Show the expectation value is non negative

[bobred]

Homework Statement

The kinetic energy is given by \left\langle E_{kin} \right\rangle = \frac{\left\langle \widehat{p}^2 \right\rangle}{2m}

In Dirac notation we have

\left\langle E_{kin} \right\rangle = \frac{1}{2m} \left\langle \widehat{p}\Psi | \widehat{p}\Psi \right\rangle

Homework Equations

We are asked to give this in integral form and involving \left| \frac{\partial \Psi}{\partial x} \right|

\left\langle E_{kin}\right\rangle =\dfrac{1}{2m} \displaystyle\int_{-\infty}^{\infty}\widehat{\textrm{p}}_{x}^{*}\Psi^{*}(x,t)\widehat{ \textrm{p}}_{x} \Psi(x,t) dx

Which then becomes

\dfrac{\hbar^{2}}{2m} \displaystyle\int_{-\infty}^{\infty}\left|\dfrac{\partial\Psi}{\partial x}\right|^{2} dx

The Attempt at a Solution

This is not a problem but we are also asked to confirm explicitly that this cannot be negative in value

\dfrac{\hbar^{2}}{2m} \displaystyle\int_{-\infty}^{\infty} \frac{\partial^2}{\partial x^2} \left| \Psi \right|^{2} dx

Is this valid? As the constants on the left are positive and the square of the modulus of a complex function is non negative.

Thanks

 

[DrClaude]

The square of a first derivative of a function is not the second derivative of that function squared.


Isn't the answer obtained directly from:

\dfrac{\hbar^{2}}{2m} \displaystyle\int_{-\infty}^{\infty}\left|\dfrac{\partial\Psi}{\partial x}\right|^{2} dx

 

[bobred]

Yes, thanks.