# Homework Help: Deriving <v> from <x> - Introductory Quantum Mechanics

1. Jun 12, 2013

### rem1618

(First post, hi everyone.)
I'm reading Griffith's textbook on intro to quantum mech right now. After establishing the following equation

$$\left\langle f(x) \right\rangle = \int f(x)p(x)\,dx$$

And thus

$$\left\langle x \right\rangle = \int x|\Psi|^2\,dx$$

He goes on to find $\left\langle v \right\rangle$ by taking the derivative with respect to time, and he writes

$$\frac{d\left\langle x \right\rangle}{dt} = \int x\frac{\delta}{\delta t}|\Phi|^2\,dx$$

which I don't completely understand. Why isn't the x inside the partial derivative? I'm not sure if it's a physical or mathematical concept I'm not grasping here.

I first thought x is the position, so it could be a function of time and written as x(t), but then I figure maybe x(t) doesn't really have any meaning from the statistical perspective. If it can be pulled out of that derivative, then it's just a...constant? Is there a physical meaning to that x or is it just an operator tool that makes the math work out?

2. Jun 13, 2013

### MisterX

Each state of definite x has an associated value of x, which is the x position that the state represents. It is this value of x which appears in the integral to find the average (weighted mean) of the observable x. It's something that is fixed - each definite x position state has an associated x value, which is the value of x that this state represents. The changes in state are changes in $\psi$ - but the correspondence between eigenvectors and eigenvalues remains fixed.

Here are some examples

$\mid x_1\rangle$ represents the state of being at $x_1$. Suppose our wave function is 100% at $x_1$. Then $\mid \psi \rangle = \mid x_1 \rangle$

$$\langle \hat{x}\rangle =x_1 \langle x_1 \mid x_1 \rangle = x_1$$

if instead we had $\mid \psi \rangle = a_1(t)\mid x_1 \rangle + a_2(t)\mid x_2 \rangle$

$$\langle \hat{x}\rangle = x_1|a_1(t)|^2 + x_2|a_2(t)|^2$$

So we have a weighted average of positions $x_1$ and $x_2$. I wouldn't make sense to have $x_1$ and $x_2$ change with time.

The integral $\left\langle x \right\rangle = \int x|\Psi(x)|^2\,dx$ is like this only instead of two positions there are infinitely many.

The eigenvalue (which is the value of position) $x_1$ will always correspond to the state $\mid x_1 \rangle$. Hopefully this will make sense and be intuitive.

Last edited: Jun 13, 2013
3. Jun 13, 2013

### rem1618

That helped a lot. Thank you :]