Calculate the commutator [ p, e^ik.x ] where the function e^ik.x may be expanded as:
e^ik.x = ∑infinite,n=0 (ik.x^n/n!)
To calculate the commutator [p, e^ik.x], we need to apply the commutator formula and compute it step by step. The commutator is defined as:
[p, e^ik.x] = p(e^ik.x) - (e^ik.x)p
First, let's start with the expression p(e^ik.x). To do this, we need to apply the momentum operator p on the exponential function e^ik.x.
To apply the momentum operator p on e^ik.x, we need to use the Taylor expansion of e^ik.x and then act with the momentum operator on each term individually. The Taylor expansion of e^ik.x is given as:
e^ik.x = ∑(infinite, n=0) [(ik.x)^n/n!]
So, let's calculate p(e^ik.x) using the Taylor expansion:
p(e^ik.x) = p[∑(infinite, n=0) [(ik.x)^n/n!]]
To simplify this expression, we need to distribute the momentum operator p inside the summation and act on each term individually. The momentum operator p acts on the variable x, while keeping the terms with k and i constant. Thus, we get:
p(e^ik.x) = ∑(infinite, n=0) [(ik)^n/n!] p(x^n)
Now, let's consider the term p(x^n). The momentum operator p acts on x^n by differentiating it n times with respect to x. Therefore, we get:
p(x^n) = (d^n/dx^n) x^n
Since the derivative of x^n with respect to x gives n*(x^(n-1)), we can simplify this as:
p(x^n) = n(x^(n-1))
Now, substituting this back into our previous expression for p(e^ik.x), we have:
p(e^ik.x) = ∑(infinite, n=0) [(ik)^n/n!] n(x^(n-1))
Simplifying further, we can bring n outside the summation:
p(e^ik.x) = ∑(infinite, n=0) [(ik)^n/n!] * n * x^(n-1)
Next, let's compute the term (e^ik.x)p. We can again use the Taylor expansion to rewrite this expression:
(e^ik.x)p = ∑(infinite, n=0) [(ik.x)^n/n!] * p
Here, we keep the terms involving ik.x constant and act with the momentum operator p. Therefore:
(e^ik.x)p = ∑(infinite, n=0) [(ik)^n/n!] * p(x^n)
Applying the same steps as before, where p(x^n) = n(x^(n-1)), we get:
(e^ik.x)p = ∑(infinite, n=0) [(ik)^n/n!] * n * x^(n-1) * p
Now, we can substitute both expressions back into the original commutator formula:
[p, e^ik.x] = p(e^ik.x) - (e^ik.x)p
= ∑(infinite, n=0) [(ik)^n/n!] n * x^(n-1) - ∑(infinite, n=0) [(ik)^n/n!] n * x^(n-1) * p
Since both terms have the same series of [(ik)^n/n!] * n * x^(n-1), they cancel each other out. The remaining term in the commutator is:
[p, e^ik.x] = - ∑(infinite, n=0) [(ik)^n/n!] n * x^(n-1) * p
This is the final expression for the commutator [p, e^ik.x] using the given Taylor expansion of e^ik.x.