# How do you expand $\langle x'-\Delta x'\rvert \alpha\rangle$?

In my textbook (Sakurai) the following identity is often used:

$$\left< x'-\Delta x' \, \middle| \, \alpha\right>~=~\left< x' \, \middle| \, \alpha \right> - \Delta x'\frac{\partial}{\partial x'} \left< x' \, \middle| \, \alpha \right> \,.$$

How can this identity be derived?

• Let $\alpha(x) = \langle x | \alpha \rangle$ be the wavefunction of $|\alpha \rangle$. This just says $\alpha(x - \Delta x) \approx \alpha(x) - \Delta x \, \alpha'(x)$, which is the first part of a Taylor expansion. – knzhou May 15 '18 at 10:17

you have to derive the taylors expansion. I wont give the full proof but I will give how you proceed. $$f(x)-f(x_0)=\int^{x}_{x_{0}}f'dx$$ and $$f'(x)-f'(x_0)=\int^{x}_{x_{0}} f''dx$$ then if you substitue latter to former you would have
$$f(x)-f(x_0)=\int^{x}_{x_{0}}f'(x_0)dx+\int^{x}_{x_{0}}\int f''(x)dx$$ and $$f(x)=f(x_0)+f'(x_0)(x-x_0)+\int^{x}_{x_{0}}\int f''(x)dx$$
by induction you can find more terms but in your question the first two is enough. then you have to treat your inner product as $f(x)$ that will give you the answer.
• I'd like to add that this holds for small $\Delta x'$ if you expand about $x_0 = x'$. – lmr May 15 '18 at 9:02