# Sakurai pr. 1.18 - bra-c-ket sandwiches

• bjnartowt
\int {\left\langle {x'|p|x'} \right\rangle \cdot \left\langle {x'|\alpha } \right\rangle \cdot dx'} = p\left\langle {x'|\alpha } \right\rangle seems bogus to me...is it?
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?

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?

argh! try this

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

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

Your approach is correct, but there seems to be a mistake in your source's calculation. The correct expression should be:

⟨x'|p|α⟩=−iℏ∂⟨x'|α⟩∂x'−⟨p⟩⟨x'|α⟩

This can be easily verified by applying the momentum operator to the Gaussian wave function:

p|α⟩=−iℏ∂∂x'|α⟩=−iℏ(2πd2)−1/4exp(i⟨p⟩x'ℏ−(x'−⟨x⟩)24d2)

Then, taking the inner product with ⟨x'| gives:

⟨x'|p|α⟩=−iℏ⟨x'|∂∂x'|α⟩−⟨p⟩⟨x'|α⟩=−iℏ∂⟨x'|α⟩∂x'−⟨p⟩⟨x'|α⟩

which is the same as the expression you obtained using the identity operator. Therefore, your calculation is correct and your source's calculation is incorrect.

## 1. What is Sakurai pr. 1.18?

Sakurai pr. 1.18 refers to the Sakurai perturbation theory, which is a mathematical tool used in quantum mechanics to calculate the energy levels of a quantum system.

## 2. What does the number 1.18 represent in Sakurai pr. 1.18?

The number 1.18 represents the order of the perturbation theory. Sakurai pr. 1.18 is the first-order perturbation theory, which means it takes into account the first-order corrections to the energy levels of a system.

## 3. What are "bra-c-ket sandwiches" in Sakurai pr. 1.18?

"Bra-c-ket sandwiches" are a visual representation of the mathematical notation used in Sakurai pr. 1.18. The "bra" and "ket" symbols represent the bra and ket vectors in quantum mechanics, while the "c" represents the complex conjugate operation and the "-" represents the inner product operation.

## 4. How are "bra-c-ket sandwiches" used in Sakurai pr. 1.18?

In Sakurai pr. 1.18, "bra-c-ket sandwiches" are used to represent the matrix elements of a perturbation operator. These matrix elements are then used to calculate the first-order corrections to the energy levels of a system.

## 5. Why is Sakurai pr. 1.18 important in quantum mechanics?

Sakurai pr. 1.18 is important in quantum mechanics because it provides a way to calculate the energy levels of a quantum system by taking into account the effect of external perturbations. This is essential in understanding and predicting the behavior of quantum systems in real-world applications.

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