Question

Wikipedia indicates that the following relation is "easily shown": [xi,F(p⃗ )]=iℏ∂F(p⃗ )∂pi

, however I'm having some trouble showing it. I think I'm just messing up the multivariable Taylor expansion (of F(p⃗ )
). Can one of you walk me through it or link me to site that will? Thanks.

Edit: Here's what I get (without using x=iℏ∂∂p
which I haven't proven yet):

F(p⃗ )=F(0⃗ )+∑j=13∂F(0⃗ )∂pjpj+12∑k=13∑j=13∂2F(0⃗ )∂pk∂pjpjpk+…
so
(xiF(p⃗ )−F(p⃗ )xi)ψ
=xi[F(0⃗ )ψ−iℏ∑j=13∂F(0⃗ )∂pj(∇ψ)j+ℏ212∑k=13∑j=13∂2F(0⃗ )∂pk∂pj(∇ψ)j(∇ψ)k+…]
−[F(0⃗ )xiψ−iℏ∑j=13∂F(0⃗ )∂pj(∇xiψ)j+ℏ212∑k=13∑j=13∂2F(0⃗ )∂pk∂pj(∇xiψ)j(∇xiψ)k+…]

where
(∇xiψ)j=∂xi∂xjψ+xi∂ψ∂xj=δijψ+xi∂ψ∂xj
and
(∇xiψ)j(∇xiψ)k=(δijψ+xi∂ψ∂xj)(δikψ+xi∂ψ∂xk)=δjkψ2+xjψ∂ψ∂xk+xk∂ψ∂xjψ+x2i∂2ψ∂xj∂xk Thus:
(xiF(p⃗ )−F(p⃗ )xi)ψ
=xi[F(0⃗ )ψ−iℏ∑j=13∂F(0⃗ )∂pj∂ψ∂xj+ℏ212∑k=13∑j=13∂2F(0⃗ )∂pk∂pj∂ψ∂xj∂ψ∂xk+…]
−[F(0⃗ )xiψ−iℏ∑j=13∂F(0⃗ )∂pj(δijψ+xi∂ψ∂xj)+ℏ212∑k=13∑j=13∂2F(0⃗ )∂pk∂pj(δjkψ2+xjψ∂ψ∂xk+xk∂ψ∂xjψ+x2i∂2ψ∂xj∂xk)+…]

From here it doesn't look like those higher order terms are all going to cancel out.

Answer 1

The student is grappling with quantum mechanics and the relation between the commutator of position and a function of momentum. Clarifying this requires understanding the momentum operator and the linearity of differentiation in the context of the Taylor expansion of the function involved.

Step-by-step explanation:

The student is trying to demonstrate an important relation in quantum mechanics, the commutator of position x_i with a function of momentum F(⎛ p ⎴). This commutation relation is one of the fundamental principles underlying quantum mechanics and relates to the operation of measuring position and momentum.

The expression [x_i, F(⎛ p ⎴)] = iħ ∂F(⎛ p ⎴)/∂p_i showcases the non-commuting nature of these quantities, which is central to Heisenberg's uncertainty principle. To show this relation, one may need to utilize the definition of the momentum operator in quantum mechanics and the property that differentiation is a linear operator, meaning that the derivative of a sum is equal to the sum of the derivatives. Only the first-order terms in the Taylor expansion of F(⎛ p ⎴) are necessary for the calculation since higher-order terms do not contribute to the commutator at first order.