Questions: Klein-Gordon Field (Canonical Quantization)

4 questions to test your understanding

Score: 0 / 4
Question 1 Multiple Choice

In canonical quantization of the Klein-Gordon field, one imposes [phi(x,t), pi(y,t)] = i delta^3(x-y). A student asks: 'Why is this a delta function rather than a Kronecker delta?' What is the correct explanation?

AThe delta function is an approximation that becomes a Kronecker delta on a lattice
BBecause x and y are continuous labels for the infinite degrees of freedom of the field — the commutation relation is the continuum generalization of [q_i, p_j] = i delta_{ij}
CThe delta function ensures that the commutator is Lorentz invariant
DThe Kronecker delta only applies to fermionic fields
Question 2 Multiple Choice

After quantization, the Klein-Gordon field phi(x) can be expanded as phi(x) = integral [a_p e^{ipx} + a_p-dagger e^{-ipx}] d^3p / (2pi)^3 (2E_p). The operator a_p-dagger creates a particle with momentum p. What happens if you try to define a position-space 'particle creation operator' by Fourier transforming a_p-dagger?

AYou obtain a well-defined operator that creates a particle localized at a point
BThe resulting operator is phi-dagger(x) itself, which creates a particle at position x — but the particle cannot be perfectly localized due to the energy-momentum relation, and the state is not a position eigenstate in the non-relativistic sense
CThe Fourier transform diverges and no such operator exists
DThe operator creates an antiparticle rather than a particle
Question 3 True / False

The vacuum energy of the quantized Klein-Gordon field is the sum of (1/2) hbar omega_p over all momentum modes, which diverges. This infinity is physically meaningful and must be included in all calculations.

TTrue
FFalse
Question 4 Short Answer

Explain why canonical quantization of the Klein-Gordon field produces a theory of particles, even though the starting point is a continuous classical field.

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