Question

Calculate the expectation value of `x^3` for a particle in the state n=2 in a square well potential?

Answer 1

Well, after several integration by parts...

`<< x^3 >> = ((4pi^2 - 3)L^3)/(16pi^2)`

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DISCLAIMER: REALLY LONG ANSWER!

I assume you mean the one with

`V(x) = { (0, 0 < x < L),(oo, "otherwise"):}`

For this, I assume you already have the known wave function solution as:

`psi_n(x) = sqrt(2/L) sin((npix)/L)`

In that case, for `n = 2` we have `psi_2(x) = sqrt(2/L) sin((2pix)/L)`.

The `x^3` expectation value for a normalized wave function (which this is) is:

`<< x^3 >> = << psi_2 | x^3 | psi_2 >>`

`= 2/L int_(0)^(L) x^3sin^2((2pix)/L)dx`

Use the identity that `1/2 (1 - cos(2u)) = sin^2u` so that:

`= 1/L int_(0)^(L) x^3(1 - cos((4pix)/L))dx`

`= 1/L int_(0)^(L) x^3dx - 1/L int_(0)^(L) x^3cos((4pix)/L)dx`

Unfortunately there is not much we can do except integration by parts...

Let:

`u = x^3`
`dv = cos((4pix)/L)dx`
`v = L/(4pi) sin((4pix)/L)`
`du = 3x^2dx`

First we get:

`1/L int_(0)^(L) x^3cos((4pix)/L)dx`

`= 1/L [L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)int_(0)^( L)x^2sin((4pix)/L)dx]`

Repeat again. Let:

`u = x^2`
`dv = sin((4pix)/L)dx`
`v = -L/(4pi)cos((4pix)/L)`
`du = 2xdx`

Then we get:

`= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) int_(0)^(L)xcos((4pix)/L)dx]}`

And we have one more integration by parts... Let:

`u = x`
`dv = cos((4pix)/L)dx`
`v = L/(4pi)sin((4pix)/L)`
`du = dx`

Then we get:

`= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) {L/(4pi)xsin((4pix)/L) - L/(4pi)int_(0)^(L)sin((4pix)/L)dx}]}`

Evaluating the innermost integral,

`= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) {L/(4pi)xsin((4pix)/L) - L/(4pi)(-L/(4pi)cos((4pix)/L))}]}`

And now we simplify it, before evaluating. Factor in the `L/(4pi)` on the inside:

`= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) {L/(4pi)xsin((4pix)/L) + L^2/(16pi^2)cos((4pix)/L)}]}`

Factor in the `(2L)/(4pi)`:

`= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L^2)/(16pi^2)xsin((4pix)/L) + (2L^3)/(64pi^3)cos((4pix)/L)]}`

Factor in the `-(3L)/(4pi)`:

`= 1/L {L/(4pi) x^3sin((4pix)/L) + (3L^2)/(16pi^2)x^2cos((4pix)/L) - (6L^3)/(64pi^3)xsin((4pix)/L) - (6L^4)/(256pi^4)cos((4pix)/L)}`

Lastly, factor in the `1/L`:

`= 1/(4pi) x^3sin((4pix)/L) + (3L)/(16pi^2)x^2cos((4pix)/L) - (6L^2)/(64pi^3)xsin((4pix)/L) - (6L^3)/(256pi^4)cos((4pix)/L)`

And now we evaluate this from `0` to `L`... Lots of stuff goes to `0`.

`= [cancel(1/(4pi) L^3sin(4pi))^(0) + (3L)/(16pi^2)L^2cancel(cos(4pi))^(1) - cancel((6L^2)/(64pi^3)Lsin(4pi))^(0) - (6L^3)/(256pi^4)cancel(cos(4pi))^(1)] - [cancel(1/(4pi) 0^3sin((4pi cdot0)/L))^(0) + cancel((3L)/(16pi^2)0^2cos((4picdot0)/L))^(0) - cancel((6L^2)/(64pi^3)cdot0cdot sin((4picdot0)/L))^(0) - (6L^3)/(256pi^4)cancel(cos((4picdot0)/L))^(1)]`

`= [(3L)/(16pi^2)L^2 - cancel((6L^3)/(256pi^4))] - [- cancel((6L^3)/(256pi^4))]`

`= (3L^3)/(16pi^2)`

Now to finish this off:

`color(blue)(<< x^3 >>) = << psi_2 | x^3 | psi_2 >>`

`= 1/L int_(0)^(L) x^3dx - (3L^3)/(16pi^2)`

`= 1/L L^4/4 - (3L^3)/(16pi^2)`

`= L^3/4 - (3L^3)/(16pi^2)`

`= color(blue)(((4pi^2 - 3)L^3)/(16pi^2))`