H Bar And No Bar

$h=x.mv$

In the case of a circular path, $n=1$

$h=2\pi a_\psi.mv$

$\hbar=\cfrac{h}{2\pi}= a_\psi.mv$

in this case $a_\psi$ is parallel to $x$, but perpendicular to $\lambda$,

When $n=2$,

$\Delta s=\sqrt { a^{ 2 }_{ \psi  }sin^{ 2 }(2\theta )(\Delta \theta )^{ 2 }+a^{ 2 }_{ \psi  }cos^{ 2 }(2\theta )(\Delta \theta )^{ 2 } }$

$\Delta s=a_{ \psi  }\Delta \theta \sqrt { sin^{ 2 }(2\theta )+cos^{ 2 }(2\theta ) }$

$\Delta s=a_{ \psi  }\Delta \theta$

$s=2\int _{ 0 }^{ \pi  }{ a_{ \psi  } } d\theta =2\pi a_{ \psi  }$

Strangely both paths have the same length.  In fact,

$\Delta s=\sqrt { a^{ 2 }_{ \psi  }sin^{ 2 }(n\theta )(\Delta \theta )^{ 2 }+a^{ 2 }_{ \psi  }cos^{ 2 }(n\theta )(\Delta \theta )^{ 2 } }$

$\Delta s=a_{ \psi  }\Delta \theta \sqrt { sin^{ 2 }(n\theta )+cos^{ 2 }(n\theta ) }$

$\Delta s=a_{ \psi  }\Delta \theta$

Which means,

$\hbar=a_\psi.mv$

is valid for all $n$.  The path length being the same for all $n$ also means that for all $n$, given a constant velocity at light speed, the period, $T$ to traverse all paths is the same, hence, $\psi$ has only one fundamental frequency, $f_{n=1}$.

$f_{n=1}=\cfrac{1}{T}$

This is consistent with the fact that,

$p=mc$    and    $E=mc^2$

which are both constants, independent of $n$.  The amplitude of the wave changes such that the energy of the wave is fully defined by,

$E=hf_{n=1}$ --- (*)

For all cases of $n$,

$c=nf_{n=1}.\cfrac{\lambda_{n=1}}{n}=f_{n=1}.\lambda_{n=1}$

Both $p$ and $E$ remain constant.  In this scenario where $n$ wavelengths are packed into a perimeter of $2\pi a_\psi$ all harmonics generated has the same energy given by (*).

The paths for $n=1$ and $n=2$ and the reference circle is generated using the following code in Scilab.

    t = 0:0.01:(2.0*%pi);
    x = 3*cos(t);
    y = 3*sin(t);
    z = 10*0.2*sin(t.*2);
 
    param3d(x,y,z,45,60,"",[2,0]);
    p=get("hdl"); //get handle on current entity (here the polyline entity)
    p.foreground=2;

    t = 0:0.01:(2.0*%pi);
    x = 3*cos(t);
    y = 3*sin(t);
    z = 10*0.2*sin(t.*1);
 
    param3d(x,y,z,45,60,"",[2,0]);
    p=get("hdl"); //get handle on current entity (here the polyline entity)
    p.foreground=5;  
 
    t = 0:0.01:(2.0*%pi);
    x = 3*cos(t);
    y = 3*sin(t);
    z = 0.000002*sin(t.*1);
 
    param3d(x,y,z,45,60,"",[2,0]);
    p=get("hdl"); //get handle on current entity (here the polyline entity)
    p.foreground=1;

Have a nice day.