# Knowledge and understanding of QM

1. Mar 14, 2013

### LagrangeEuler

In problem of infinite potential well particle can't be in a region where $V=\infty$. How we know that in case of finite potential well that particle is in the region where $V>E$. How we know what is energy of the particle in moment when we localize it in the region with potential $V$. This is very hard for me to understand. Can u help me?

2. Mar 14, 2013

### Staff: Mentor

It is an experimental result that you can find the particles in those "forbidden" regions with some probability.
You can use quantum mechanics to calculate that probability.

Do you want a good position or a good energy measurement? Those will lead to different results.

3. Mar 14, 2013

### LagrangeEuler

That is not my question. I asked how we know energy of the particles in different region?
http://en.wikipedia.org/wiki/Finite_potential_well
"There are two possible families of solutions, depending on whether E is less than (the particle is bound in the potential) or E is greater than (the particle is free)."

How we know from measurement is it particle free or not?

I want to know more about this measurements. I know how to solve this problem with use og mathematics.

4. Mar 14, 2013

### Staff: Mentor

That does not exist. The particle has a unique [expectation value of the] energy, which does not depend on the position.

Measure the energy of the particle ;). Or just see if it stays in the well (and in its direct vicinity) - if it does, it is not free, otherwise it is free.

5. Mar 15, 2013

### LagrangeEuler

I have a problem with this measurements. How they measure those quantities. In real experiment you will need to think about temperature, all other kinds of noise...

What about bond states in the well. If finite potential well is symmetric then for $E<V$, there are bond states. If isn't symmetric then could be bound states. Why is that physically?

6. Mar 15, 2013

### Staff: Mentor

Temperature is a property of many particles at the same time, a single particle does not have a (meaningful) temperature.

What about bound states? Well, those are the states with E<V, what exactly are you asking?
And which symmetry do you mean?

7. Mar 15, 2013

### LagrangeEuler

If I have finite potential well with $0>E>V_1<V_2$ I don't have bond states for every $V_1$ and $V_2$. Why I don't have it always.

8. Mar 15, 2013

### Staff: Mentor

What are V1, V2 (or where are they), and why do you have an E in the description of your potential?
In more than 1 dimension, some potential wells are so shallow that they do not have bound states. In 1 dimension, you always have a bound state.

9. Mar 15, 2013

### LagrangeEuler

It is one dimension square well. $V_1$ is potential for $x<0$, $V_2$ is potential for $x>a$ and for $0<x<a$ potential is $0$. Symbol for $E$ in the question isn't potential, it defines bond energy states. Can you please give me some answer?

No you don't have always bond state! That is the problem!

10. Mar 21, 2013

### Staff: Mentor

That is not a potential well, that is a function with two steps (where the direction of the right step depends on the sign of V2). You need some global minimum in a limited range for a (proper) potential well.