Energy in Schrodinger’s Equation

[exmarine]

I have dozens of physics textbooks, but still can’t get a decisive fix on the energies in Schrodinger’s equation. Start with the "total energy" E. It seems to usually be defined as the sum of the kinetic and the potential.
1. Then why can’t I subtract the potential from both sides of the equation. That doesn’t make any sense.
2. It is often displayed as a constant, i.e., not as a function of position like the potential V(x). Again it doesn’t seem like E could then be the sum of kinetic and potential.
3. In tunneling situations, the potential V(x) is said to be LARGER than the total E. So how can that be true?

My second question is about the reference point for the potential energy. In mechanics, we set zero potential wherever it is convenient, as the change seems to be the only important feature. Is the same true for the potential in Schrodinger’s equation? I notice that in electrostatics, the potential energy is usually set to zero at infinity. So are all the potentials between like charges (repulsive forces) positive and all the potentials between opposite charges (attractive forces) negative?

Someone please write a paragraph (or two or three!) on this? Thanks, I appreciate those on here who have the knowledge and patience to answer questions.

BB

 

[zhermes]

..."total energy" E. It seems to usually be defined as the sum of the kinetic and the potential.
1. Then why can’t I subtract the potential from both sides of the equation. That doesn’t make any sense.

You can...

2. It is often displayed as a constant, i.e., not as a function of position like the potential V(x). Again it doesn’t seem like E could then be the sum of kinetic and potential.

Generally, conservation of energy does apply.

3. In tunneling situations, the potential V(x) is said to be LARGER than the total E. So how can that be true?

Classically you would say it has a negative potential energy. This is just one of the myriad wonders of quantum mechanics---wonders which our minds just weren't built for.

In mechanics, we set zero potential wherever it is convenient, as the change seems to be the only important feature. Is the same true for the potential in Schrodinger’s equation?

For the most part, yes. There are some additional things that need to be considered: e.g. the square of the wave-function must be normalizable (thus the potential generally needs to approach zero at infinity), etc etc.

So are all the potentials between like charges (repulsive forces) positive and all the potentials between opposite charges (attractive forces) negative?

It is the gradient of the potential that determines the direction of the force. $$F = - \nabla U$$

 

[sweet springs]

Hi, BB.

1. Then why can’t I subtract the potential from both sides of the equation. That doesn’t make any sense.

You can, but it is not eigenvalue equation any more.

2. It is often displayed as a constant, i.e., not as a function of position like the potential V(x). Again it doesn’t seem like E could then be the sum of kinetic and potential.

E is eigenvalue. E should be value so that the equation is eigenvalue equation.

3. In tunneling situations, the potential V(x) is said to be LARGER than the total E. So how can that be true?

For that kinetic energy should be negative, i.e. exponential damping of wave function occurs.

Regards.

 

[exmarine]

If one can subtract the potential from both sides of the equation, then it can have no influence on the solution - is that not correct? Somehow I don't believe that.

 

[naima]

In mechanics, we set zero potential wherever it is convenient, as the change seems to be the only important feature. Is the same true for the potential in Schrodinger’s equation?

You can add to the potential any function that will not change
the Euler Lagrange equation of the motion.

 

[K^2]

Full equation

$$\hat{H}\Psi(x,t) = \hat{E}\Psi(x,t)$$

Where hats denote operators. If H is time-independent, we look for solutions of the form

$$\Psi(x,t) = \psi(x)\phi(t)$$

So then we have

$$\left(\hat{H}\psi\right)\phi = \psi\left(\hat{E}\phi\right) = E\psi\phi$$

Here, E is just a number. This gives us two individual equations to be solved.

$$\hat{E}\phi = E\phi$$

$$\hat{H}\psi = E\psi$$

Keeping in mind that operator E is always the time derivative with some factors, it's easy to solve the first equation.

$$\hat{E}\phi = i\hbar \frac{\partial}{\partial t}\phi = E\phi$$

$$\phi(t) = e^{-i \frac{E}{\hbar}t}$$

The second equation depends on specific form for H, and is usually written in this form.

$$\hat{H}\psi = \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right)\psi = E\psi$$

And that's your time-independent Shroedinger equation as you are used to seeing it.