# Transmission and Reflection amplitude

1. Jun 24, 2015

### hokhani

To calculate Transmission and Reflection amplitudes (t and r), say in the figure below which is composed of three regions, first we have the solution of Schrodinger equation in each region as: $$\psi _i=A_i |i> +B_i |-i> ; i=I,II,III$$. It seems that by matching the three wave functions and their derivatives, the coefficients A and B can be obtained. However, there are only four equations obtained at the two borders while there are six unknown coefficients! In addition, as far as I know, all the texts consider $$B_{III} =0$$ to calculate t and r ! Could anyone please help me?

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2. Jun 24, 2015

### Staff: Mentor

The incoming wave has to come from either the left or the right, and we have to choose which direction, as an initial condition. If the wave enters from the left, then AI is nonzero and BIII is zero. If the wave enters from the right, then AI is zero and BIII is nonzero.

Regardless of the direction of the incoming wave, its amplitude is arbitrary. We treat it as an arbitrary constant, then divide it into other coefficients in order to find r and t.

3. Jun 24, 2015

### hokhani

Thanks. But I don't know what you mean by incoming waves. In time-independent Schrodinger Equation the wave function is spread throughout the system and considering incoming or outgoing wave does't seem reasonable to me.

If the amplitude in each region were arbitrary, then we would have different wave functions with the same energy (degeneracy even in a 1-D system) while it is not possible.
Please let me know if I am making any mistakes.

4. Jun 24, 2015

### Staff: Mentor

In $\psi=Ae^{ikx} + Be^{-ikx}$ the positive exponential corresponds to waves/particles/whatever traveling to the right (+x direction), and the negative exponential, to the left.

To get the time-dependent wave function, your solution for $\psi$ gets multiplied at the end by an overall factor of $e^{-iEt/\hbar}$ which gives you traveling waves in one direction or the other depending on the sign of the x-exponential.

I didn't say that the amplitude in each region is arbitrary. The coefficient corresponding to the incoming waves/particles is arbitrary because it relates to e.g. the number of incoming particles per second (if you think in terms of particles); or perhaps better, to the intensity of the original source of the particles/waves. All the other coefficients are proportional to this one, except for the coefficient that you set to zero because of the boundary conditions (incoming waves/particles from the left or right).

5. Jun 24, 2015

### Staff: Mentor

You're misunderstanding what we're doing here, but in view of your question below, you may want to review your understanding of differential equations before you go any further (I will provide a more detailed explanation in another post in a moment, but it won't be a substitute for understanding how eigenfunctions of differential equations are used in general).

That's not how it works. If $\psi(x)$ is an eigenfunction with eigenvalue $E$, then we have $H\psi=E\psi$; but if we set $\phi(x)=\lambda\psi(x)$ we have $H\phi=H(\lambda\psi)=\lambda{H}\psi=\lambda{E}\psi=E\lambda\psi=E\phi$ so any constant multiple of $\psi(x)$ is also a solution. That doesn't make for a different solution, it's just an example of multiplying both sides of an equation by a constant without changing anything. You will note that $\psi(x)$ and $\phi(x)$ are not linearly independent, so this is not an example of degeneracy. No matter what constant we use, we'll find that the quantity that we care about
$$\frac{\psi^*(x)\psi(x)}{\int_{-\infty}^{+\infty}\psi^*(x')\psi(x')dx'}$$
comes out the same, so we choose the amplitude $\lambda$ such that the integral is equal to one and we can ignore it. This process is called "normalization" and when we choose $\lambda$ accordingly we say that $\psi$ is "normalized".

6. Jun 24, 2015

### hokhani

My problem is with the essence of incoming or outgoing particle. A particle Incoming (left to right) is one which at a time, say t=0, is at x but by passing the time, t>0, is at y>x. For a quantum particle whose wave function is spread throughout the system all the time, how do you consider coming from left to right or vice versa?

7. Jun 24, 2015

### Staff: Mentor

You can't, if you're just looking at the energy eigenstates you find by solving the time-independent Schrodinger equation; the expectation value of the position is a constant with respect to time so there's no motion going on. All that proves is that the state of a moving particle is not an energy eigenstate.
OK, here we go....

Suppose that $\psi_E(x)$ is an eigenfunction of the time-independent Schrodinger equation with eigenvalue $E$. There are, of course, many such eigenfunctions corresponding to different values of $E$.

You can verify by substitution that $\Psi_E(x,t)=e^{-iEt/\hbar}\psi_E(x)$ is a solution to the time-dependent Schrodinger equation (as long as the potential is independent of time, as it is here). Note also that $\Psi_E(x,0)=\psi(x)$ so we can interpret $\psi_E(x)$ as the state of a particle in that energy eigenstate at time zero.

You can also verify by substitution that any linear combination of these solutions to the time-dependent equation is also a solution of the time-dependent equation. You will have to take my word for it (or take my advice about reviewing differential equations in general) that not only is every such linear combination a solution, but also that every solution can be written as such a linear combination. Thus, once we know the eigenfunctions of the time-dependent equation, we can construct all the solutions of the time-dependent equation.

A moving particle isn't in an energy eigenstate, but you can write its state $\Psi(x,t)$ as a sum of the various $\Psi_E(x,t)$ for various values of $E$. Pick the right ones with the right amplitudes and the sum will give you a Gaussian-looking wave packet, the center of which is moving with time and being reflected and transmitted when it reaches the barriers. Furthermore, the initial state of that wave packet can be written as a sum of the $\psi_E(x)$ solutions to the time-independent equation; once we do that we have the complete time-dependent wave function of the moving particle.

Thus, the eigenvalues of the time-independent equation tell us what the possible energies are; and the eigenfunctions tell us how to build the solution to the time-dependent equation.

8. Jun 25, 2015

### Staff: Mentor

When I was an undergraduate more than 40 years ago, in one of my classes I saw a film which showed an animation of a Gaussian wave packet passing through a barrier. Back in those days, it took a very expensive computer and a film-output system to produce such animations. Nowadays you can probably do it in Mathematica , and there should be examples of them floating around the web, but I can't find any with a quick search.

I expect that the transmission and reflection probabilities for a wave packet with a very narrow spread in momentum and energy would be closely approximated by considering a plane wave with the momentum and energy of the "central" wave of the packet. Therefore, introductory QM textbooks analyze barrier penetration using monochromatic plane waves for simplicity.

Last edited: Jun 25, 2015
9. Jun 25, 2015

### DrDu

In a time independent setting, you can fix the meaning of ingoing and outgoing by calculating the expectation value of the current operator which is basically either +k or -k in your example.

10. Jun 25, 2015

### Staff: Mentor

Quite true... The only thing I might add (because OP may not realize this) is that if $\psi_E(x)$ is the solution to the time-independent Schrodinger equation (with $V=0$) with eigenvalue $E$, then the functions $e^{\pm{iE}t/\hbar}\psi_E(x)$ are solutions of the time-dependent equation corresponding to plane waves travelling left and right... a bit of algebraic massage with many examples on the web will get this into the traditional travelling wave form.

11. Jun 25, 2015

### DrDu

Only $e^{-{iE}t/\hbar}\psi_E(x)$ is an eigenvector to eigenvalue E. But, as I said, you have two degenerate eigenvectors corresponding to the same E, namely those with +k and -k. So the exponent becomes $i(\pm kx -Et)$. Points x(t) where this phase is constant will move with speed $v=\pm E/k$. That's the phase velocity.

12. Jun 26, 2015

This is the main point of the argument: to obtain the energy eigenstates according to what was said earlier, we first suppose an incoming particle (moving particle) from left to right and so put $$B_{III}=0[\tex]. However you said "a moving particle is not an energy eigenstate". So how one can calcualate the energy eigenstates? 13. Jun 26, 2015 ### Nugatory ### Staff: Mentor You calculate the energy eigenstates from the time-independent equation. That doesn't give you the wave function of the particle, it gives you the eigenfunctions that can be selectively combined according to the initial conditions (in this case, particle is known to be moving from left to right) to give you the time-dependent wave function of the particle. However, in this problem you don't need to do all that work. The plane wave is (the time evolution of) an energy eigenstate, and as jtbell explained, you can use that to calculate the transmission and reflection coefficients. You can do this even though the plane wave doesn't localize the particle at all. However, you asked above: "In time-independent Schrodinger Equation the wave function is spread throughout the system and considering incoming or outgoing wave doesn't seem reasonable to me". You are right that each individual eigenfunction of the time-independent equation describes a particle that is spread out through all of space; I was explaining how these can be linearly combined to produce solutions of the time-dependent equation that do describe a moving particle that is at some known place at a given time. 14. Jun 26, 2015 ### jtbell ### Staff: Mentor Also, even though the eigenfunctions are each "spread out" over all space and not localized, they are nevertheless traveling-waves that have a "direction of motion" as described by DrDu in posts 9 and 11. 15. Jun 29, 2015 ### hokhani Let me explain what I understand of this argument and please correct me if I am wrong; We can not distinguish the direction of movement considering only time-independent Schrodinger equation but by considering the time-dependent behavior, we know the direction the particle is coming from ( for example left to right) and hence to solve the time-independent Schrodinger Equation we set (in my example) [tex] B_{III} =0.$$
Then we obtain the eigenstates of time-independent Schrodinger Equation, and by linearly combining them we can obtain the travelling wave packet.

16. Jun 29, 2015

### hokhani

How can we physically distinguish that the "direction of motion" of the traveveling-wave $$expi(kx-\omega t)$$ is towards right? In the viewpoint of probability of finding, we know that the probability of finding the particle all the time is constant throughout the system and hence we can not imagine the movement.

17. Jun 29, 2015

### DrDu

You could measure the Doppler shift of light being reflected off the electrons .(just like your local police does with your car).

18. Jun 29, 2015

### Staff: Mentor

You can recognize it as rightwards movement by considering that t is constantly increasing. As t increases, what happens to the position of any point of constant kx−ωt? You may find this easier to visualize if you consider an ordinary classical wave moving across the surface of a body of water - it obeys the same equation.

And although you are right that the plane wave tells us nothing about the position, that doesn't mean that can't have a speed and direction; in fact the plane wave is the wave function of a particle of completely unknown position and completely known energy and direction. The right-moving wave impinging on the barrier from the left is the wave function of a particle about which we know nothing except that it is somewhere to the left of the barrier and moving right with energy E.
It's also an idealization - we always know at least a little bit about the position of the particle, give or take a few light-years - but it's quite good enough to calculate the transmission and reflection coefficients.

19. Jun 30, 2015

### DrDu

You should start to think quantum mechanically. Velocity and position cannot be measured at the same time and in fact you don't have to.
You can regard velocity as asking about a correlation function, if the particle is at point x at time t=0 then it is at point x+vt at time t. To get the velocity, you subtract the positions at the two time points and position x drops out altogether.
As you are doing QM now, you should have a very good understanding of classical mechanics right now. So you should be able to provide a list of, say, 10 different ways to measure velocity.