به انگلیسی

Particles as field quanta

Science · Quantum & Modern Physics · ages 23-24

Name ______________________ Date ____________

For a bosonic mode, N = a†a and a†|n⟩ = √(n+1)|n+1⟩. The state a†|1⟩ equals √2|2⟩. What is the N eigenvalue of this state?
Answer: ______________
A free bosonic field mode has ħω = 0.20 eV. Which statement about its levels is correct?
- E_n = (n + ½)ħω for n = 0, 1, 2, …; n counts quanta
- E_n = n ħω with no ground energy at all, for any n
- Each level holds at most one quantum, like a fermion
A free bosonic field mode is a harmonic oscillator with levels E_n = (n + ½)ħω. What does n count?
- How many times it has swung back and forth
- The spin carried by the oscillator, always
- How many quanta of ħω sit above ground
What is a photon in the language of field modes?
- One quantum of the electromagnetic field
- A particle trapped inside an oscillator
- Half of a ground state
In a free bosonic field the number operator commutes with the Hamiltonian. What changes when an interaction is added?
- Number is then conserved in every single process, always, no matter what
- An interaction can change the count, subject to conservation laws
- The field stops being quantised in any sense at all
For a bosonic field mode with a† the creation operator, a the annihilation operator, and number operator N = a†a, which action is correct?
- a†|n⟩ = |n+1⟩ and a|n⟩ = |n−1⟩, with no factors
- a|0⟩ = |1⟩
- a†|n⟩ = √(n+1)|n+1⟩, a|n⟩ = √n|n−1⟩, a|0⟩ = 0
For a free bosonic field mode with ħω = 0.50 eV, what is the energy of level n = 2 in eV?
Answer: ______________
A free bosonic harmonic mode with ħω = 0.20 eV is in the superposition (|0⟩ + |1⟩)/√2. What is the expectation value of its energy in eV?
Answer: ______________
For two bosonic field modes i and j, why is a two-quantum state a_i†a_j†|0⟩ automatically symmetric under exchange?
- Because each particle gets its own brand new field
- Commuting creators give a symmetric built state
- Because particles are painted with ID numbers
Rae claims particle number can never change. What is wrong?
- Nothing: the number operator always commutes with any Hamiltonian, in every theory
- Free number commutes with H, but interaction terms can change it
- Creation operators do not exist, so number is fixed

Answer key

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Particles as field quanta W1-mt_BJyZWQA5cb-s1

2 · a†|1⟩ = √2|2⟩, and N|2⟩ = 2|2⟩, so the eigenvalue is 2.
E_n = (n + ½)ħω for n = 0, 1, 2, …; n counts quanta · A free bosonic mode is an oscillator with levels (n + ½)ħω; n is the occupation number and the ground energy ½ħω is not zero.
How many quanta of ħω sit above ground · Each step up adds one quantum of ħω above the ground state; the ground energy ½ħω is zero-point energy, not half a particle.
One quantum of the electromagnetic field · A photon is one quantum of the electromagnetic field, in one of its modes; the mode's occupation count is the photon number.
An interaction can change the count, subject to conservation laws · Free number is conserved because N commutes with H; an interaction coupling modes can create or destroy quanta, provided energy, momentum, and charges are conserved.
a†|n⟩ = √(n+1)|n+1⟩, a|n⟩ = √n|n−1⟩, a|0⟩ = 0 · The normalized ladder actions include the square-root factors, and the annihilator kills the vacuum.
1.25 · (2 + ½)(0.50 eV) = 1.25 eV: ground 0.25 eV plus two quanta of 0.50 eV each.
0.2 · The average is ½(0.10 + 0.30) = 0.20 eV, but a measurement returns 0.10 or 0.30 eV, never the average.
Commuting creators give a symmetric built state · Bosonic creation operators commute, [a_i†, a_j†] = 0, so the built state is symmetric under exchange of modes; fermions anticommute, {c_i†, c_j†} = 0, and give antisymmetric states instead.
Free number commutes with H, but interaction terms can change it · In the free theory N commutes with H and number is conserved; an interaction coupling modes can create or destroy quanta, subject to energy, momentum, and charge conservation.