Quantum and classical uncertainties

Simulation

Challenges

Stern-Gerlach apparatus (SGA)

Choose input state:

Spin-up
(spin 1/2) Spin-up
(spin 1)

Outcome distribution

–ħ

–ħ/2

0

ħ/2

ħ

Sθ

Quantum uncertainty: ΔSθ = 0

Spot size due to instrumental imperfections: 0

New draft template. The text on start-up appears here.

Quantum uncertainty

Outcome probabilities:

Prob(+ħ/2) = 1

Prob(–ħ/2) = 0

Prob(+ħ) = 1

Prob(0) = 0

Prob(-ħ) = 0

Certain of outcome +ħ/2 prior to measurement

ΔSθ = 0

Change quantum uncertainty: SGA angle θ = 0°

▼

Instrumental imperfections

Perfect apparatus and detector

Spread (standard deviation σ) of each spot after

many measurements: 0

Change instrumental imperfections: σ = 0

▲

Statistical uncertainty

No statistical uncertainty: infinite counts

1.000 of total counts in +ħ/2 spot (expected: 1.000)

0.000 of total counts in –ħ/2 spot (expected: 0.000)

1.000 of total counts in +ħ spot (expected: 1.000)

0.000 of total counts in 0 spot (expected: 0.000)

0.000 of total counts in –ħ spot (expected: 0.000)

Change statistical uncertainty: Number of counts N = ∞

Your score:

0/100

For the potential energy V(x)=kx^2/2, is the normalized wave function shown an energy eigenfunction or an energy superposition state? What can you therefore say about ... compared with the ground state energy E_0?

label radio 0

label radio 1

label radio 2

label radio 3

label radio 4

Correct, congratulations! A superposition state will have contributions of the ground state as well as excited states. Thus, the average energy ... will be greater than E0. Mathematically, for an energy superposition state ... = c0 E0 + c1 E1 + c2E2 + ..., with at least two non-zero c_i and where the modulus squared of these coefficients adds to 1 due to normalization. This sum is always greater than E0.

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Histogram A

–ħ

–ħ/2

0

ħ/2

ħ

Sθ

Histogram B

–ħ

–ħ/2

0

ħ/2

ħ

Sθ