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

#### fluidistic

Gold Member
If one consider a 1 dimensional potential of the form $$V(x)=0$$ for $$x<0$$ and $$V(x)=V_0$$ for $$x \geq 0$$.
The corresponding Schrödinger's equation for the region greater than 0, is $$\Psi '' (x)+ \Psi (x) \frac{(E-V_0)2m}{\hbar ^2}=0$$.
Now if $$E=V_0$$, the solution to this equation is a straight line $$\Psi (x)=ax+b$$. However since the region is infinitely large, there is no values for a and b other than 0 that will normalize $$\Psi$$, hence we must conclude that $$\Psi (x)=0$$ in this region.
Fine, that mean for $$E=V_0$$, there is a total reflection of a particle coming from the left (from negative x).
However if $$E\neq V_0$$ (for both $$E<V_0$$ and $$E>v_0$$), 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 $$V(x)=\infty$$ for $$x<0$$, $$V(x)=V_0$$ for $$0 \leq x \leq a$$ and $$V(x)=\infty$$ for $$x>a$$, and if $$E=V_0$$, then I think there's an infinity of possible solutions. I mean there are an infinite number of constants A and B such that $$\int_0 ^a |Ax+B|^2 dx =1$$. I don't think it's possible physically, so what's going on?
By the way I get the condition $$\frac{A^2 a^3}{3}+ a^2|AB| + B^2 a =1$$.

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

What's happening?

#### Joseph14

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
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.

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