# Homework Help: Commutator in the Darwin Term

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1. Jun 25, 2017

### Sigma057

1. The problem statement, all variables and given/known data
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.

$$H_D=\frac{1}{8 m^2 c^2}\left(-2\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]+\left[\overset{\rightharpoonup }{P}\cdot \overset{\rightharpoonup }{P},V\right]\right) =-\frac{1}{8 m^2 c^2}\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]$$

2. Relevant 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
$$\left[p_x,f (x)\right]=\text{-i\hbar }\frac{df}{dx}$$

3. 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.

2. Jun 25, 2017

### TSny

The notation is a bit confusing in the text. On the right side of the above equation, the text actually writes
$$-\frac{1}{8 m^2 c^2}\left[\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right] \right]$$
I think this is to be interpreted as a double commutator.
$$\left[\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right] \right] = \overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right] - \left[\overset{\rightharpoonup }{P},V\right] \cdot \overset{\rightharpoonup }{P}$$

They should probably have included another comma in the double commutator
$$\left[\overset{\rightharpoonup }{P}\cdot , \left[\overset{\rightharpoonup }{P},V\right] \right]$$

3. Jun 28, 2017

### Sigma057

Thank you so much for your reply! Without it I would have probably skipped to the next equality and missed a valuable learning opportunity. I'll post my solution to the problem to assist future readers.

With your clarification I'll rewrite the problem statement as

$$H_D=\frac{1}{8 m^2 c^2}\left(-2\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]+\left[\overset{\rightharpoonup }{P}\cdot \overset{\rightharpoonup }{P},V\right]\right) =-\frac{1}{8 m^2 c^2}\left[\overset{\rightharpoonup }{P},\left[\overset{\rightharpoonup }{P},V\right]\right]$$

Using the convention
$$\overset{\rightharpoonup }{A} \overset{\rightharpoonup }{B}\equiv \overset{\rightharpoonup }{A}\cdot \overset{\rightharpoonup }{B}$$

I will also make use of the chain rule for commutators of vector operator products, which follows easily from the chain rule for scalar operator products as a consequence of the definition of the dot product and the linearity of the commutator.

$$\left[\overset{\rightharpoonup }{A}\cdot \overset{\rightharpoonup }{B},C\right]=\overset{\rightharpoonup }{A}\cdot \left[\overset{\rightharpoonup }{B},C\right]+\left[\overset{\rightharpoonup }{A},C\right]\cdot \overset{\rightharpoonup }{B}$$

I can now finally fill in the steps between these equations.
$$H_D=\frac{1}{8 m^2 c^2}\left(-2\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]+\left(\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]+\left[\overset{\rightharpoonup }{P},V\right]\cdot \overset{\rightharpoonup }{P}\right)\right) =\frac{1}{8 m^2 c^2}\left(-\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]+\left[\overset{\rightharpoonup }{P},V\right]\cdot \overset{\rightharpoonup }{P}\right)=-\frac{1}{8 m^2 c^2}\left(\overset{\rightharpoonup }{P}\cdot \left[\overset{\rightharpoonup }{P},V\right]-\left[\overset{\rightharpoonup }{P},V\right]\cdot \overset{\rightharpoonup }{P}\right)=-\frac{1}{8 m^2 c^2}\left[\overset{\rightharpoonup }{P},\left[\overset{\rightharpoonup }{P},V\right]\right]$$