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17 Cards in this Set
- Front
- Back
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Why is it impossible for the ground state of the quantum harmonic oscillator to be zero?
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Oscillators will have a residual energy at absolute zero: the zero-point energy.
The energy must be higher than the Heisenberg uncertainty value: 1/2 ħω. The annihilation operator terminates at zero, meaning that no negative values of energy can be generated. |
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The annihilation and creation operators
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a+ = (2mħω)^1/2 (mωx + ip)
a- = (2mħω)^1/2 (mωx - ip) a+† = a- and a-† = a+ |
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What is the position operator x in terms of the ladder operators a+ and a-?
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Calculate (a+ + a-)
multiply out and rearrange: x = (a+ + a-) √(ħ/2mω ) |
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What is the momentum operator p in terms of the ladder operators a+ and a-?
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Calculate (a+ - a-)
multiply out and rearrange: p = (a+ - a-) i√(mωħ/2 ) |
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Hamiltonian for the quantum harmonic oscillator
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H = p^2/2m + 1/2 mω^2x
H = ħω(a+a- +1/2) |
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Prove that H = ħω(a+a- +1/2)
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* Multiply the definitions of a+, a- including the commutator term [x,p]
* Rearrange to H and substitute [x, p] = iħ Obtain H = ħω (a+a- +1/2) |
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How do the annihilation and creation operators commute with H?
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[H, a+] = ħωa+
[H, a-] = - ħωa- thus [a+, H] = - ħωa+ [a-, H] = ħωa- |
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How do the raising and lowering operators commute with each other?
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[a-, a+] = 1
[a+, a-] = -1 |
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Show that if |n> is an eigenvector of H having eigenvalue E then a+ |n> is an eigenvector of H having eigenvalue E + ħω
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From the definition of the commutator:
Ha+|n> = a+H|n> - [a+, H]|n> = a+H |n> + ħωa+ |n> = a+ E|n> ħωa+|n> = (E + ħω) a+ |n> therefore a+ |n> is an eigenvector of H having eigenvalue (E + ħω) |
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Show that if |n> is an eigenvector of H having eigenvalue E then a- |n> is an eigenvector of H having eigenvalue E - ħω
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From the definition of the commutator:
Ha-|n> = a-H|n> - [a-, H]|n> = a-H |n> - ħωa- |n> = a-E|n> - ħωa-|n> = (E - ħω) a+ |n> therefore a+ |n> is an eigenvector of H having eigenvalue (E- ħω) |
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The number operator.
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N = (a+a-)
N |n> = n|n> |
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Find the energy eigenvalues of H using the ladder operator method.
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H = ħω (a+a- + 1/2)
N = a+a- therefore H = ħω (N = 1/2) H |n> = ħω (N + 1/2) |n> = ħωn |n> +ħω/2 |n> = ħω (n+1/2) therefore the energy eigenvalues of H are ħω(n +1/2) |
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Show that the commutator of a+ and a-, [a-, a+] = 1
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Using the full definitions of a+ and a-
* calculate [a+, a-] = 1/2ħωm [mωx + ip, mωx - ip] *multiply out such that the terms [x, x], [p, p], [x, p], [p, x] remain * Apply known commutators and cancel out to 1 |
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What are the eigenvalues and eigenvectors of the annihilation and creation operators?
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a+|n> = √(n+1)|n>
a-|n> = √(n)|n-1> |
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Find the eigenvalues and eigenstates of a+
(LONG) |
a+|n> is an eigenstate of N with eigenvalue (n+1)
N|n+1> = (n+1)|n+1> but N(a+)|n> = (n+1) (a+|n>) Therefore a+|n> must be a multiple of |n+1>: a+|n> = c|n+1> Since a+† = a- : <n|a-a+|n> = |c|^2 <n+1|n+1> = |c|^2 However [a-, a+] = 1, a-a+ = a+a- +1 = N+1 <n|N+1|n> = |c|^2 thus c = √(n+1) |
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Find the eigenvalues and eigenstates of a-
(LONG) |
a-|n> is an eigenstate of N with eigenvalue (n-1)
N|n-1> = (n-1)|n-1> but N(a-)|n> = (n-1) (a-|n>) Therefore a-|n> must be a multiple of |n-1>: a-|n> = c|n-1> a+a-|n> = ca+|n-1> = c√(n) |n> Since a+†=a- : <n|a+a-|n> = c√(n) <n|n> = c√(n) <n|N|n> = n = c√(n) therefore c = √(n) |
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What is the action of the raising and lowering operators?
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From a given state |n>, an eigenstate of N with eigenvalue λ, a sequence of eigenstates is generated by repeated application of the raising and lowering operators.
Repeated application of a+ gives a sequence of eigenstates of N with eigenvalues λ+1, λ+2 etc. Repeated application of a- gives a series of eigenstates of N with eigenvalues λ-1, λ-2, etc. The lowering sequence must terminate, since eigenstates of the QHO cannot be negative. Hence a-|0> = |0> and |n> = 1/ √(n !) (a+)^2|0> generates all the eigenstates of N. |
