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Derivation Problems

These problems combine established results to obtain a new diagnostic. They use natural units ℏ=c=1\hbar=c=1 and the mostly-minus metric. Treat the linked spectra, normalization rules and free-field algebra as inputs; the purpose is to learn what additional conclusions follow and under which assumptions.

Helpful background. Relativistic Normalization supports problem 1; Relativistic Landau Levels supports problem 2; Invariant Phase Space supports problem 3; and Locality and Causality Warnings supplies the causal commutator distinction for problem 4.

A rapidity packet and two different acceptance regions

Section titled “A rapidity packet and two different acceptance regions”

Problem 1. Work explicitly in 1+11+1 dimensions with a massive spin-zero particle, m>0m>0. Set p=msinh⁡yp=m\sinh y, E=mcosh⁡yE=m\cosh y, and use covariant normalization ⟨p′∣p⟩C=2Ep(2π)δ(p′−p)\langle p'|p\rangle_C=2E_p(2\pi)\delta(p'-p). Let

F(y)=(2πs2)−1/4exp⁡ ⁣[−(y−y0)24s2],f(p(y))=4π F(y),s>0.\begin{aligned} F(y)&=(2\pi s^2)^{-1/4} \exp\!\left[-\frac{(y-y_0)^2}{4s^2}\right],\\ f(p(y))&=\sqrt{4\pi}\,F(y),\qquad s>0. \end{aligned}

The state is ∣f⟩=∫dp f(p)∣p⟩C/(2π 2Ep)|f\rangle=\int dp\,f(p)|p\rangle_C/(2\pi\,2E_p).

  1. Show it has unit norm and find its coefficient a(p)a(p) in the basis with overlap (2π)δ(p′−p)(2\pi)\delta(p'-p).
  2. Apply a passive boost of rapidity ξ\xi, so y′=y−ξy'=y-\xi. Derive the transformation of aa and check its norm.
  3. Compare the probability of p′>0p'>0 with the probability of the transformed image of the original region p>0p>0.
  4. Compute ⟨E⟩\langle E\rangle, ⟨p⟩\langle p\rangle, ⟨P⟩2\langle P\rangle^2 and ⟨P2⟩\langle P^2\rangle. Interpret the last two.
Solution

The mass-shell measure simplifies to

dp2π 2Ep=dy4π.\frac{dp}{2\pi\,2E_p}=\frac{dy}{4\pi}.

Hence the norm is ∫dy ∣F(y)∣2=1\int dy\,|F(y)|^2=1. The coefficient in the stated NN basis is a(p)=f(p)/2Epa(p)=f(p)/\sqrt{2E_p} and obeys ∫dp ∣a(p)∣2/(2π)=1\int dp\,|a(p)|^2/(2\pi)=1. This is a 1+11+1 measure; no hidden transverse integration is present.

For a scalar covariant coefficient, f′(y′)=f(y′+ξ)f'(y')=f(y'+\xi). Therefore

a′(p′)=EpEp′ a(p),dp′=Ep′Ep dp.\begin{aligned} a'(p')&=\sqrt{\frac{E_p}{E_{p'}}}\,a(p),\\ dp'&=\frac{E_{p'}}{E_p}\,dp. \end{aligned}

The square-root and Jacobian cancel in the norm. The coefficient transformation has the inverse energy factor to the corresponding normalized ket transformation.

Let ΦN(z)\Phi_{\rm N}(z) denote the standard normal cumulative distribution function. Since p′>0p'>0 means y′>0y'>0,

Pr⁡(p′>0)=ΦN ⁣(y0−ξs).\Pr(p'>0)= \Phi_{\rm N}\!\left(\frac{y_0-\xi}{s}\right).

The transformed original region is y′>−ξy'>-\xi, equivalently p′>−msinh⁡ξp'>-m\sinh\xi. Its probability is

Pr⁡(y′>−ξ)=ΦN(y0/s).\Pr(y'>-\xi)=\Phi_{\rm N}(y_0/s).

The first expression uses a different laboratory acceptance region. The second describes the same projector after transformation. Lorentz covariance requires the latter equality, not equality of two different experiments.

The Gaussian moment formula ⟨ety⟩=ety0+t2s2/2\langle e^{t y}\rangle= e^{t y_0+t^2s^2/2} gives

⟨E⟩=mes2/2cosh⁡y0,⟨p⟩=mes2/2sinh⁡y0,⟨Pμ⟩⟨Pμ⟩=m2es2,⟨PμPμ⟩=m2.\begin{aligned} \langle E\rangle&=m e^{s^2/2}\cosh y_0,\\ \langle p\rangle&=m e^{s^2/2}\sinh y_0,\\ \langle P^\mu\rangle\langle P_\mu\rangle &=m^2e^{s^2},\\ \langle P^\mu P_\mu\rangle&=m^2. \end{aligned}

The representation’s Casimir is unchanged. The mean vector includes the energy spread of the packet and its square is not a new irreducible mass label.

For m=1m=1, y0=0.7y_0=0.7, s=0.4s=0.4, ξ=1.1\xi=1.1, useful numerical checks are

Pr⁡(p′>0)≈0.1586552539,Pr⁡(y′>−ξ)≈0.9599408431,⟨E⟩≈1.3597083515,⟨p⟩≈0.8217639140.\begin{aligned} \Pr(p'>0)&\approx0.1586552539,\\ \Pr(y'>-\xi)&\approx0.9599408431,\\ \langle E\rangle&\approx1.3597083515,\\ \langle p\rangle&\approx0.8217639140. \end{aligned}

These are momentum-space probabilities; the calculation does not define a relativistic position Born density.

The lowest Landau level controls a continuum check

Section titled “The lowest Landau level controls a continuum check”

Problem 2. For a massive Dirac particle in a uniform magnetic field, put b=∣q∣B>0b=|q|B>0. Import the positive-energy spectrum and its physical multiplicity:

EN2=m2+pz2+2bN,dN=2−δN0.E_N^2=m^2+p_z^2+2bN,\qquad d_N=2-\delta_{N0}.

The orbital degeneracy per transverse area is b/(2π)b/(2\pi). For τ>0\tau>0, evaluate the convergent spectral trace per volume

Kb(τ)=b2π∑N=0∞dN∫dpz2πe−τ(EN2−m2).K_b(\tau)=\frac{b}{2\pi} \sum_{N=0}^{\infty}d_N \int\frac{dp_z}{2\pi} e^{-\tau(E_N^2-m^2)}.

Find the zero-field limit and the first field correction. Diagnose the effect of assigning multiplicity two to N=0N=0, or adding the negative-energy sector to this definition.

Solution

The longitudinal Gaussian integral and geometric level sum are

∫dpz2πe−τpz2=12πτ,1+2∑N=1∞e−2bτN=coth⁡(bτ).\begin{aligned} \int\frac{dp_z}{2\pi}e^{-\tau p_z^2} &=\frac1{2\sqrt{\pi\tau}},\\ 1+2\sum_{N=1}^{\infty}e^{-2b\tau N} &=\coth(b\tau). \end{aligned}

Thus

Kb(τ)=bcoth⁡(bτ)4π3/2τ.K_b(\tau)= \frac{b\coth(b\tau)}{4\pi^{3/2}\sqrt{\tau}}.

At zero field the two positive-energy spin states give

K0(τ)=2∫d3p(2π)3e−τp2=14π3/2τ3/2.K_0(\tau)= 2\int\frac{d^3p}{(2\pi)^3}e^{-\tau\mathbf p^2} =\frac1{4\pi^{3/2}\tau^{3/2}}.

With y=bτy=b\tau the ratio is

KbK0=ycoth⁡y=1+y23−y445+O(y6).\frac{K_b}{K_0} =y\coth y =1+\frac{y^2}{3}-\frac{y^4}{45} +O(y^6).

The first correction is quadratic. Incorrectly doubling the lowest level adds yy to this ratio and creates a spurious linear term. Adding the negative-energy branch doubles the entire trace: that counts a second sector absent from the problem’s definition.

This is a positive-band spectral diagnostic. It is not a thermal partition function, a vacuum effective action or an anomaly calculation. Its ingredients come from Relativistic Landau Levels; no second derivation of that spectrum is needed.

A recoil bound for replacing a target by a source

Section titled “A recoil bound for replacing a target by a source”

Problem 3. A massless projectile p=(E,En)p=(E,E\mathbf n) scatters elastically from P=(M,0)P=(M,\mathbf0) into p′=(E′,E′n′)p'=(E',E'\mathbf n'). The target remains on its original mass shell, and n⋅n′=cos⁡θ\mathbf n\cdot\mathbf n'=\cos\theta. Derive E′/EE'/E and the condition for the fractional recoil loss to be at most ε\varepsilon, with 0<ε<10<\varepsilon<1. Explain why a weak coupling does not by itself satisfy this condition.

Solution

Momentum conservation gives P′=P+p−p′P'=P+p-p'. Expanding P′2=M2P'^2=M^2 yields

2M(E−E′)−2EE′(1−cos⁡θ)=0,2M(E-E')-2EE'(1-\cos\theta)=0,

so

E′E=11+(E/M)(1−cos⁡θ).\frac{E'}{E} =\frac1{1+(E/M)(1-\cos\theta)}.

Let a=(E/M)(1−cos⁡θ)a=(E/M)(1-\cos\theta). Then

E−E′E=a1+a≤ε⟺a≤ε1−ε.\begin{gathered} \frac{E-E'}{E}=\frac{a}{1+a}\le\varepsilon\\ \Longleftrightarrow\quad a\le\frac{\varepsilon}{1-\varepsilon}. \end{gathered}

Forward scattering has no recoil energy loss in this massless elastic kinematics; the largest loss at fixed E/ME/M occurs at θ=π\theta=\pi. For E/M=0.1E/M=0.1, losses at π/2\pi/2 and π\pi are 1/111/11 and 1/61/6.

The bound is kinematic and independent of the interaction’s Born expansion. Small recoil and weak coupling must be checked separately before comparing with fixed-source Mott scattering. For a massive projectile use ∣p∣=E2−m2|\mathbf p|=\sqrt{E^2-m^2}; the massless relation cannot be reused unchanged.

Problem 4. Let ϕ\phi be a free real scalar field. Take real smooth compactly supported spacetime test functions gA,gBg_A,g_B whose supports are everywhere spacelike separated. Write ϕ(g)=∫d4x g(x)ϕ(x)\phi(g)=\int d^4x\,g(x)\phi(x). Use the free-field Weyl algebra with causal commutator support.

For λ∈R\lambda\in\mathbb R, set UA=eiλϕ(gA)U_A=e^{i\lambda\phi(g_A)} and WB=eiϕ(gB)W_B=e^{i\phi(g_B)}. Show that the expectation of WBW_B is unchanged by the local unitary UAU_A in every state. Does this require ⟨ϕ(gA)ϕ(gB)⟩=0\langle\phi(g_A)\phi(g_B)\rangle=0 or a factorized vacuum?

Solution

The smeared commutator vanishes because its distribution has no support on these pairs of points. The free-field Weyl relations therefore give UAWB=WBUAU_AW_B=W_BU_A. Both exponentials are bounded unitaries, and

UA†WBUA=WB.U_A^\dagger W_B U_A=W_B.

Taking an expectation in any initial state proves the claim. The real and imaginary Hermitian parts of WBW_B are corresponding bounded observables. The argument uses the Weyl relation, so it does not rely on manipulating unbounded operators without their domains.

The vacuum cross-correlation may be nonzero. Correlation is compatible with invariance under this spacelike local control; no product-state assumption was made. See Propagators to Correlators.

Exact compact spacelike separation matters. Gaussian smearings have tails, and a finite mode cutoff does not preserve the exact local field algebra. Neither may be substituted silently into this proof. The result is a free-field checkpoint, not a derivation of the general interacting theory’s locality axioms.

  • Greiner, Walter. Relativistic Quantum Mechanics: Wave Equations. Third edition. Springer, 2000. doi:10.1007/978-3-662-04275-5. External-field spectra.
  • Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. doi:10.1017/9781139540940. Scattering kinematics and field commutators.
  • Tong, David. Lectures on Quantum Field Theory. University of Cambridge, 2006–2007, section 2. Free scalar fields and causality.
  • Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995, chapters 2–3. doi:10.1017/CBO9781139644167. Relativistic state normalization and scattering.