Sakurai page 54: Is this a Taylor expansion?

[omoplata]

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

Obtaining equation (1.7.15),<br /> \begin{eqnarray}<br /> \left(1- \frac{ip\Delta x&#039;}{\hbar} \right) \mid \alpha \rangle &amp; = &amp; \int dx&#039; \mathcal{T} ( \Delta x&#039; ) \mid x&#039; \rangle \langle x&#039; \mid \alpha \rangle \\<br /> &amp; = &amp; \int dx&#039; \mid x&#039; + \Delta x&#039; \rangle \langle x&#039; \mid \alpha \rangle \\<br /> &amp; = &amp; \int dx&#039; \mid x&#039; \rangle \langle x&#039; - \Delta x&#039; \mid \alpha \rangle \\<br /> &amp; = &amp; \int dx&#039; \mid x&#039; \rangle \left( \langle x&#039; \mid \alpha \rangle - \Delta x&#039; \frac{\partial}{\partial x&#039;} \langle x&#039; \mid \alpha \rangle \right)<br /> \end{eqnarray}<br />How does the last line come about? Is it a Taylor expansion?

 

[mfb]

It looks like a Taylor expansion, indeed.

 

[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.

 

[mfb]

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

 

[fzero]

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.

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.

 

[omoplata]

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.

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,<br /> \int dx&#039; \mid x&#039; \rangle \langle x&#039; - \Delta x&#039; \mid \alpha \rangle = \int dx&#039; \mid x&#039; \rangle \left( \langle x&#039; \mid \alpha \rangle - \Delta x&#039; \frac{\partial}{\partial x&#039;} \langle x&#039; \mid \alpha \rangle \right)<br /> and not, <br /> \int dx&#039; \mid x&#039; \rangle \langle x&#039; - \Delta x&#039; \mid \alpha \rangle = \int dx&#039; \mid x&#039; \rangle \left( \langle x&#039; \mid \alpha \rangle - \Delta (x&#039;)^* {\frac{\partial}{\partial x&#039;}}^{\dagger} \langle x&#039; \mid \alpha \rangle \right)<br />?

 

[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 into play?

 

[Fightfish]

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 into play?

Because position is a real variable?