به انگلیسی

One bracket for motion and conservation

Science · Forces & Motion · ages 22-23

Name ______________________ Date ____________

How does a position coordinate change with time?
- As {H, H}
- As the Poisson bracket of q with q
- As {q, H}
A has no explicit time dependence, and {A, H} = 0 along the motion. What does that say about A?
- A is conserved
- A is zero
- A is the Hamiltonian
For the canonical pair (q, p), quantization turns the classical bracket {q, p} = 1 into the commutator relation [q̂, p̂] = iħ.
Circle one: True False
For H = p²/(2m) + ½kx², compute {x, H}.
- p/m
- −kx
- p/m − kx
A has no explicit time dependence, and {A, H} ≠ 0 at one instant. What follows?
- It stays exactly constant for all time
- Its rate is nonzero at that instant
- It vanishes at once
For the same H = p²/(2m) + ½kx², compute {H, H}.
- kx · p/m
- 0
- 2kx · p/m
Two independent Cartesian positions x and y have vanishing classical brackets, {x, y} = 0 and {x, x} = {y, y} = 0. What follows after standard canonical quantization for this pair?
- Their operators commute and are compatible
- They can never be measured at all, ever
- Their commutator must diverge without bound
A = A(q, p) has no explicit time dependence, and {A, H} = 0 all along the motion. A student says this means A is zero. What is wrong?
- Vanishing brackets are impossible in physics
- Zero bracket means constant, not zero
- Only H may appear inside a bracket

Answer key

For grown-ups. Fold this page away before handing over the rest.

As {q, H} · The canonical equations are brackets with H.
A is conserved · With no explicit time dependence, zero bracket means zero rate, so A is conserved; its value may be zero or nonzero.
True · For the canonical pair, {q, p} = 1 becomes [q̂, p̂] = iħ, with hats marking operators and ħ the reduced Planck constant; the map is exact for such stated canonical examples, not for arbitrary classical functions.
p/m · The signed sum of partial-derivative products gives {x, H} = p/m, which is ẋ.
Its rate is nonzero at that instant · A nonzero bracket gives a nonzero rate at that instant; whether A keeps drifting depends on the rate along the whole motion.
0 · The two products are identical with opposite signs, so {H, H} = 0; with no explicit t, the energy is conserved.
Their operators commute and are compatible · For this stated pair, {x, y} = 0 becomes [x̂, ŷ] = 0, so the operators commute and can be known together; the map is not exact for arbitrary classical functions.
Zero bracket means constant, not zero · With no explicit time dependence, a zero bracket along the whole motion gives zero rate, so A is constant; its value may be zero or nonzero. A = t would break the no-explicit-t condition.