# 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.