Momentum operator identity

1. Aug 22, 2007

malawi_glenn

1. The problem statement, all variables and given/known data
I wanna show:

$$\langle x' - \triangle x' \vert \alpha \rangle = \langle x' \vert \alpha \rangle - \triangle x' \dfrac{\partial}{\partial x'}\langle x' \vert \alpha \rangle$$

2. Relevant equations

$$\vert \alpha \rangle$$ is a state.

3. The attempt at a solution

i have no clue, can anyone give me a hint? Is the bra being splitted in to two, or what?

This is a part of eq (1.7.15) in Sakurai mordern quantum mechanics, 2ed.

Last edited: Aug 22, 2007
2. Aug 22, 2007

Dick

It's just the first term in a power series expansion. Like f(x-a)=f(x)-a*df(x)/dx.

3. Aug 22, 2007

malawi_glenn

aha okay I'll try that one, thanks alot!

4. Jan 4, 2009

crazypoets

But I think the first term of the series expansion must be a constant but not a function of x.

So I'm wondering..

5. Jan 4, 2009

Dick

In f(x-a)=f(x)-a*df(x)/dx I'm thinking of 'x' as the constant (point to expand around) and 'a' as the expansion variable.

6. Jan 5, 2009

crazypoets

But f is differentiated by 'x' in f(x-a)=f(x)-a*df(x)/dx.

I think if 'a' is the variable the series expansion must be

$$f(x-a)=f(x)-a*\frac{df(x-a)}{da} |_{a=0}$$

I'm studying the part of QM and have difficulty in understanding the eq (1.7.15).

I wanna verify that eq,

but I think there may be some mistakes in the explanation by series expansion.

7. Jan 5, 2009

Dick

Now that has the wrong sign in the derivative if you are taking it d/da. It will give you -f'(x). If you want to be super precise how about (to first order in a),

$$f(x-a)=f(x)-a \frac{df(z)}{dz} |_{z=x}$$

the variable in the differentiation symbol is really just a dummy, not x or a. Or just f(x-a)=f(x)-a*f'(x) is also clear.