Sakurai pr. 1.18 - bra-c-ket sandwiches

[bjnartowt]

Homework Statement

c) explicit calculations, using the usual rules of wave mechanics, show that the wave function for a Gaussian wave packet given by

$$\left\langle {x'|\alpha } \right\rangle = {(2\pi {d^2})^{ - 1/4}}\exp \left( {{\bf{i}}{\textstyle{{\left\langle p \right\rangle x'} \over \hbar }} - {\textstyle{{{{(x' - \left\langle x \right\rangle )}^2}} \over {4{d^2}}}}} \right)$$

…satisfies the minimum uncertainty relation,
$$\sqrt {\left\langle {\Delta {x^2}} \right\rangle } \sqrt {\left\langle {\Delta {p^2}} \right\rangle } = {\textstyle{1 \over 2}}\hbar$$

Prove that the requirement,
$$\left\langle {x'|\Delta x|\alpha } \right\rangle = \lambda \cdot \left\langle {x'|\Delta p|\alpha } \right\rangle {\rm{ Re(}}\lambda ) = 0$$

…is indeed satisfied for such a Gaussian wavepacket, in agreement with b, which said,
$${\left\langle {\Delta {A^2}} \right\rangle \left\langle {\Delta {B^2}} \right\rangle = {\textstyle{1 \over 4}}{{\left| {\left\langle {\left[ {\Delta A,\Delta B} \right]} \right\rangle } \right|}^2}}$$

Homework Equations

dirac's bra-c-kets

The Attempt at a Solution

well, I'm fine chunking through this stuff, but i reached this step, where i am comparing my result to someone else's, and they claim,
$$\left\langle {x'|\Delta p|\alpha } \right\rangle = \left\langle {x'|(p - \left\langle p \right\rangle )|\alpha } \right\rangle = \left\langle {x'|p|\alpha } \right\rangle - \left\langle p \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle = ... = - {\bf{i}}\hbar {\textstyle{\partial \over {\partial x'}}}\left\langle {x'|\alpha } \right\rangle - \left\langle p \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle$$

i recognize -i*hbar*(d/dx) as the momentum operator, of course. but i am at odds with the "..." i indicated between equals signs above. i tried inserting an identity operator,
$$\left\langle {x'|p|\alpha } \right\rangle = \left\langle {x'|p{\bf{I}}|\alpha } \right\rangle$$

and chunked out,
$$\left\langle {x'|p{\bf{I}}|\alpha } \right\rangle = \left\langle {x'|p\left( {\int {\left| {x'} \right\rangle \left\langle {x'} \right|} \cdot dx'} \right)|\alpha } \right\rangle = \int {\left\langle {x'|p|x'} \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle \cdot dx'}$$

and i found this implied (i think),
$$\int {\left\langle {x'|p|x'} \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle \cdot dx'} = p\left\langle {x'|\alpha } \right\rangle$$

which looks not right to me.

suggestions? or is my source incorrect?

 

[bjnartowt]

well, I'm fine chunking through this stuff, but i reached this step, where i am comparing my result to someone else's, and they claim,

$\left\langle {x'|\Delta p|\alpha } \right\rangle = \left\langle {x'|(p - \left\langle p \right\rangle )|\alpha } \right\rangle = \left\langle {x'|p|\alpha } \right\rangle - \left\langle p \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle = ... = - {\bf{i}}\hbar {\textstyle{\partial \over {\partial x'}}}\left\langle {x'|\alpha } \right\rangle - \left\langle p \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle$

did that go through?

 

[bjnartowt]

argh! try this

$\left\langle {x'|\Delta p|\alpha } \right\rangle = \left\langle {x'|(p - \left\langle p \right\rangle )|\alpha } \right\rangle = \left\langle {x'|p|\alpha } \right\rangle - \left\langle p \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle = - {\bf{i}}\hbar {\textstyle{\partial \over {\partial x'}}}\left\langle {x'|\alpha } \right\rangle - \left\langle p \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle$

 

[bjnartowt]

sigh...how obnoxious. well, "i found this implied (i think)" is where my real question is...