Energy on the RHS of Schrodinger wave equation

1. Mar 1, 2015

Urmi Roy

Hi,

So in the steady state Schrodinger equation (SE), the E on the RHS (see http://scienceworld.wolfram.com/physics/SchroedingerEquation.html)
is the sum of the kinetic and potential energy of the electron.

However, is this the same as the 'internal energy'? The statistical distribution used to describe electrons is the Fermi-Dirac, where the exponential terms contain 'E' and 'mu'. So is this E the same as that in the SE equation?

Also, if E is the total sum of the KE and PE of the electron, what aspect of the energy of the electron is described by the chemical potential 'mu'?

2. Mar 2, 2015

Staff: Mentor

It's in the Feynman lectures somewhere, but he explains it clearly. All forms of energy are related to fundamental forces that are conservative:
http://en.wikipedia.org/wiki/Force#Nonconservative_forces

In fact the form of the Schroedinger equation (ie is the sum of a kinetic energy term and a potential term) is determined by probabilities being frame independent which is rather obvious - we would not expect different observers to disagree on that. Something very drastically would need to be wrong with our conception of nature if that's not true. You can find the detail in Chapter 3 of Ballentine.

Thanks
Bill

3. Mar 2, 2015

matteo137

It's very delicate to mix terms from statistical mechanics (in which in general the temperature is non-zero and the total number of particles is not fixed), and quantum mechanics (in which usually the temperature is assumed to be zero and the total number of particles is fixed).

Example.
We have $N$ non-interacting particles (e.g. bosons, fermions, ...) in a potential. You can write down a many-body Schroedinger equations (SE), which in the case of non-interacting particles factorizes in $N$ times a single-particle SE, and find the energies $E_i$ of all allowed configurations of this particles in the potential.

If you want to prepare this $N$ particles in a thermal state, their chemical potential $\mu$ is fixed by the constraint that the total number of particles must be $N$.
In the case of bosons, for example, $\sum_i \left( e^{-(E_i-\mu)/(k_B T)} - 1\right)^{-1} = N$, where you fix $T$ and $N$, and you extract $\mu$.

The internal energy, $U$, can be defined as the expectation value $U=\langle H\rangle$. Now you can see that if the $N$ particle (bosons) system is prepared as an eigenstate of the Hamiltonian, it's (internal) energy is simply the eigenstate associated. While a state prepared in a thermal state has internal energy $U=\sum_i E_i \left( e^{-(E_i-\mu)/(k_B T)} - 1\right)^{-1}$.

P.S. if you have internal (spin) degrees of freedom, you might need to take into account some multiplicity.

Last edited: Mar 2, 2015
4. Mar 2, 2015

Urmi Roy

Hi matteo137 and bhobba. Thanks for your replies. There are however certain aspects of your replies that are still not clear to me.

So this energy E you're mentioning is not the KE because you said in quantum mechanics you usually assume 0K...Is it then a potential energy?

So now you've increase the temperature from 0K to say T [K]. You're saying the chemical potential is a value that doesn't having any meaning on its own (like you can't define the chemical potential of a box of gas only) so in the case you've mentioned above, you just assign a value that fits well with other restrictions?
In this case how do we know values of Ei?

5. Mar 2, 2015

matteo137

Let's forget about temperature, for the moment.
The Schroedinger equation include both the kinetic and the potential energy (as the Hamiltonian does). Each eigenvalue $E_i$ is the sum of the two energies.
If you consider a free particle (i.e. without a potential $V$) you have zero potential energy and any possible kinetic energy, including zero.

Now let's consider the concrete example of $N$ non-interacting bosons in a harmonic potential. Each boson can be in one of the states of the harmonic oscillator (i.e. Fock states): $\vert 0\rangle, \vert 1\rangle, \vert 2\rangle, ...$, which are associated to the energies (kin+pot) $e_0, e_1, e_2, ...$.
If we prepare the system with, for example, all $N$ particles in state $\vert 2\rangle$, the total energy of the system will be $N e_2$.

You could now ask.... what is the temperature of this system?
And I would say that there is no clear answer. The best you can do is to express the total energy as a temperature, using the relation $T=E/k_B$. But this $T$ is not a temperature in the usual sense, because a temperature in the framework of statistical mechanics gives different values for the velocity distribution, for the populations in the energy levels, etc .... let me make this more clear.
In statistical mechanics, the populations in the energy levels $\vert 0\rangle, \vert 1\rangle, \vert 2\rangle, ...$ are given by $p_i = (e^{-(E_i-\mu)/(k_BT)}-1)^{-1}$, and $\mu$ is fixed by requiring $\sum_i p_i =N$. And this is because you are assuming that your system is in equilibrium (there is no time dependence), and that the number of particles is $N=$constant.
You can understand that there is no choice of $\mu, T$ which leads to our example (which can be realized!) of having all particles in state $\vert 2\rangle, ...$, and zero in $\vert 0\rangle, \vert 1\rangle, \vert 3\rangle, ...$. The same inconsistency appears if you try to evaluate the expectation value of the energy or of other quantities.

Conversely, you can say that your system of $N$ particles is in thermal equilibrium at the temperature $T$ (in it's usual interpretation), and extract the populations in the states $\vert 0\rangle, \vert 1\rangle, \vert 2\rangle, ...$ using the formula $p_i=...$.

Note the very important fact that AS $T$ TENDS TO ZERO, $\mu$ TENDS TO $e_0$, in order to keep $\sum_i p_i =N$.

6. Mar 2, 2015

Urmi Roy

Thanks matteo137, I think things are clear for me now. Thanks for your patience!