# Potential step with Infinite negative potential

When a particle with Energy $$E>0$$ approaches from $$x<0$$ an infinite negative potential with following conditions:

$$V(y)= \begin{cases} 0 &\text{if}\,& x\leq 0 \\ -\infty&\text{if}\,& x>0 \\ \end{cases}$$

What is the probability its going to be refelcted?

So when I look at this my first Idea is that its goint to have $$0$$ probability of being refelcted because it would just gain infinite energy by passing the barrier. However my answer is wrong and it turns out it has a probability of $$1$$ to be reflected. Im not sure how this can be explained.

Quick edit to show my work:

$$k_1=\sqrt{\frac{2m}{h^2}(E)}$$

$$k_2=\sqrt{\frac{2m}{h^2}(E-V_0)} = \sqrt{\frac{2m}{h^2}(E+\infty)} \rightarrow \infty$$,

Transmission coefficient:

$$T \propto \frac{k_2}{k_1} = \infty$$ so there cant be any reflection.

• Do you know the deflection probability for a potantial of the form $V'(x) = \begin{cases} 0 &\text{if}\,& x\leq 0 \\ -V_0 &\text{if}\,& x>0 \\ \end{cases},~ V_0 > 0$ ? You could look at the result and take the limit $V_0 \to \infty$. – Cream Apr 22 at 13:20
• Thats what I was using, I edited my question with how I found my solution. I just dont understand how it wouldnt just gain the energy as it passes into the new area – Butty Apr 22 at 14:13
• Ah, very good! Where is the claim coming from, that there can't be any reflection? That would also be good to know. This is not a realistic scenario, more of a mathematical couriosity, right? – Cream Apr 22 at 14:29
• This question came up in an old homework of mine that I'm using for revision and I was just a bit confused. I assume there can't be any reflection because the transmission is infinite. – Butty Apr 22 at 14:31
• But both reflection and transmission coefficients should always satisfy $0 < R,T < 1$ and $(R + T = 1)$. How can it be larger than 1 already at finite $V_0$? – Cream Apr 22 at 15:12

Because there is a change in potential, and you must account for probability current. I'm guessing that's where you got your assumption for transmission coefficient condition. You didn't include, however, the fact the solution amplitudes cannot be ignored, and the boundary conditions at $$x=0$$ force the amplitudes themselves to be dependent on $$k_1$$ and $$k_2$$. Your $$k_2/k_1$$ factor is actually an additional multiplication factor.
The transmission coefficient in this case turns out to be $$T=\frac{4|k_1|^2}{|k_1+k_2|^2}\frac{k_2}{k_1},$$ which goes to zero in the limit that $$k_2\to\infty$$.
• Thank you for the answer! So the way I understand what your saying, is that a particle can't gain any energy by passing into a negative potential. It just acts the same at any potential step. Wouldnt the $k$ vector have to look something like this then: $k_2=\sqrt{\frac{2m}{h^2}(E- | V_0 | )}$ – Butty Apr 22 at 18:30
• @Butty No, I didn't say that. I'm saying that the transmission probability is not merely $k_2/k_1$ because you must consider that amplitudes of the solutions. They depend on the $k$ values, too. If $V_0$ is finite, there will be a non-zero transmission probability. Also, the $E-|V_0|$ is not right.if $V(x\gt 0)=-V_0$ then you have $E+V_0$. – Bill N Apr 22 at 18:56