Dirac notation - expectation value of kinetic energy

[Clarky48]

It's my first post so big thanks in advance :)

1. Homework Statement
So the question states "By interpreting <p_xΨ|p_xΨ> in terms of an integral over x, express <E_kin> in terms of an integral involving |∂Ψ/∂x|. Confirm explicitly that your answer cannot be negative in value." ##The 'p_x's should have hats to indicate an operator##
If it helps, it's part of a question where Ψ is the wavefunction of a particle in a 1D pot energy well and the question is testing on a chapter about introducing and working with Dirac notation.

Homework Equations

None that I can think of

The Attempt at a Solution

The maths of my solution so far is

So, the part I'm stuck with is proving explicitly that it can't be negative in value. Obviously the factor to the left is always positive but it's the integral that's the problem. I know the integrand is always real and positive as it's a modulus squared, I just can't see any way to prove that the integral of that, over infinity, is never negative.

It's due in for next Monday so any help would be greatly appreciated!

 

[TSny]

Looks good.

You are integrating from $- \infty$ to $+ \infty$. So, what is the sign of $dx$ in the integral?

 

[Clarky48]

I'm not sure what you mean - I'd presume $\text{d}x$ is always positive, as it's effectively a small interval $\delta x$ centered on $x$.

So are you saying if I take the integral to be the infinite sum of small sections $| \partial \Psi / \partial x|^2 ~ \delta x$, then each section must be positive, as both $| \partial \Psi / \partial x|^2$ and $\delta x$ are always positive?

Thanks

 

[TSny]

So are you saying if I take the integral to be the infinite sum of small sections $| \partial \Psi / \partial x|^2 ~ \delta x$, then each section must be positive, as both $| \partial \Psi / \partial x|^2$ and $\delta x$ are always positive?

Thanks

Yes.

Another way to think about it is that $|\partial \Psi / \partial x|^2$ is a non-negative function of x. So, the integral is the "area under" the graph of a function that never goes below the x axis. So, the area is positive according to the sign conventions of area-under-a-graph as used in calculus.

 

[Clarky48]

Ah right, I think I get you.

So, to check I understand it - if the limits of integration were reversed $\int^{-\infty}_{\infty}$ (so you're integrating across the graph in the negative x direction), would the integral be negative?

 

[TSny]

So, to check I understand it - if the limits of integration were reversed $\int^{-\infty}_{\infty}$ (so you're integrating across the graph in the negative x direction), would the integral be negative?

Yes, that's right.

 

[Clarky48]

Awesome. Thanks for all the help