- #1
Sigma057
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Homework Statement
I am trying to fill in the steps between equations in the derivation of the coordinate representation of the Darwin term of the Dirac Hamiltonian in the Hydrogen Fine Structure section in Shankar's Principles of Quantum Mechanics.
$$
H_D=\frac{1}{8 m^2 c^2}\left(-2\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]+\left[\overset{\rightharpoonup }{P}\cdot \overset{\rightharpoonup }{P},V\right]\right)
=-\frac{1}{8 m^2 c^2}\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]
$$
Homework Equations
What Shankar calls the "chain rule for commutators of product" I think he means
$$[\text{AB},C]=A[B,C]+[A,C]B$$.
On the same page he mentions the identity
$$
\left[p_x,f (x)\right]=\text{-i$\hbar $}\frac{df}{dx}
$$
The Attempt at a Solution
One way this equality could be satisfied is if
$$\left[\overset{\rightharpoonup }{P}\cdot \overset{\rightharpoonup }{P},V\right]=\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]$$
In component form this means
$$
\left[P_x^2+P_y^2+P_z^2,V\right]=\left(\overset{\wedge }{x}P_x+\overset{\wedge }{y}P_y+\overset{\wedge }{z}P_z\right)\cdot \left[\overset{\wedge }{x}P_x+\overset{\wedge }{y}P_y+\overset{\wedge }{z}P_z,V\right]
$$
$$
=\left(\overset{\wedge }{x}P_x+\overset{\wedge }{y}P_y+\overset{\wedge }{z}P_z\right)\cdot \left(\overset{\wedge }{x}\left[P_x,V\right]+\overset{\wedge }{y}\left[P_y,V\right]+\overset{\wedge }{z}\left[P_z,V\right]\right)
$$
Or
$$
\left[P_x^2,V\right]+\left[P_y^2,V\right]+\left[P_z^2,V\right]=P_x\left[P_x,V\right]+P_y\left[P_y,V\right]+P_z\left[P_z,V\right]
$$
One way this equality could be satisfied is if
$$
\left[P_i^2,V\right]=P_i\left[P_i,V\right]
$$
WLOG let's compute ##\left[P_x^2,V\right]## in the coordinate basis acting on a test function ##\phi(x)##
$$
\left[p_x^2,V\right]\phi =\left(p_x\left[p_x,V\right]+\left[p_x,V\right]p_x\right)\phi =p_x\left[p_x,V\right]\phi +\left[p_x,V\right]p_x\phi
$$
$$
=\text{-i$\hbar $}\frac{d}{dx}(\text{-i$\hbar $}\frac{dV}{dx})\phi+\text{-i$\hbar $}\frac{dV}{dx}(\text{-i$\hbar $}\frac{d}{dx})\phi
=
\text{-$\hbar ^2$}\frac{d}{dx}(\frac{dV}{dx}\phi)\text{-$\hbar ^2$}\frac{dV}{dx}\frac{d\phi}{dx}
$$
$$
=
\text{-$\hbar ^2$}(\frac{d^2V}{dx^2}\phi+\frac{dV}{dx}\frac{d\phi}{dx})
\text{-$\hbar^2$}\frac{dV}{dx}\frac{d\phi}{dx}
=\text{-$\hbar^2$}\frac{d^2V}{dx^2}\phi-2\text{$\hbar^2$}\frac{dV}{dx}\frac{d\phi}{dx}
$$
In comparison
$$
p_x\left[p_x,V\right]\phi = (\text{-i$\hbar$}\frac{d}{dx})(\text{-i$\hbar$}\frac{dV}{dx})\phi
=
\text{-$\hbar^2$}\frac{d}{dx}(\frac{dV}{dx}\phi)
=\text{-$\hbar^2$}(\frac{d^2V}{dx^2}\phi+\frac{dV}{dx}\frac{d\phi}{dx})
=\text{-$\hbar^2$}\frac{d^2V}{dx^2}\phi\text{-$\hbar^2$}\frac{dV}{dx}\frac{d\phi}{dx}
$$
I can't figure out where I've gone wrong.
