# Very basic question on Hamiltonian representation?

1. Jul 6, 2015

### askhetan

I am trying to teach myself DFT (yet again) from books and my maths is only improving at a modest pace to understand how people calculate using QM. So a very basic question now. When a Hamiltonian for a many body system is written as given in page 8 on this presentation.

https://www.uam.es/personal_pdi/ciencias/jcuevas/Talks/JC-Cuevas-DFT.pdf

Where i represents a certain electron, then what is(are) the meanings of x1, x2, etc (and likewise R1, R2, etc.). I know, they represent the coordinates of electron1, electron2, etc.. (and likewise nucleus1, nucleus2, etc) But those electrons do not have any coordinates in the first place. They have wave functions. Does it implicitly mean that x1 actually stands for the 'space dependent wave function of electron1' , and so on and so forth? I know its rather too simple but i can't get my head round what do the coordinates mean when there are no particles really which possess fixed coordinates!

Thanks for your help in advance

2. Jul 6, 2015

### Dr. Courtney

I think your interpretation is right, but the presentation is ambiguous.

3. Jul 6, 2015

### askhetan

Thanks for your reply. Just to confirm, do you mean the presentation of this idea in the link is ambiguous ? Its strange because almost every book on DFT uses similar presentation.

4. Jul 6, 2015

### Dr. Courtney

Books usually take greater care to define variables and symbols than presentations. I think your interpretation of the coordinates is correct, but since they are not defined explicitly, I would regard the presentation as ambiguous.

5. Jul 6, 2015

### Avodyne

Let's do a single electron with normalized wave function $\psi(\vec x)$. This means that the probability to find the electron in a small volume $d^3\!x$ centered on a particular point $\vec x$ is given by $|\psi(\vec x)|^2 d^3\!x$.

Now there is a long story about measurement theory and what it means to "find" the electron, etc, but there is no need to go into it in order to understand the basic conceptual point: $\vec x$ is a place where the electron might be found, and $|\psi(\vec x)|^2 d^3\!x$ gives us the probability that the electron will in fact be found there if we look.

6. Jul 8, 2015

### askhetan

Oh nice! exactly what I wanted to bring up next. So my new understanding is rather that when we say $\psi(\vec x_1)$, it actually means $\psi$ defined in the position-space that we call $\vec x_1$. This brings me now to another perplexing question. What does it mean when someone writes the Slater determinant form of the antisymmetric wave function for a two electron system in terms of the imaginary single electron wave-functions, $\psi_1$ for electron 1 and $\psi_2$ for electron 2, and says that the sign of the total wave-function $\psi_o$ changes when the two electrons are exchanged, as given by:

$\psi_o = \psi_1(\vec x_1)\psi_2(\vec x_2) - \psi_1(\vec x_2)\psi_2(\vec x_1)$

At this juncture, I have no reason to believe why the position space defined by $\vec x_1$ will be different from $\vec x_2$. Lets assume for a moment the electrons have ownership of their 'own' positions spaces.

What does electron exchange then really mean if they are indistinguishable ? To clarify better - even if the total wave-function changes sign (a postulate, I'm not contesting its validity), the probabilities and the energy do not change because they are dependent on its square. It is always argued in the Hartree-Fock theory that this representation, although not complete, can account for the exact exchange energy, if we can come down to the exact single electron wave-functions some way or the other.

What I'm trying to understand is this - if Pauli's exclusion principle says no two electrons (which are fermionic and indistinguishable) can have the same quantum state, then does exchange simply mean the representation of the imaginary wave function of electron1 in the position space of electron2 defined by $\vec x_2$ and vice versa. This hardly makes sense, even mathematically. Whats a position space ? an electron actually doesn't have the ownership of a position space and therefore to me space defined by $\vec x_1$ is equal to space defined by $\vec x_2$. If I am able to take electron 1 and electron 2 and exchange their positions instantaneously, then just having the possibility of doing so lowers the energy of the system. This is what I cannot understand in the physical sense - why will the possibility of exchanging two electrons in a system lower its energy?

Once we enforce this anti-symmetrization in some way, we then have a wave-function that represents a system with lowered ground state energy due to the possibility of exchange - would this be correct way to look at it? Mathematically, its seems so. I know a lot of questions - would be great if someone could point out the mistakes in my arguments.

Thanks

Last edited: Jul 8, 2015
7. Jul 8, 2015

### Avodyne

OK, I'm only going to try to explain the basic point here.

For simplicity, I will ignore spin (but really we need to include spin for a fully correct description).

Suppose (counter to the actual facts of the real world) that we had a red electron and a blue electron. That is, if I handed you the red electron, you could look at it and say, "yes, this electron is the red one".

Then we could write down a probability that the red electron is near point $\vec x_1$ and the blue electron is near point $\vec x_2$ in the form $\rho(\vec x_1,\vec x_2)d^3\!x_1 d^3\!x_2$, where $\rho(\vec x_1,\vec x_2)$ is a probability density. This could be different from the probability of finding the blue electron near $\vec x_1$ and the red electron near point $\vec x_2$, which would be given by $\rho(\vec x_2,\vec x_1)d^3\!x_1 d^3\!x_2$.

However, as far as we can tell, electrons are indistinguishable; they all look exactly the same. In this case, we cannot ask for the probability that the "red" electron is near $\vec x_1$ and the "blue" electron is near $\vec x_2$. All we can ask for is the probability that there is one electron near $\vec x_1$ and another electron near $\vec x_2$. If we say that this probability is given by an expression of the form $\rho(\vec x_1,\vec x_2)d^3\!x_1 d^3\!x_2$, then consistency with the words ("one electron near $\vec x_1$ and another near $\vec x_2$") requires that we have $\rho(\vec x_2,\vec x_1)=\rho(\vec x_1,\vec x_2)$, because the words do not allow the ordering of the arguments of $\rho$ to make any difference.

In quantum mechanics, the probability density is the square of the position-space wave function, $\rho(\vec x_1,\vec x_2)=|\psi(\vec x_1,\vec x_2)|^2$. So in order to have $\rho(\vec x_2,\vec x_1)=\rho(\vec x_1,\vec x_2)$, we must have $\psi(\vec x_2,\vec x_1)=e^{i\theta}\psi(\vec x_1,\vec x_2)$, where $\theta$ is a real phase. Now it turns out that consistency conditions in three or more dimensions imply that the only possibilities are $\theta=0$ or $\theta=\pi$, or equivalently $e^{i\theta}=1$ or $e^{i\theta}=-1$. (In two dimensions, there are more possibilities; see https://en.wikipedia.org/wiki/Anyon .)

Then, it further turns out that whether we use $+1$ or $-1$ depends on the total spin of the particle. For spin-one-half particles (which include electrons), we must use $-1$. (This is the spin-statistics theorem of relativistic quantum field theory, which is fully consistent with experiment.)

So, for two electrons, we must use a position-space wave function that obeys $\psi(\vec x_2,\vec x_1)=-\psi(\vec x_1,\vec x_2)$. Energies and all other physical quantities must be computed with this restriction in place. States that do not obey this restriction do not exist. It's not that these putative other states have higher or lower energies; they simply do not exist.

Last edited: Jul 8, 2015
8. Jul 8, 2015

### askhetan

Thank you so much. A great explanation.

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