Special case of energy value for a particle in a non zero potential region

  • Thread starter fluidistic
  • Start date

fluidistic

Gold Member
3,543
72
If one consider a 1 dimensional potential of the form [tex]V(x)=0[/tex] for [tex]x<0[/tex] and [tex]V(x)=V_0[/tex] for [tex]x \geq 0[/tex].
The corresponding Schrödinger's equation for the region greater than 0, is [tex]\Psi '' (x)+ \Psi (x) \frac{(E-V_0)2m}{\hbar ^2}=0[/tex].
Now if [tex]E=V_0[/tex], the solution to this equation is a straight line [tex]\Psi (x)=ax+b[/tex]. However since the region is infinitely large, there is no values for a and b other than 0 that will normalize [tex]\Psi[/tex], hence we must conclude that [tex]\Psi (x)=0[/tex] in this region.
Fine, that mean for [tex]E=V_0[/tex], there is a total reflection of a particle coming from the left (from negative x).
However if [tex]E\neq V_0[/tex] (for both [tex]E<V_0[/tex] and [tex]E>v_0[/tex]), there's no total reflection, quite strange. I'm trying to understand the physical meaning of that. It's like there's a well definite energy that isn't allowed for a particle to pass the potential.

If we now consider a region [tex]V(x)=\infty[/tex] for [tex]x<0[/tex], [tex]V(x)=V_0[/tex] for [tex]0 \leq x \leq a[/tex] and [tex]V(x)=\infty[/tex] for [tex]x>a[/tex], and if [tex]E=V_0[/tex], then I think there's an infinity of possible solutions. I mean there are an infinite number of constants A and B such that [tex]\int_0 ^a |Ax+B|^2 dx =1[/tex]. I don't think it's possible physically, so what's going on?
By the way I get the condition [tex]\frac{A^2 a^3}{3}+ a^2|AB| + B^2 a =1[/tex].

And also in the case of having a region of potential of the form [tex]V(x)=0[/tex] for [tex]x<0[/tex], [tex]V(x)=V_0[/tex] for [tex]0 \leq x \leq a[/tex] and [tex]V(x)=0[/tex] for [tex]x>a[/tex], the same problem appears in the central region. It seems like the particle can have infinitely many different [tex]\Psi (x)[/tex] which doesn't make sense to me.

What's happening?
 
Your problem is that you are not treating the case of a free particle correctly.

I'm just learning tex and by browser seems to have a bug when giving me the preview so please excuse me.

Consider a free particle where [tex]V(x)=0[\tex]. The solutions will be a basis consisting of sines and cosines at every frequency (your linear solution is actually a special case where [tex]f=0 so \lambda=\inf[\tex]). Sine is not normalizable however it is still useful because it forms a basis that using fourier (adding together a combination of sine functions) can create arbitrary wave packets. The same thing applies when [tex] V(x)=V_0[\tex].

For the case with in infinite bounding potentials you have given the problem of a particle in a box which is well documented. Given that this is quantum mechanics it should not be a surprise that for a bounded particle like this the allowed energy of solutions is quantized. [tex] E=V_0 [\tex] is not one of the allowed solutions.
 

fluidistic

Gold Member
3,543
72
Thanks for all!
For the case with in infinite bounding potentials you have given the problem of a particle in a box which is well documented. Given that this is quantum mechanics it should not be a surprise that for a bounded particle like this the allowed energy of solutions is quantized. [tex] E=V_0 [\tex] is not one of the allowed solutions.
Oh nice, I totally missed that part.
Problem solved.
 

Want to reply to this thread?

"Special case of energy value for a particle in a non zero potential region" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top