Commutator of r.p with H=p²/2m+V(r)

[Trinitiet]

Homework Statement

Let H be the hamiltonian H = p²/2m+V(r)
Let r.p be the scalar product between the vector r and p.

Calculate the Commutator [r.p , H]

(Commutator of [A,B]=AB-BA )

Homework Equations

The equations citated we should be using are:

[x_i, p_i]=i $$\hbar$$

And commutatoralgebra equations:
[A,B]=-[B,A]
[A,B+C]=[A,B]+[A,C]
[A,BC]=[A,B]C+B[A,C]
[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0

The momentumoperator in configuration space:
p=-i$$\hbar \nabla$$

The Attempt at a Solution

The given solution in my course is:

[r.p, H] = [ (xp_x, yp_y, zp_z), 1/2m (p_x²+p_y²+p_z²)+V(x,y,z) ]
= ihbar / m (p_x²+p_y²+p_z²) - i hbar (x dV/dx + y dV/dy + z dV/dz)
= 2 i hbar T - i hbar (r. nabla V)

In the second rule, I used dV/dx where I should have used partial differentials.
In the last rule, T stands for the kinetic energy of the particle

I understand how we get from the second to the third rule, but the first to the second rule is a complete mystery.

Anyone in for some help? Thanks

Trinitiet

 

[kuruman]

How you get from the "first to the second rule" is not obvious. You need to provide some intermediate steps. I would start from
[r.p, H] = [ (xp_x, yp_y, zp_z), 1/2m (p_x²+p_y²+p_z²)+V(x,y,z) ]
=(1/2)m[ (xp_x, yp_y, zp_z),(p_x²+p_y²+p_z²)]+[ (xp_x, yp_y, zp_z),V(x,y,z)]
and calculate each of the two commutators separately. Of course, they need to be split into more sums of commutators. Don't forget to simplify things using the relevant equations that you posted. They are very handy.

 

[Trinitiet]

How you get from the "first to the second rule" is not obvious. You need to provide some intermediate steps. I would start from
[r.p, H] = [ (xp_x, yp_y, zp_z), 1/2m (p_x²+p_y²+p_z²)+V(x,y,z) ]
=(1/2)m[ (xp_x, yp_y, zp_z),(p_x²+p_y²+p_z²)]+[ (xp_x, yp_y, zp_z),V(x,y,z)]
and calculate each of the two commutators separately. Of course, they need to be split into more sums of commutators. Don't forget to simplify things using the relevant equations that you posted. They are very handy.

Oh :P I thought there was "some" easy trick I missed that allowed us to do it all in once. Thanks, I'll work it out

 

[vela]

I noticed one mistake: The dot product isn't a vector.

It might also be less tedious to do the problem using Einstein's summation convention: repeated indices imply summation.

$$[\vec{r}\cdot\vec{p},H] = [r_i p_i,\frac{p_j^2}{2m}+V(\vec{r})]$$

 

[paranormal]

,

The first step in solving this problem is to rewrite the commutator in terms of the individual components of the vectors r and p. Using the definition of the commutator, we can write:

[r.p, H] = [xp_x, H] + [yp_y, H] + [zp_z, H]

Next, we need to use the commutator algebra rules to simplify this expression. Using the rule [A,B]=AB-BA, we can write:

[r.p, H] = xp_xH - Hxp_x + yp_yH - Hyp_y + zp_zH - Hzp_z

Now, we can use the fact that the momentum operator in configuration space is given by p=-i\hbar \nabla to substitute in for the individual components of p:

[r.p, H] = x(-i\hbar\nabla_x)H - H(-i\hbar\nabla_x)x + y(-i\hbar\nabla_y)H - H(-i\hbar\nabla_y)y + z(-i\hbar\nabla_z)H - H(-i\hbar\nabla_z)z

Next, we can use the commutator algebra rule [A,BC]=[A,B]C+B[A,C] to simplify this expression:

[r.p, H] = -i\hbar[x\nabla_xH - H\nabla_xx + y\nabla_yH - H\nabla_yy + z\nabla_zH - H\nabla_zz]

Finally, we can use the fact that the Hamiltonian H = p²/2m+V(r) to substitute in for the individual components of p squared:

[r.p, H] = -i\hbar[x\nabla_x(p_x²/2m+V(x,y,z)) - (p_x²/2m+V(x,y,z))\nabla_xx + y\nabla_y(p_y²/2m+V(x,y,z)) - (p_y²/2m+V(x,y,z))\nabla_yy + z\nabla_z(p_z²/2m+V(x,y,z)) - (p_z²/2m+V(x,y,z))\nabla_zz]

Simplifying this expression further will result in the given solution