- #1

- 68

- 0

SOLVED

(Example 6.15 from Modern Physics 3e- Serway)

Compute the average position <x> for the particle in a box assuming it is in the ground state

[tex]

|\Psi|^2=(2/L)\sin^2{(\pi x/L)}

[/tex]

[tex]

<x> = \int^{x_0+L}_{x_0}x|\Psi|^2dx

[/tex]

[tex]

<x>=x_0+L/2-\frac{L}{2\pi}\sin{\frac{2\pi x_0}{L}}

[/tex]

I'm pretty sure this is the answer, however, I don't understand why I get that last term, I mean, the average position should be [tex] x_0 + L/2 [/tex] right?

If I take [tex]x_0=0[/tex] then the answer is what I was hoping for (Indeed this is the original procedure in the book), but in the more general expression with [tex] x_0 \neq 0 [/tex] I get the previous answer.

(Example 6.15 from Modern Physics 3e- Serway)

## Homework Statement

Compute the average position <x> for the particle in a box assuming it is in the ground state

## Homework Equations

[tex]

|\Psi|^2=(2/L)\sin^2{(\pi x/L)}

[/tex]

[tex]

<x> = \int^{x_0+L}_{x_0}x|\Psi|^2dx

[/tex]

## The Attempt at a Solution

[tex]

<x>=x_0+L/2-\frac{L}{2\pi}\sin{\frac{2\pi x_0}{L}}

[/tex]

I'm pretty sure this is the answer, however, I don't understand why I get that last term, I mean, the average position should be [tex] x_0 + L/2 [/tex] right?

If I take [tex]x_0=0[/tex] then the answer is what I was hoping for (Indeed this is the original procedure in the book), but in the more general expression with [tex] x_0 \neq 0 [/tex] I get the previous answer.

Last edited: