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There's a particle that's going towards a potential step

$U(x) = \begin{cases} 0, & \text{if } x < 0 \\ U_0, & \text{if } x \geq 0 \end{cases}$

we supposed $U_0>0$ and the energy of the particle $E<U_0$

Now resolving the Schroedinger equation we obtain: $\psi(x) = \begin{cases} A e^{ikx} + B e^{-ikx}, & \text{if } x < 0 \\ C e^{-ibx} , & \text{if } x \geq 0 \end{cases}$

With $k = \frac{\sqrt{2mE}}{\hbar}$ and $b = \frac{\sqrt{2m(U_0-E)}}{\hbar}$

Now my book (Quantum processes systems & information of Benjamin Schumacher) say that we can put the coefficient $A$ equals to $1$, why is it possible without losing of generality?

And I can use the same tecnique for solve all the problem with an incident particle like the delta function potential barrier?

daniele
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2 Answers2

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In these types of problems, we are modeling a scattering experiment. In such an experiment, the scientist shoots particles at an object and detects what comes out. In these 1D quantum scattering problems, we can think of the particles as electrons in a beam moving to the right and scattering off of the object creating the potential. Crucially, the scientist has control of two things in the experiment. First, she can choose the kinetic energy that each electron has; that's what is called $E$ in this model. Second, she can choose the flux of particles, i.e., she can choose how many particles to shoot at the scattering center per unit time; this quantity is proportional to the amplitude of the incoming wave, i.e., this quantity is determined by $A$.

For these reasons, both $E$ and $A$ have to be free parameters so that we are faithfully modeling the experimental fact that the scientist can control both the energy and flux of incoming particles. This is reflected in the mathematics in the following way. After taking care of the "boundary conditions at infinity"$^1$ to the right of $x=0$, we are left with four unknowns (the energy $E$, the amplitudes $A$ and $B$ on the waves for $x<0$, and the amplitude $C$ on the wave function for $x>0$). Continuity of the wave function and its derivative at $x=0$ is two conditions, which then allows us to solve for two out of the four remaining unknowns. This is as it should be, because there are two remaining unknowns: the energy $E$ and the incoming flux, which is proportional to $A$ (or, perhaps, $|A|^2$).

Finally, here is a question we can ask: "What proportion of particles are reflected at the boundary at $x=0$?" To answer this question, we have to take a ratio of the outgoing and incoming fluxes, i.e., we compute the reflectivity, equal to $|B/A|^2$. Since everything is linear, you'll find that $B$ is proportional to $A$, and so $A$ cancels out in the calculation $|B/A|^2$. This is the real reason you are allowed to set $A=1$.

This same sort of structure works for all quantum 1D scattering problems.

$^1$ If $E>V_0$, then this "boundary condition" is that there should be only right-moving waves in the region $x>0$, i.e., $\psi_{x>0}(x) = Ce^{ibx}$. If $E<V_0$, then this boundary condition is that particles can't "pile up" at $x\to\infty$, which forces us to choose only the decaying exponential in this region, i.e., $\psi_{x>0}(x) = Ce^{-bx}$.

march
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why is it possible without losing of generality?

You are solving the time independent Schrodinger equation, so it means that you are finding a set of solutions of the equation. Maybe not the most general one.

Different set of solutions could have different physical meanings. As @CosmasZachos said, for $k>0$, if we consider the temporal dependence

$$ \Psi(x,t) = e^{-iE_k t/\hbar}\psi(x) = Ae^{i(kx-w_k t)} + Be^{-i(Kx+w_k t)} $$

The first term represents a planar wave traveling to the right, and the second another plane wave going to the left. Both are perfect solutions in the mathematical sense, but only the first correspond to the physical situation that you want.

It is not question related, but your solution for $x\geq 0$ should be an evanescent wave, so a real exponential (I don't know if it's a typo or not).

Ruffolo
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