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Chapter 1: Problem 11

Evaluate the commutators (a) \(\left[H, p_{x}\right]\) and (b) \([H, x]\) where \(H=p_{x}^{2} / 2 m+V(x) .\) Choose (i) \(V(x)=V,\) a constant, (ii) \(V(x)=\frac{1}{2} k_{t} x^{2},\) (iii) \(V(x) \rightarrow V(r)=e^{2} / 4 \pi \varepsilon_{0} r .\) Hint. For part (b), case (iii), use \(\left(\partial r^{-1} / \partial x\right)=-x / r^{3}\)

Short Answer

Expert verified
The results of calculations are that for constant potential \(V\), \([H, p_{x}]\) is 0 and \([H, x]\) is not 0. These results change for \(V(x)=1/2 k_{t} x^{2}\) and \(V(x) = e^{2} / (4 \pi \varepsilon_{0} r)\), having not null results for both commutators which also depend on the specifics of potential function.

Step by step solution

01

Calculate first commutator for constant potential

Let's start with the first commutator \( \left[H, p_{x}\right] \) and for \(V(x) = V\) constant. Substituting the given Hamiltonian \(H = p_{x}^{2} / 2m + V\) into the commutator, we get \([H, p_{x}] = [p_{x}^{2} / 2m + V, p_{x}]\). Using the properties of the commutator this can be simplified as:\([H, p_{x}] = [p_{x}^{2} / 2m, p_{x}] + [V, p_{x}]\). Since \(p_{x}\) commutes with itself and any constant, both terms will be zero.
02

Calculate first commutator for the other potentials

For the next potential \(V(x) = 1/2 k_{t} x^{2}\), the same process can be repeated but this time also \(p_{x}\) and \(x\) do not commute. You'd find out that it's not zero, same goes for the third potential \(V(x) = e^{2} / (4 \pi \varepsilon_{0} r)\).
03

Calculate second commutator for constant potential

Next, let's calculate the second commutator \([H, x]\) for the constant potential \(V(x)=V\). Using the defined Hamiltonian \(H = p_{x}^{2} / 2m + V\) and substituting into \([H, x] = [p_{x}^{2} / 2m + V, x]\), by separating it becomes: \([H, x] = [p_{x}^{2} / 2m, x] + [V, x]\). Because \(x\) does not commute with \(p_{x}\), but does with constant \(V\), the first term would not vanish, while the second one will.
04

Calculate second commutator for the other potentials

For the next potential \(V(x) = 1/2 k_{t} x^{2}\), the process can be repeated observing that \(x^2\) commutes with \(x\). And the other part \(V, x\) of the commutator would vanish since \(V(x) = 1/2 k_{t} x^{2}\), hence \([V, x] = 0\). The results for the third case \(V(x) = e^{2} / (4 \pi \varepsilon_{0} r)\) are obtained similarly, but be sure to use the hint \(\left(\partial r^{-1} / \partial x\right)=-x / r^{3}\).

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Most popular questions from this chapter

Evaluate the commutators (a) \([x, y]\) (b) \(\left[p_{x}, p_{y}\right]\) (c) \(\left[x, p_{x}\right]\) (d) \(\left[x^{2}, p_{x}\right]\) (e) \(\left[x^{n}, p_{x}\right]\)

(a) Show that \(\mathrm{e}^{A} \mathrm{e}^{B}=\mathrm{e}^{A+B}\) only if \([A, B]=0 .\) (b) If \([A, B] \neq 0\) but \([A,[A, B]]=[B,[A, B]]=0,\) show that \(\mathrm{e}^{A} \mathrm{e}^{B}=\) \(\mathrm{e}^{A+B} \mathrm{e}^{f},\) where \(f\) is a simple function of \([A, B] .\) Hint. This is another example of the differences between operators \((q-\text { numbers })\) and ordinary numbers (c-numbers). The simplest approach is to expand the exponentials and to collect and compare terms on both sides of the equality. Note that \(\mathrm{e}^{A} \mathrm{e}^{B}\) will give terms like \(2 A B\) while \(\mathrm{e}^{A+B}\) will give \(A B+B A .\) Be careful with order.

Find the operator for position \(x\) if the operator for momentum \(p\) is taken to be \((\hbar / 2 m)^{1 / 2}(A+B),\) with \([A, B]=1\) and all other commutators zero. Hint. Write \(x=a A+b B\) and find one set of solutions for \(a\) and \(b\)

The ground-state wavefunction of a hydrogen atom has the form \(\psi(r)=N \mathrm{e}^{-b r}, b\) being a collection of fundamental constants with the magnitude \(1 / a_{0},\) with \(a_{0}=53 \mathrm{pm}\) Normalize this spherically symmetrical function. Hint. The volume element is \(\mathrm{d} \tau=\sin \theta \mathrm{d} \theta \mathrm{d} \varphi r^{2} \mathrm{d} r,\) with \(0 \leq \theta \leq \pi\) \(0 \leq \varphi \leq 2 \pi,\) and \(0 \leq r<\infty,\) 'Normalize' always means 'normalize to 1 ' in this text.

Evaluate the expectation values of the operators \(p_{x}\) and \(p_{x}^{2}\) for a particle with wavefunction \((2 / L)^{1 / 2} \sin (\pi x / L)\) in the range 0 to \(L\)
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