How to Derive the Momentum Operator Commutation in Position-Dependent Potentials

Understanding the Momentum-Potential Commutator in Quantum Mechanics

What We're Solving:

We need to derive the commutation relation [p̂, V(x̂)] = (ℏ/i) dV(x̂)/dx, which tells us how the momentum operator and a position-dependent potential energy function behave when applied in different orders to a quantum wavefunction.

The Approach:

This is a fundamental calculation in quantum mechanics! We're going to use the definition of what these operators actually do when they act on a wavefunction. The key insight is that we'll apply the commutator [p̂, V(x̂)] = p̂V(x̂) - V(x̂)p̂ to an arbitrary wavefunction ψ(x), then use the product rule of differentiation. This approach shows us why the commutator has this specific form.

Step-by-Step Solution:

Step 1: Set up the commutator acting on a test function Let's apply [p̂, V(x̂)] to an arbitrary wavefunction ψ(x): [p̂, V(x̂)]ψ(x) = [p̂V(x̂) - V(x̂)p̂]ψ(x) = p̂[V(x̂)ψ(x)] - V(x̂)[p̂ψ(x)]

Step 2: Substitute the momentum operator Remember that p̂ = (ℏ/i)(d/dx), so: [p̂, V(x̂)]ψ(x) = (ℏ/i)(d/dx)[V(x)ψ(x)] - V(x)[(ℏ/i)(d/dx)ψ(x)]

Step 3: Apply the product rule Here's the crucial step! When we differentiate V(x)ψ(x), we must use the product rule: (d/dx)[V(x)ψ(x)] = (dV/dx)ψ(x) + V(x)(dψ/dx)

Step 4: Substitute back [p̂, V(x̂)]ψ(x) = (ℏ/i)[(dV/dx)ψ(x) + V(x)(dψ/dx)] - V(x)[(ℏ/i)(dψ/dx)]

Step 5: Expand and simplify [p̂, V(x̂)]ψ(x) = (ℏ/i)(dV/dx)ψ(x) + (ℏ/i)V(x)(dψ/dx) - (ℏ/i)V(x)(dψ/dx)

Notice that the last two terms cancel out!

Step 6: Final result [p̂, V(x̂)]ψ(x) = (ℏ/i)(dV/dx)ψ(x)

Since this must be true for any wavefunction ψ(x), we can write:

The Answer:

[p̂, V(x̂)] = (ℏ/i) dV(x̂)/dx

This beautiful result shows that the commutator of momentum with a potential depends on how rapidly the potential changes with position!

Memory Tip:

Think of it this way: "Momentum cares about how the potential changes" - that's why we get the derivative dV/dx. The momentum operator "picks out" the spatial variation of the potential when they don't commute. When V(x) is constant, dV/dx = 0, so momentum and constant potentials do commute - which makes physical sense!

This derivation is a perfect example of how the mathematical structure of quantum mechanics emerges naturally from the fundamental definitions of the operators. Great work tackling this important relationship!