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Question:

Prove that $p^2$ and ${\bf r}\cdot {\bf p}$ commute with every component of ${\bf L}$ using the identity $$[{\bf p},{\bf e}\cdot {\bf L}]=i\hbar\, {\bf e}\times {\bf r} $$ where ${\bf e}$ is a unit vector given by ${\bf e}=a\hat{i}+b\hat{j}+c\hat{k}$. where $\sqrt{a^2+b^2+c^2}=1$

As well, prove that ${\bf L}$ commutes with any function $f(r^2)$

Attempt: $$ [p^2, {\bf e}\cdot {\bf L}]=[{\bf p}\cdot {\bf p},aL_x+bL_y+cL_z]=$$ $$[{\bf p} \cdot {\bf p},aL_x]+[{\bf p} \cdot {\bf p},aL_y] +[{\bf p} \cdot {\bf p},aL_z]$$ $$a[{\bf p} \cdot {\bf p},L_x]+b[{\bf p} \cdot {\bf p},L_y]+c[{\bf p} \cdot {\bf p},L_x] $$

I know that

$$ [AB,C]=A[B,C]+[B,C]A$$

but does that mean that

$$[{\bf A}\cdot{\bf B},{\bf C}]={\bf A}\cdot [{\bf B},{\bf C}]+[{\bf B}\cdot {\bf C}]\cdot {\bf A}$$

Is this the correct procedure? If it is, then I only need to show via the vector identity

$${\bf A}\cdot {\bf B}\times {\bf C}={\bf B}\cdot {\bf C}\times {\bf A}={\bf C}\cdot {\bf A}\times {\bf B}$$

that the commutator above is zero, but I am unsure if the above identity holds for dot products. Furthermore, how do I show that ${\bf L}$ commutes with any function $f(r^2)$? I'm a little confused how to get started on this part of the question. Any help would be greatly appreciated.

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but does that mean that $$[{\bf A}\cdot{\bf B},{\bf C}]={\bf A}\cdot [{\bf B},{\bf C}]+[{\bf B}\cdot {\bf C}]\cdot {\bf A}$$ Is this the correct procedure?

I think not. You are looking for an equation of the form $$[{\bf A}\cdot{\bf B}, C]={\bf A}\cdot [{\bf B}, C]+[{\bf B}, C]\cdot {\bf A}$$ because in the right argument of the commutator there is only a $C = L_x, L_y$ or $L_z$ and not a vector ${\bf C}$ of operators. And yea, this equation holds. Just write the left hand side out in components, use the known equation for the commutator for each components and recollect the correct terms to get the right hand side of the equation. And I think it is better not to write out $\bf e \cdot \bf L$ in components because then you can use the identity provided in the exercise.

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You can do it by brute force too. Take the first addend, I will show what I mean with brute force, and you can do the other two.

$$[\vec{p}\cdot\vec{p},L_x]=[p_x^2+p_y^2+p_z^2,L_x]$$ Now since $\vec{L}=\vec{r}\times\vec{p}$ you have that $L_i=\epsilon^{ijk}r_jp_k$ where $r_1=x$, $r_2=y$, $r_3=z$.

So you have $L_x=\epsilon^{1ij}r_jp_k=\epsilon^{123}r_2p_3+\epsilon^{132}r_3p_2=yp_z-zp_y$.

Therefore you have $$[\vec{p}\cdot\vec{p},L_x]=[p_x^2+p_y^2+p_z^2,yp_z-zp_y]=[p_y^2, y]p_z-[p_z^2,z]p_y$$.

Now you can work out the commutators of $[p_i^2, r_i]=p_i[p_i, r_i]+[p_i, r_i]p_i=-2i\hbar p_i$.

So you have $$[\vec{p}\cdot\vec{p},L_x]=-2i\hbar p_y-2i\hbar p_z$$

Work on the same way on the other two addeds and you will see that they sum to zero.

As for the second question, start with showing which is the commutator of $\vec{L}$ with $r^2$. Then expand $f(r^2)$ in Taylor series, work on each addend and then you are done.

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  • $\begingroup$ I know how to do the problem in general--but I think that I have to use the identity provided to solve the problem. $\endgroup$ – Bronzeclocksofbenin Oct 31 '13 at 23:38

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