Expectation value of position

[Doitright]

Homework Statement

Show that, for a general one-dimensional free-particle wave packet

$$\psi (x,t) = (2 \pi h)^{-1/2} \int_{-\infty}^{\infty} exp [i (p_x x - p_x^2 t / 2 m)/h] \phi (p_x) dp_x$$

the expectation value <x> of the position coordinate satisfies the equation

$$<x> = <x>_{t=t_0} + \frac{<p_x>}{m}(t - t_0)$$

Hints:

Use the fact that

$$\frac{\partial}{\partial p_x} exp[i (p_x x - p_x^2 t / 2 m)/h] = \frac{i}{h} (x - p_x t / m) exp[i (p_x x - p_x^2 t / 2 m)/h]$$

to show that

$$<x> = \int_{-\infty}^{\infty} \phi^* (p_x) [ih \frac{\partial}{\partial p_x} +\frac{p_x}{m}t]\phi(p_x)dp_x$$

Homework Equations

The Attempt at a Solution

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I am struggling to prove what is mentioned in hints, ie:

$$<x> = \int_{-\infty}^{\infty} \phi^* (p_x) [ih \frac{\partial}{\partial p_x} +\frac{p_x}{m}t]\phi(p_x)dp_x$$

$$<x> = \int_{-\infty}^{\infty} \psi^*(x,t)x\psi(x,t)dx$$
$$= (2 \pi h)^{-1} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} exp [-i (p'_x x - p_x^{'2} t / 2 m)/h] \phi (p'_x) dp'_x (ih \frac{\partial}{\partial p_x}) $$
$$\int_{-\infty}^{\infty} exp [i (p_x x - p_x^2 t / 2 m)/h] \phi (p_x) dp_x dx$$

I am able to get

$$<x> = \int_{-\infty}^{\infty} \phi^* (p_x) [ih \frac{\partial}{\partial p_x} +\frac{p_x}{m}t - x]\phi(p_x)dp_x$$

I get one more term $$-\phi^*(p_x) x \phi(p_x)$$ than what is shown in the hints. I am wondering whether this extra term should be zero.

 

[Doitright]

Hi, I am wondering the extra term I get, which is different from what is asked to prove in the hints, ie,

$$\int_{-\infty}^{\infty} -\phi^*(p_x) x \phi(p_x) dp_x$$

is equal to zero?

 

[Doitright]

The steps:
$$<x> = \int_{-\infty}^{\infty} \psi^*(x,t)x\psi(x,t)dx$$
$$= (2 \pi h)^{-1} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} exp [-i (p'_x x - p_x^{'2} t / 2 m)/h] \phi^* (p'_x) dp'_x (ih \frac{\partial}{\partial p_x})$$
$$\int_{-\infty}^{\infty} exp [i (p_x x - p_x^2 t / 2 m)/h] \phi (p_x) dp_x dx$$
$$= (2 \pi h)^{-1} i h \int_{-\infty}^{\infty} dp'_x \int_{-\infty}^{\infty} dp_x \int_{-\infty}^{\infty} dx exp [-i (p'_x x - p_x^{'2} t / 2 m)/h] \phi^* (p'_x)$$
$$<\frac{i}{h} (x - \frac{p_x t}{m}) exp [i (p_x x - p_x^2 t / 2 m)/h] \phi (p_x) + exp [i (p_x x - p_x^2 t / 2 m)/h] \frac{\partial \phi(p_x)}{\partial p_x}>$$
$$= i h \int_{-\infty}^{\infty} dp'_x \int_{-\infty}^{\infty} dp_x \delta(p_x - p'_x) exp [\frac{i p_x^{'2} t}{2 h m}] exp [\frac{-i p_x^{2} t}{2 h m}] \phi^* (p'_x) $$
$$<\frac{i}{h} (x - \frac{p_x t}{m}) + \frac{\partial}{\partial p_x}> \phi(p_x)$$
$$= i h \int_{-\infty}^{\infty} \phi^* (p_x) <\frac{i}{h} (x - \frac{p_x t}{m}) + \frac{\partial}{\partial p_x}> \phi(p_x) dp_x$$
$$= \int_{-\infty}^{\infty} \phi^* (p_x) (- x + \frac{p_x t}{m} + i h \frac{\partial}{\partial p_x}) \phi(p_x) dp_x$$

And I get the extra term,
$$\int_{-\infty}^{\infty} -\phi^*(p_x) x \phi(p_x) dp_x$$
which I am not sure if it's zero.