Tunnelling

[A23]

I thought bout this but don't know if it is right ?:

 

let an infinite potential [math]V_{x_0}(x)=\left\lbrace\begin{array}{ll} -\infty & \textrm{ if }x\in\{x_0-l,x_0+l\}\\0 & else\end{array}\right.[/math]

 

suppose L>l, E<0 and the Schrödinger equation

 

[math]-\frac{\hbar^2\partial^2}{2m\partial x^2}\Psi_0(x)+V_0(x)\Psi_0(x)=E\Psi_0(x)[/math]

 

add [math]\underbrace{V_L(x)\Psi_0(x)}_{=0}[/math]

 

[math]-\frac{\hbar^2\partial^2}{2m\partial x^2}\Psi_L(x)+V_L(x)\Psi_L(x)=E\Psi_L(x)[/math]

 

add [math]\underbrace{V_0(x)\Psi_L(x)}_{=0}[/math]

 

gives:

[math]-\frac{\bar{h}^2\partial^2}{2m\partial x^2}(\Psi_0(x)+\Psi_L(x))+(V_0(x)+V_L(x))(\Psi_0(x)+\Psi_L(x))=E\underbrace{(\Psi_0(x)+\Psi_L(x))}_{\Psi(x)}[/math]

 

hence [math]\Psi(x)[/math] is made of 2 separate blobs.

 

If this wavefunction was constructed this way : put 1 particle and increase the central potential to infinity,

could the particle tunnel from one to the other?

[SamSpeedo]

Looks good, don't cha haf ta close de bracket?

[A23]

Sorry, but what means 'bracket' in your msg:

a) some kind of parentheses ?

:shrug: in some trousers a zipper ?

[sanctus]

QM is long time back for me, but I would say that Psi defined your way is kind of a superposition of two states (or better set of states). If this is right, then there would be no need for tunneling, I think.

 

But, I repeat, QM is many years back, it may be complete BS what I said ;-)

[Pyrotex]

Sorry, but what means 'bracket' in your msg:

a) some kind of parentheses ?

:naughty: in some trousers a zipper ?

Speedo was referring to the leftbrace and underbrace you used in your equations.

And no, you do not have to "close" them with rightbrace's and overbrace's.