# Taylor Series/Groups

1. Dec 1, 2011

### limddavid

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

A translation operator T(a) coverts ψ(x) to ψ(x+a),

T(a)ψ(x) = ψ(x+a)

In terms of the (quantum mechanical) linear momentum operator p_x = -id/dx, show that T(a) = exp(iap_x), that is, p_x is the generator of translations. Hint. Expand ψ(x+a) as a Taylor series.

2. Relevant equations

Groups?

3. The attempt at a solution

I'm lost at the hint. To expand ψ(x+a) as a Taylor series, don't I need a point around which to expand it?

2. Dec 1, 2011

Yes- a.

3. Dec 1, 2011

### limddavid

Ok. so would the taylor series be:

ψ(a+a)+ψ'(a+a)(x-a)+ψ''(a+a)*(x-a)^2/2!+... ? and maybe disregard the higher order terms O(3)? Or would it be

ψ(a)+ψ'(a)*(x)+ψ''(a)*(x)^2/2!+... ? Either way, I'm not sure how I would prove that T(a) is the given exponential function..

4. Dec 1, 2011

### vela

Staff Emeritus
You want to expand about x=x, i.e., ψ(x+a) = ψ(x)+....

5. Dec 1, 2011

### limddavid

ok.. so i expanded that, and got ψ(x)+aψ'(x)+a^2*ψ''(x)/2!+ ...

But the LHS gives me:

e^(i*a*px)[ψ(x)]=e^(a*dψ/dx)=e^(-iak*ψ(x)), which is clearly not the left hand side. Am I interpreting the operator px wrong? This class is not a quantum dynamics class, so I'm having difficulty figuring out what i'm missing.

6. Dec 2, 2011

### vela

Staff Emeritus
You can't pull ψ(x) into the exponent like that because ψ(x) isn't an eigenfunction of the operator $\hat{p}_x$.

It's not be exactly obvious what the expression $e^{ia\hat{p}_x}$ means. How do you exponentiate an operator? The answer is that it's defined by the Taylor series for ex:
$$e^{ia\hat{p}_x} \equiv 1+ia\hat{p}_x+\frac{(ia\hat{p}_x)^2}{2!}+\cdots$$What do you get when you apply the righthand side to ψ(x)?