Matrix elements of one-body operator

[Bill Foster]

Homework Statement

Consider two fermion states:

|n_1 n_2\rangle = a_{n_1}^\dagger a_{n_2}^\dagger|0\rangle

and

|n_1 n_3\rangle = a_{n_1}^\dagger a_{n_3}^\dagger|0\rangle

where a_n^\dagger denotes the fermion creation operator in the single-particle state u_n\left(\vec{r},s\right). Evaluate the matrix elements of the one-body operator \hat{O}=\sum_{nn'}\langle n|O|n'\rangle a_n^\dagger a_{n'}^\dagger between these two states.

The Attempt at a Solution

Not sure if I should start out like this:

\langle 0| a_{n_1}^\dagger a_{n_2}^\dagger \sum_{nn'} \langle n|O|n'\rangle a_n^\dagger a_{n'}^\dagger a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle

of if I should start out like this:

\langle n_1 n_2 | \sum_{nn'} \langle n|O|n'\rangle a_n^\dagger a_{n'}^\dagger|n_1 n_3\rangle

 

[fzero]

You'll need to use the anticommutation relations to compute the matrix element, so you might as well start with the form with all of the operators

\langle 0| a_{n_1}^\dagger a_{n_2}^\dagger \sum_{nn'} \langle n|O|n'\rangle a_n^\dagger a_{n'}^\dagger a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle

However, there is an error with this formula, since taking the conjugate of |n_1n_2\rangle leads to

\langle n_1 n_2 | = \langle 0 | a_{n_2} a_{n_1}.

I'd also question whether or not there's a typo in that one-body operator, since it doesn't conserve the fermion number and it's matrix element between even particle states vanishes.

\sum_{nn'} \langle n|O|n'\rangle a_n^\dagger a_{n'}

would make more sense.

 

[Bill Foster]

You'll need to use the anticommutation relations to compute the matrix element, so you might as well start with the form with all of the operators
However, there is an error with this formula, since taking the conjugate of |n_1n_2\rangle leads to

\langle n_1 n_2 | = \langle 0 | a_{n_2} a_{n_1}.

I'd also question whether or not there's a typo in that one-body operator, since it doesn't conserve the fermion number and it's matrix element between even particle states vanishes.

\sum_{nn'} \langle n|O|n'\rangle a_n^\dagger a_{n'}

would make more sense.

You are correct. The typo was on my part.

So I start with this:

<br /> \langle 0| a_{n_1} a_{n_2} \sum_{nn&#039;} \langle n|O|n&#039;\rangle a_n^\dagger a_{n&#039;} a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle<br />

And then start applying the anti-commutation relations. I'll give that a shot.

Thanks.

 

[Bill Foster]

\{a_{n&#039;},a_{n}^\dagger\}=a_{n&#039;}a_{n}^\dagger + a_{n}^\dagger a_{n&#039;}=\delta_{nn&#039;}

a_{n}^\dagger a_{n&#039;}=\delta_{nn&#039;}-a_{n&#039;}a_{n}^\dagger

Plugging that in...

\langle 0| a_{n_1} a_{n_2} \sum_{nn&#039;} \langle n|O|n&#039;\rangle \left(\delta_{nn&#039;}-a_{n&#039;}a_{n}^\dagger\right) a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle
=\langle 0| a_{n_1} a_{n_2} \sum_{nn&#039;} \langle n|O|n&#039;\rangle \delta_{nn&#039;} a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle-\langle 0| a_{n_1} a_{n_2} \sum_{nn&#039;} \langle n|O|n&#039;\rangle a_{n&#039;}a_{n}^\dagger a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle
=\langle 0| a_{n_1} a_{n_2} \langle n|O|n\rangle a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle-\langle 0| a_{n_1} a_{n_2} \sum_{nn&#039;} \langle n|O|n&#039;\rangle a_{n&#039;}a_{n}^\dagger a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle

So my question now is this:

Is the following true?

\langle n|O|n\rangle = O

If so, then I have this:

=\langle 0| a_{n_1} a_{n_2} O a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle-\langle 0| a_{n_1} a_{n_2} \sum_{nn&#039;} \langle n|O|n&#039;\rangle a_{n&#039;}a_{n}^\dagger a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle
=\langle n_1 n_2| O |n_1 n_3\rangle-\langle 0| a_{n_1} a_{n_2} \sum_{nn&#039;} \langle n|O|n&#039;\rangle a_{n&#039;}a_{n}^\dagger a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle

Now I'm not quite sure how to deal with that second term.

 

[fzero]

Is the following true?

\langle n|O|n\rangle = O

No, but \langle n|O|n&#039;\rangle is a c-number and can be brought out of the expectation value that you're computing.

As for how to do these computations in general, note that

a_i |0\rangle =0.

Therefore, we should use the anticommutation relations to move the a^\daggers to the left.

 

[Bill Foster]

No, but \langle n|O|n&#039;\rangle is a c-number and can be brought out of the expectation value that you're computing.

I don't know what you mean by "c-number".

 

[fzero]

I don't know what you mean by "c-number".

A c-number is an ordinary scalar quantity. It doesn't depend on the state vectors or other operators in an expectation value, so

\langle \alpha | c A | \beta \rangle = c \langle \alpha | A | \beta \rangle .

 

[Bill Foster]

Which is true:

\langle 0|a_{n_3}^\dagger \langle n_2|O|n_1\rangle a_{n_1}|0\rangle = 0

or


\langle 0|a_{n_3}^\dagger \langle n_2|O|n_1\rangle a_{n_1}|0\rangle = \langle n_2|O|n_1\rangle

 

[fzero]

Which is true:

\langle 0|a_{n_3}^\dagger \langle n_2|O|n_1\rangle a_{n_1}|0\rangle = 0

or


\langle 0|a_{n_3}^\dagger \langle n_2|O|n_1\rangle a_{n_1}|0\rangle = \langle n_2|O|n_1\rangle

Those are both zero because the a annihilates the vacuum. On the other hand,


\langle 0|a_{n_3} \langle n_2|O|n_1\rangle a_{n_1}^\dagger|0\rangle = \delta_{n_1 n_3}\langle n_2|O|n_1\rangle .

 

[Bill Foster]

So once I get one creation operator all the way to the left (or one annihilation operator all the way to the right), then that entire term goes to zero, correct?

For example, suppose I have:

\langle 0|a_{n_1} \sum_{nn&#039;}\langle n|O|n&#039; \rangle a_n^\dagger a_{n_2} a_{n&#039;} a_{n_1}^\dagger a_{n_3}^\dagger |0\rangle

Then all I need to do is swap a_{n_1} with a_n^\dagger and then that term will go to zero? (Except, of course), the additional term resulting from the delta function).

 

[fzero]

Yes, you just have to get one annihilation operator to the right (or vice versa). The remaining delta terms contain all of the information about the original matrix element.

 

[Dickfore]

Consider the product:

<br /> \left\langle p_{1}, \ldots, p_{m} | q_{1}, \ldots q_{n} \right\rangle \equiv\left\langle 0 | a_{p_{1}} \, \ldots a_{p_{m}} \, a^{\dagger}_{q_{n}} \ldots a^{\dagger}_{q_{1}} | 0 \right\rangle<br />

Let us start moving a_{p_{m}} to the right. With each creation operator it gives a contraction:

<br /> a_{p_{m}} \, a^{\dagger}_{q_{k}} = \delta_{p_{m}, q_{k}} + \zeta \, a^{\dagger}_{q_{k}} \, a_{p_{m}}<br />

and one factor of \zeta when the flipping occurs. When it reaches the furthest right, it annihilates the vacuum giving zero a_{p_{m}} \, | 0 \rangle = 0. Finally, we are left with the following sum:

<br /> \sum_{k_{m} = 1}^{n}{\zeta^{k_{m}} \, \delta_{p_{m}, q_{k}} \, \langle 0 | a_{p_{1}} \, \ldots a_{p_{m - 1}} \,\ldots a^{\dagger}_{q_{k_{m} + 1}} \, a^{\dagger}_{q_{k_{m} - 1}} \ldots | 0 \rangle}<br />

The brackets in the sum differ from the original one by a a_{p_{m}} and a a_{q_{k_{m}}} and each one has a factor of \zeta^{k_{m}}, where \zeta = +1 for bosons and \zeta = -1 for fermions.

We may repeat this procedure for a_{p_{m - 1}} for each of the terms in the sum. Then, we will get a double sum with one pair of creation-annihilation operators less and some power of \zeta. Continuing this procedure, we will be left with either a bracket of the form:

<br /> \langle 0 | a_{p_{1}} | 0 \rangle = 0, \; m &gt; n<br />

or:

<br /> \langle 0 | a^{\dagger}_{q_{j_{n - m}}} \, a^{\dagger}_{q_{j_{n - m - 1}}} \, a^{\dagger}_{q_{j_{1}}} | 0 \rangle = 0, \; n &gt; m<br />

So, we can only get a non-zero matrix if m = n, i.e. we have the same number of annihilation operators as creation operators. Furthermore, due to the Kronecker deltas, we must have the single-particle states p_{1}, \ldots , p_{n} be some permutation of the states q_{1}, \ldots q_{n}. Finally, the power of \zeta is equal to the number of transopsitions necessary to bring the sequence \{q_{1}, \ldots , q_{n}\} to be identical to the sequence \{p_{1}, \ldots, p_{n} (notice that \zeta^{2} = 1 for any kind of particle).

You could have used this rule if you wrote the one-body operator as:

<br /> O = \sum_{m, m&#039;}{(m | O | m&#039;) a^{\dagger}_{m} \, a_{m&#039;}} = \zeta \, \sum_{m, m&#039;}{(m&#039; | O | m) a^{\dagger}_{m} \, a_{m&#039;}} - \zeta \, \sum_{m}{(m | O |m)}<br />

EDIT:
I noticed that you wrote the one-body operator with two creation operators. If this is so, then the matrix element is zero.

 

[Bill Foster]

Here's my final answer:

\langle 0|\langle n_2|O|n_1\rangle \delta_{n_1 n_3}|0\rangle - \langle 0|\langle n_1|O|n_1\rangle \delta_{n_2 n_3}|0\rangle - \langle 0|\langle n_1|O|n_3\rangle \delta_{n_1 n_1}|0\rangle + \langle 0|\langle n_1|O|n_3\rangle \delta_{n_2 n_1}|0\rangle

 

[Bill Foster]

I noticed that you wrote the one-body operator with two creation operators. If this is so, then the matrix element is zero.

That was a typo.

I don't suppose you all could help me with this problem: https://www.physicsforums.com/showthread.php?t=431118

It's another matrix element problem.

 

[fzero]

Here's my final answer:

\langle 0|\langle n_2|O|n_1\rangle \delta_{n_1 n_3}|0\rangle - \langle 0|\langle n_1|O|n_1\rangle \delta_{n_2 n_3}|0\rangle - \langle 0|\langle n_1|O|n_3\rangle \delta_{n_1 n_1}|0\rangle + \langle 0|\langle n_1|O|n_3\rangle \delta_{n_2 n_1}|0\rangle

The 3rd term should have \langle n_2|O|n_3\rangle. I'd simplify this by pulling the O matrix elements out and using \langle 0 | 0 \rangle=1.

 

[fzero]

That was a typo.

I don't suppose you all could help me with this problem: https://www.physicsforums.com/showthread.php?t=431118

It's another matrix element problem.

I don't understand what that problem is trying to get at. The Coulomb potential commutes with the spin operators, so it doesn't make any sense for the matrix element to depend on spin. To see spin effects in a two electron system, you have to consider the magnetic fields of the electrons, which results in the spin-orbit coupling. The Coulomb potential is electrostatic.

There's a discussion of the expectation value of the Coulomb Hamiltonian in Baym's QM book, but I can't view the entire argument on Google Docs. He certainly doesn't have any spin dependence in his result.

Edit: I figured it out. There's an energy splitting due to Fermi statistics. I'll post in the other thread.