# Sakurai page 54: Is this a Taylor expansion?

1. Apr 27, 2013

### omoplata

From page 54 of 'Modern Quantum Mechanics, revised edition" by J. J. Sakurai.

Obtaining equation (1.7.15),$$\begin{eqnarray} \left(1- \frac{ip\Delta x'}{\hbar} \right) \mid \alpha \rangle & = & \int dx' \mathcal{T} ( \Delta x' ) \mid x' \rangle \langle x' \mid \alpha \rangle \\ & = & \int dx' \mid x' + \Delta x' \rangle \langle x' \mid \alpha \rangle \\ & = & \int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle \\ & = & \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta x' \frac{\partial}{\partial x'} \langle x' \mid \alpha \rangle \right) \end{eqnarray}$$How does the last line come about? Is it a Taylor expansion?

2. Apr 27, 2013

### Staff: Mentor

It looks like a Taylor expansion, indeed.

3. Apr 27, 2013

### omoplata

Can a Bra be Taylor expanded like this? I guess it is like Taylor expanding a vector? I know that Kets are elements of a vector space, and Bras are the elements of the corresponding dual space.

4. Apr 27, 2013

### Staff: Mentor

If the result is differentiable in some variable, and behaves nicely (insert strict mathematical formulation here): why not?

5. Apr 27, 2013

### fzero

It's not quite a bra that is being expanded here. $\langle x | \alpha \rangle = \psi_\alpha(x)$ is the position space wavefunction corresponding to the state $|\alpha\rangle$. What appears in the expression above is $\psi_\alpha(x'-\Delta x')$, which is usually a nice enough function to have a well-defined Taylor expansion. It may be the case that you can also manipulate the states themselves in some way, but I expect that if the procedure is to be well-defined, there must be a way to express the same manipulations in terms of appropriate wavefunctions.

6. Apr 27, 2013

### omoplata

Oh, OK. That's easier to understand.

I was about to ask since all operators $A$ should be $A^{\dagger}$ and all constants $c$ should be $c^*$ in the Bra space, why is it,$$\int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle = \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta x' \frac{\partial}{\partial x'} \langle x' \mid \alpha \rangle \right)$$ and not, $$\int dx' \mid x' \rangle \langle x' - \Delta x' \mid \alpha \rangle = \int dx' \mid x' \rangle \left( \langle x' \mid \alpha \rangle - \Delta (x')^* {\frac{\partial}{\partial x'}}^{\dagger} \langle x' \mid \alpha \rangle \right)$$?

7. Apr 27, 2013

### omoplata

I mean, I understand that what's given in the book easily follows if you think of it as a wavefunction. But if you think of it as Taylor expanding a Bra, why don't the complex and hermitian conjugates don't come in to play?

8. Apr 28, 2013

### Fightfish

Because position is a real variable?