |2⟩.

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   1. 1

      From the first row of (H − 5.0 meV I)c = 0: (4.0 − 5.0)c₁ + 3.0i c₂ = 0, so c₂ = −ic₁/3

   2. 2

      Normalising with |c₁|² + |c₂|² = 1 gives |E₊⟩ = (3|1⟩ − i|2⟩)/√10, to within an overall phase

3. (c)

   Calculate Calculate the probability that a measurement of the energy at t = 0 gives +5.0 meV.

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   1. 1

      P(+5.0 meV) = |⟨E₊|1⟩|² = |3/√10|² = 0.90

4. (d)

   Derive Starting from the time-dependent Schrödinger equation, derive an expression for the probability P₂(t) that the electron is found in |2⟩ at time t.

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   1. 1

      Because H is time independent, iℏ d|ψ⟩/dt = H|ψ⟩ gives |ψ(t)⟩ = Σ e^(−iE±t/ℏ) |E±⟩⟨E±|1⟩, with ⟨E₊|1⟩ = 3/√10, ⟨E₋|1⟩ = 1/√10 and |E₋⟩ = (|1⟩ + 3i|2⟩)/√10 (or, since H² = (5.0 meV)² I, uses e^(−iHt/ℏ) = cos(Et/ℏ) I − i sin(Et/ℏ) H/E)

   2. 2

      ⟨2|ψ(t)⟩ = (3/√10)(−i/√10)e^(−iEt/ℏ) + (1/√10)(3i/√10)e^(iEt/ℏ) = (3i/10)(e^(iEt/ℏ) − e^(−iEt/ℏ)) = −(3/5) sin(Et/ℏ), where E = 5.0 meV

   3. 3

      P₂(t) = (9/25) sin²(Et/ℏ) = 0.36 sin²[(5.0 meV)t/ℏ]

5. (e)

   Calculate Calculate the earliest time after t = 0 at which P₂ is greatest, and state that greatest value.

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   1. 1

      P₂ is greatest, at 0.36, when (5.0 meV)t/ℏ = π/2

   2. 2

      t = πℏ/(2 × 5.0 meV) = π × 6.582 × 10⁻¹⁶ eV s / (1.0 × 10⁻² eV) = 2.1 × 10⁻¹³ s

An electron of mass mₑ = 9.109 × 10⁻³¹ kg is bound in a symmetric one-dimensional square well occupying −a < x < a, with a = 0.500 nm. Its potential energy is zero inside the well and V₀ = 1.20 eV outside it. Measure the energy E from the bottom of the well and write k = √(2mₑE)/ℏ, z = ka and z₀ = (a/ℏ)√(2mₑV₀). The even bound states satisfy z tan z = √(z₀² − z²) and the odd bound states satisfy −z cot z = √(z₀² − z²). Take ℏ = 1.055 × 10⁻³⁴ J s and e = 1.602 × 10⁻¹⁹ C.

1. (a)

   Determine Calculate z₀, and hence determine how many bound states the well holds.

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   1. 1

      z₀ = (0.500 × 10⁻⁹ m / 1.055 × 10⁻³⁴ J s) × √(2 × 9.109 × 10⁻³¹ kg × 1.20 × 1.602 × 10⁻¹⁹ J), with V₀ converted to joules

   2. 2

      z₀ = 2.80 (dimensionless)

   3. 3

      An even root always exists; an odd root exists because z₀ > π/2 = 1.57; a second even root would need z₀ > π = 3.14. So the well holds 2 bound states, equivalently floor(2z₀/π) + 1 = floor(1.79) + 1 = 2

2. (b)

   Show that Show that z = 1.15 satisfies the even-parity condition to within 1%.

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   1. 1

      Left side: 1.15 × tan(1.15 rad) = 1.15 × 2.234 = 2.57

   2. 2

      Right side: √(2.80² − 1.15²) = 2.55 (2.56 with z₀ = 2.805), which differs from 2.57 by less than 1% (0.45% to 0.66%, depending on the z₀ used)

3. (c)

   Compare Calculate the ground-state energy of the electron, and compare it with the ground-state energy the electron would have in an infinitely deep well of the same width 2a.

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   1. 1

      E₁ = ℏ²z²/(2mₑa²) = V₀(z/z₀)² = 1.20 eV × (1.15/2.80)² = 0.20 eV

   2. 2

      Infinite well of width 2a = 1.00 nm: E₁,∞ = π²ℏ²/[2mₑ(2a)²] = 6.03 × 10⁻²⁰ J = 0.376 eV

   3. 3

      The finite-well value is about 0.54 of the infinite-well value because the wavefunction extends into the classically forbidden region: spread over a greater effective width, it has less curvature and less kinetic energy. Both are non-zero, as confinement to a finite region requires

4. (d)

   Sketch On one set of axes of ψ against x, from x = −2a to x = +2a, sketch the wavefunctions of the two bound states. Mark x = −a and x = +a.

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   1. 1

      Ground state: even with no node, cosine-like inside the well, joining smoothly (ψ and dψ/dx continuous) onto exponential decay outside, and non-zero at x = ±a

   2. 2

      First excited state: odd with a single node at x = 0, and with exponential tails that decay more slowly than the ground state's, because it lies nearer the top of the well (κa ≈ 1.7 against 2.6)

Slow neutrons, which have spin ½ and gyromagnetic ratio γₙ = −1.832 × 10⁸ rad s⁻¹ T⁻¹, leave a polariser in the state |+z⟩. The spin operator is S = (ℏ/2)σ; in the {|+z⟩, |−z⟩} basis σ_y has rows (0, −i) and (i, 0), and exp(−iασ_y) = cos α I − i sin α σ_y for real α. An analyser set along a unit vector n̂ transmits only neutrons found in the +ℏ/2 eigenstate of n̂⋅S, and blocks the rest.

1. (a)

   Calculate The neutrons pass first through an analyser whose n̂ lies in the xz-plane at 45° to the +z axis, and then through an analyser set along −z, which transmits only |−z⟩. Calculate the fraction of the neutrons leaving the polariser that emerge from the second analyser.

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   1. 1

      |+n̂⟩ = cos 22.5° |+z⟩ + sin 22.5° |−z⟩, so the first analyser transmits |⟨+n̂|+z⟩|² = cos² 22.5° = 0.854

   2. 2

      The second analyser transmits |⟨−z|+n̂⟩|² = sin² 22.5° = 0.146 of these, so the fraction is 0.854 × 0.146 = 0.125 (exactly 1/8)

2. (b)

   Explain With the first analyser removed, no neutrons emerge from the −z analyser. Explain why inserting an analyser that removes neutrons from the beam increases the number emerging, and identify which step in the sequence is not unitary.

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   1. 1

      |+z⟩ and |−z⟩ are orthogonal, so with no intermediate analyser the amplitude to pass the −z analyser is zero; the 45° analyser replaces the state by |+n̂⟩, which has amplitude sin 22.5° on |−z⟩

   2. 2

      Each analyser performs a projective measurement: projecting with |+n̂⟩⟨+n̂| and renormalising what survives is not unitary, since it discards part of the beam and cannot be reversed, whereas free flight between analysers is unitary

3. (c)

   Show that The 45° analyser is removed. Instead, the |+z⟩ neutrons cross a region of uniform magnetic field B along +y, where the Hamiltonian is H = −γₙB S_y, taking a time t. Show that on leaving the field the spin state is cos(ωt/2)|+z⟩ + sin(ωt/2)|−z⟩, where ω = |γₙ|B.

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   1. 1

      Because γₙ < 0, H = |γₙ|B S_y = (ℏω/2)σ_y with ω = |γₙ|B; H is time independent, so the propagator is U = exp(−iHt/ℏ) = exp(−i(ωt/2)σ_y)

   2. 2

      By the given identity, U = cos(ωt/2) I − i sin(ωt/2) σ_y

   3. 3

      σ_y|+z⟩ = i|−z⟩, so U|+z⟩ = cos(ωt/2)|+z⟩ − i sin(ωt/2)(i|−z⟩) = cos(ωt/2)|+z⟩ + sin(ωt/2)|−z⟩

4. (d)

   Calculate The neutrons travel at 2200 m s⁻¹ through a field region 5.00 cm long. Calculate the field B that turns |+z⟩ into |+x⟩ = (|+z⟩ + |−z⟩)/√2.

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   1. 1

      |+x⟩ requires ωt/2 = π/4, so ωt = π/2, with t = L/v = 0.0500 m / 2200 m s⁻¹ = 2.27 × 10⁻⁵ s

   2. 2

      B = π/(2|γₙ|t) = π / (2 × 1.832 × 10⁸ rad s⁻¹ T⁻¹ × 2.27 × 10⁻⁵ s) = 3.77 × 10⁻⁴ T

5. (e)

   Predict Predict the fraction of the neutrons that a +z analyser placed after the field region would transmit if the same field pointed along +z instead of +y, and give the reason.

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   1. 1

      All of them, a fraction of 1: with B along z, H is proportional to S_z, so |+z⟩ is a stationary state that acquires only a global phase, and the z populations do not change

A single ⁴⁰Ca⁺ ion of mass m = 6.64 × 10⁻²⁶ kg is held in an ion trap whose potential along the trap axis is harmonic with angular frequency ω = 2π × 1.20 MHz = 7.54 × 10⁶ rad s⁻¹. Along that axis the position and momentum operators are x = √(ℏ/2mω)(a + a†) and p = i√(mℏω/2)(a† − a), with [a, a†] = 1 and number states |n⟩ of energy (n + ½)ℏω. Take ℏ = 1.055 × 10⁻³⁴ J s and k_B = 1.381 × 10⁻²³ J K⁻¹.

1. (a)

   Estimate After laser cooling, the axial motion is in thermal equilibrium at T = 0.50 mK. Estimate the mean number of vibrational quanta n̄, stating the approximation you make and why it is justified.

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   1. 1

      ℏω/(k_BT) = (1.055 × 10⁻³⁴ J s × 7.54 × 10⁶ rad s⁻¹)/(1.381 × 10⁻²³ J K⁻¹ × 5.0 × 10⁻⁴ K) = 0.115, so k_BT ≫ ℏω and the high-temperature (equipartition) limit n̄ ≈ k_BT/ℏω applies

   2. 2

      n̄ ≈ 1/0.115 ≈ 9 (8.7); the Bose–Einstein value 1/(e^0.115 − 1) = 8.2 is also accepted, as is any value from 8 to 9

2. (b)

   Show that Further cooling and a laser pulse prepare the state |ψ(0)⟩ = (|0⟩ + |1⟩)/√2. Show that ⟨x⟩(t) = √(ℏ/2mω) cos ωt.

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   1. 1

      Each number state evolves as e^(−i(n + ½)ωt)|n⟩, so up to a global phase |ψ(t)⟩ = (|0⟩ + e^(−iωt)|1⟩)/√2

   2. 2

      Only ⟨0|a|1⟩ = 1 and ⟨1|a†|0⟩ = 1 contribute, giving ⟨a⟩ = ½e^(−iωt) and ⟨a†⟩ = ½e^(iωt), so ⟨x⟩ = √(ℏ/2mω)(⟨a⟩ + ⟨a†⟩) = √(ℏ/2mω) cos ωt

3. (c)

   Determine Determine the product Δx Δp at t = 0 and at ωt = π/4, expressing each as a multiple of ℏ.

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   1. 1

      (a + a†)² = a² + a†² + 2N + 1, and a² and a†² cannot connect |0⟩ with |1⟩, so ⟨x²⟩ = (ℏ/2mω)(2⟨N⟩ + 1) = ℏ/(mω) with ⟨N⟩ = ½; likewise −(a† − a)² = 2N + 1 − a² − a†² gives ⟨p²⟩ = mℏω

   2. 2

      ⟨p⟩ = i√(mℏω/2)(⟨a†⟩ − ⟨a⟩) = −√(mℏω/2) sin ωt, which equals m d⟨x⟩/dt as Ehrenfest's theorem requires

   3. 3

      (Δx)² = (ℏ/2mω)(2 − cos² ωt) and (Δp)² = (mℏω/2)(2 − sin² ωt), so Δx Δp = (ℏ/2)√[(2 − cos² ωt)(2 − sin² ωt)]

   4. 4

      At t = 0, Δx Δp = (ℏ/2)√2 = 0.71ℏ; at ωt = π/4, Δx Δp = (ℏ/2)(3/2) = 0.75ℏ; both exceed ℏ/2

4. (d)

   Explain Explain why Δx Δp exceeds ℏ/2 at every time for this state, although the ground state |0⟩ reaches ℏ/2.

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   1. 1

      Equality in Δx Δp ≥ ℏ/2 requires both inequalities of the derivation to be saturated, Cauchy–Schwarz and the dropping of Re⟨f|g⟩, which together demand (p − ⟨p⟩)|ψ⟩ = iλ(x − ⟨x⟩)|ψ⟩ with λ real, and normalisability needs λ > 0; the only solutions are Gaussian wavefunctions

   2. 2

      In position space this state is proportional to (1 + c(t)x) e^(−mωx²/2ℏ) with c(t) = e^(−iωt)√(2mω/ℏ): a Gaussian multiplied by a polynomial, which is not a Gaussian at any t, so the bound is never reached, whereas |0⟩ is a Gaussian and reaches it

A scanning tunnelling microscope tip is held above a clean metal surface at a fixed bias of 0.100 V. With the feedback loop switched off, the tip is withdrawn in steps, and the tunnelling current I is recorded against Δd, the increase in the tip–surface gap from its starting value. In the opaque-barrier limit I = I₀ exp(−2κΔd), where κ = √(2mₑφ)/ℏ and φ is the apparent barrier height. Take mₑ = 9.109 × 10⁻³¹ kg, ℏ = 1.055 × 10⁻³⁴ J s and e = 1.602 × 10⁻¹⁹ C.

1. (a)

   State State the gradient of a graph of ln(I/nA) against Δd, in terms of κ.

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   1. 1

      Gradient = −2κ, since ln(I/nA) = ln(I₀/nA) − 2κΔd

2. (b)

   Determine Determine κ from the data.

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   1. 1

      ln(I/nA) for the six readings: 0.916, 0.122, −0.753, −1.523, −2.382, −3.170

   2. 2

      The points lie on a straight line (no residual larger than about 0.03) of gradient −2.05 × 10⁻² pm⁻¹ (least squares −0.0205 pm⁻¹; accept −0.0200 to −0.0210 pm⁻¹)

   3. 3

      κ = 0.0205 pm⁻¹ / 2 = 1.03 × 10⁻² pm⁻¹ = 1.03 × 10¹⁰ m⁻¹ (accept 1.00 × 10¹⁰ to 1.05 × 10¹⁰ m⁻¹)

3. (c)

   Calculate Calculate the apparent barrier height φ in electronvolts.

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   1. 1

      φ = ℏ²κ²/(2mₑ) = (1.055 × 10⁻³⁴ J s)² × (1.03 × 10¹⁰ m⁻¹)² / (2 × 9.109 × 10⁻³¹ kg) = 6.4 × 10⁻¹⁹ J

   2. 2

      φ = 4.0 eV (accept 3.8 to 4.2 eV, following from the κ found in (b))

4. (d)

   Estimate In constant-current imaging the feedback loop moves the tip to hold I fixed. Estimate the change in gap that changes the current by 10%, stating the assumption you make.

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   1. 1

      Assumes the exponential law holds over a few picometres with the fitted κ and an unchanged prefactor I₀, so that the change in gap is ln(1.10)/(2κ) for a 10% rise in current, or ln(1/0.90)/(2κ) for a 10% fall

   2. 2

      Δd = 0.0953 / 0.0205 pm⁻¹ ≈ 5 pm (4.6 pm for a rise, 5.1 pm for a fall, and 4.9 pm from the small-change form 0.10/(2κ) are all accepted), so holding the current to 10% fixes the gap to a few picometres, about 2% of an atomic diameter

5. (e)

   Evaluate Photoemission gives the work function of this surface as 4.60 eV. A student concludes that either the STM data or the tunnelling model must be wrong. Evaluate this conclusion.

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   1. 1

      The fitted φ ≈ 4.0 eV is about 13% below 4.60 eV, but the apparent barrier height is not the work function: the image-charge potential rounds and lowers the barrier between tip and surface, and the tip's own work function also enters, so a lower apparent height is expected

   2. 2

      The ln I values lie on a straight line to within about 0.03 over the whole 200 pm, which supports the exponential tunnelling model; the conclusion is not justified, because the data fit the model and the gap from 4.60 eV is explained by what φ represents

For Hermitian operators A and B and a normalised state |ψ⟩, write ⟨A⟩ = ⟨ψ|A|ψ⟩ and (ΔA)² = ⟨ψ|(A − ⟨A⟩)²|ψ⟩. The Cauchy–Schwarz inequality states that ⟨f|f⟩⟨g|g⟩ ≥ |⟨f|g⟩|² for any kets |f⟩ and |g⟩. The components of an angular momentum obey [J_x, J_y] = iℏJ_z and its cyclic permutations, J± = J_x ± iJ_y, and the simultaneous eigenstates |j, m⟩ of J² and J_z satisfy J²|j, m⟩ = ℏ²j(j + 1)|j, m⟩, J_z|j, m⟩ = mℏ|j, m⟩ and J±|j, m⟩ = ℏ√[j(j + 1) − m(m ± 1)] |j, m ± 1⟩.

1. (a)

   Derive Starting from the Cauchy–Schwarz inequality, derive the Robertson uncertainty relation, a lower bound on ΔA ΔB in terms of the commutator [A, B].

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   1. 1

      Defines |f⟩ = (A − ⟨A⟩)|ψ⟩ and |g⟩ = (B − ⟨B⟩)|ψ⟩; because the shifted operators are Hermitian, ⟨f|f⟩ = (ΔA)² and ⟨g|g⟩ = (ΔB)²

   2. 2

      Cauchy–Schwarz gives (ΔA)²(ΔB)² ≥ |⟨f|g⟩|² ≥ (Im⟨f|g⟩)² = [(⟨f|g⟩ − ⟨g|f⟩)/2i]²

   3. 3

      ⟨f|g⟩ − ⟨g|f⟩ = ⟨ψ|[A − ⟨A⟩, B − ⟨B⟩]|ψ⟩ = ⟨[A, B]⟩, because the numbers ⟨A⟩ and ⟨B⟩ commute with every operator

   4. 4

      Hence ΔA ΔB ≥ ½|⟨[A, B]⟩|, a real bound because [A, B] is anti-Hermitian and its expectation value is purely imaginary

2. (b)

   Show that Show that in the state |j, m⟩, ⟨J_x⟩ = ⟨J_y⟩ = 0 and ⟨J_x²⟩ = ⟨J_y²⟩ = ½ℏ²[j(j + 1) − m²].

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   1. 1

      J_x = (J₊ + J₋)/2 and J_y = (J₊ − J₋)/2i; J± changes m by ±1 and ⟨j, m|j, m ± 1⟩ = 0, so ⟨J_x⟩ = ⟨J_y⟩ = 0

   2. 2

      J_x² = ¼(J₊² + J₋² + J₊J₋ + J₋J₊) and J_y² = −¼(J₊² + J₋² − J₊J₋ − J₋J₊); the J±² terms change m by ±2 and have zero expectation value, so ⟨J_x²⟩ = ⟨J_y²⟩ = ¼⟨J₊J₋ + J₋J₊⟩

   3. 3

      J₊J₋ + J₋J₊ = 2(J_x² + J_y²) = 2(J² − J_z²); equivalently ⟨J₊J₋⟩ + ⟨J₋J₊⟩ = ℏ²[j(j + 1) − m(m − 1)] + ℏ²[j(j + 1) − m(m + 1)] from the given ladder elements

   4. 4

      ⟨J_x²⟩ = ⟨J_y²⟩ = ½⟨J² − J_z²⟩ = ½ℏ²[j(j + 1) − m²]

3. (c)

   Determine Determine the states |j, m⟩ in which ΔJ_x ΔJ_y is exactly equal to the Robertson bound for J_x and J_y.

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   Mark-by-mark answer

   1. 1

      With both means zero, (b) gives ΔJ_x ΔJ_y = ½ℏ²[j(j + 1) − m²]

   2. 2

      The Robertson bound is ½|⟨[J_x, J_y]⟩| = ½ℏ|⟨J_z⟩| = ½ℏ²|m|

   3. 3

      Equality requires j(j + 1) − m² = |m|, that is (j − |m|)(j + |m| + 1) = 0, and since j + |m| + 1 > 0 this means |m| = j: only the extreme states |j, ±j⟩ reach the bound, which for spin ½ includes both S_z eigenstates

4. (d)

   Evaluate A student argues: 'In the state |j = 1, m = 0⟩ the Robertson bound for J_x and J_y is zero, so a careful enough experiment could find both J_x and J_y with no spread in that state.' Evaluate this argument.

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   1. 1

      In |1, 0⟩, (b) gives ΔJ_x = ΔJ_y = ℏ, so ΔJ_x ΔJ_y = ℏ²: the spreads belong to the prepared state, no experimental care reduces them, and a lower bound of zero is not a value the state attains

   2. 2

      Zero spread in both would need a common eigenvector of J_x and J_y; then [J_x, J_y]|ψ⟩ = 0 forces J_z|ψ⟩ = 0, the cyclic relations then force J_x|ψ⟩ = J_y|ψ⟩ = 0, so J²|ψ⟩ = 0 and j = 0

   3. 3

      Judgement: the argument fails. A vanishing bound is necessary but not sufficient for joint sharpness, and no state with j ≥ ½, |1, 0⟩ included, has both components sharp