Solving the 1-D Schrodinger equation for a free particle: Confused about 2 possible general solutions

Question

I am following Griffiths' Introduction to Quantum Mechanics, as well as an online lecture that follows a different book, and both sources give different equations for the general solution of the 1-D Schrodinger equation for a free particle.

Griffiths has it as: $$\Psi(x,t)=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^{+\infty}\phi(k) e^{ i\left(kx-\frac{hk^2}{2m}t\right)}dk\tag1$$

The online lecture has it as:$$\Psi(x,t)=\frac{1}{\sqrt {2\pi\hbar}}\int_{-\infty}^{+\infty}a(p) e^{ \frac{i}{\hbar}\left(px-\frac{p^2}{2m}t\right)}dp\tag2$$

The most glaring addition to the second equation is the $\hbar$ under the square root sign. Given this, I can't see how they're both equivalent.

The link to the online lecture is http://www.youtube.com/watch?feature=player_detailpage&v=xm-LMpkqSUQ#t=1273s

Answer 1

Consider making the substitution $k = p/\hbar$ in the first expression, while simultaneously defining $$ \sqrt{\hbar} \,a(p) = \phi(k) $$ then the first integral will become $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \sqrt{\hbar}\, a(p) e^{i\left(\frac{p}{\hbar} x - \frac{\hbar}{2m}\frac{p^2}{\hbar^2}t\right)} \frac{dp}{\hbar} $$ which is exactly the second integral. It's just a change of notation!