What is the Momentum Operator Identity in Modern Quantum Mechanics?

[malawi_glenn]

Homework Statement

I want to show:

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

Homework Equations

\vert \alpha \rangle is a state.

The Attempt at a Solution

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

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

 

[Dick]

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

 

[malawi_glenn]

aha okay I'll try that one, thanks a lot!

 

[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..

 

[Dick]

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

So I'm wondering..

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.

 

[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

<br /> <br /> f(x-a)=f(x)-a*\frac{df(x-a)}{da} |_{a=0}<br /> <br />

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

I want to verify that eq,

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

 

[Dick]

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

<br /> <br /> f(x-a)=f(x)-a*\frac{df(x-a)}{da} |_{a=0}<br /> <br />

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

I want to verify that eq,

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

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),

<br /> f(x-a)=f(x)-a \frac{df(z)}{dz} |_{z=x}<br />

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.