Anticommutation relations Fermion creation and annihilation

[spaghetti3451]

Homework Statement

This problem is from Lahiri and Pal (2nd edition) Exercise 1.4:

Suppose in a system there are operators which obey anticommutation relations

$[a_{r},a^{\dagger}_{s}]_{+}\equiv a_{r}a^{\dagger}_{s}+a^{\dagger}_{s}a_{r}=\delta_{rs}$

and

$[a_{r},a_{s}]_{+}=0,$ for $r,s=1, \dots, N.$

Construct the generic normalised excited state.

Show that no state can have two or more quanta of the same species.

Homework Equations

The Attempt at a Solution

The state $\lvert 0 \rangle$ is defined by $a_{r}\lvert 0 \rangle = 0.$

For the Hamiltonian, $H = \sum^{N}_{i=1}\hbar \omega (a^{\dagger}_{i}a_{i}+a_{i}a^{\dagger}_{i}) = \sum_{i=1}^{N}\hbar \omega = N \hbar \omega$.

I used the Hamiltonian for the SHO even though the problem did not ask for it, because the section the exercise belongs to is about SHO creation and annihilation operators.

Am I on the right track here?

 

[Orodruin]

No. The problem is only about creating the general excited state, there is no need to involve a Hamiltonian.

 

[spaghetti3451]

Assuming that the ground state is $\lvert 0 \rangle$, do I not have to know if the first excited state of the $r-$mode is obtained by acting $a_{r}$ or $a_{s}^{\dagger}$ on the ground state?

And to know that, do I not have to find out the Hamiltonian of the system and then act the Hamiltonian on $a_{r}\lvert 0 \rangle$ and $a_{r}^{\dagger}\lvert 0 \rangle$?

 

[Orodruin]

No. You assume that annihilation operators destroy the vacuum. Otherwise you can just switch their names.

 

[spaghetti3451]

Altight.

So, then, is the role of the Hamiltonian only to find out the energy levels of the system?

Also, is $a_{r}\lvert - \rangle = 0$ for some given $r$, or $a_{1}a_{2}\dots\ a_{N}\lvert - \rangle = 0$?

 

[Orodruin]

All annihilation operators destroy the vacuum.

 

[spaghetti3451]

Ok, so, I take it that the role of the Hamiltonian only to find out the energy levels of the system.

And also that $a_{r}\lvert - \rangle = 0$ for any given $r$, not just $a_{1}a_{2}\dots\ a_{N}\lvert - \rangle = 0$.
Is the unnormalised $n-$th excited state for a given mode defined by

$\lvert n \rangle = (a^{\dagger})^{n}\lvert 0 \rangle$.

In this case, $\langle 0 \lvert (a)^{3}(a^{\dagger})^{3} \lvert 0 \rangle = \langle 0 \lvert (a)^{2}(a^{\dagger})^{2} \lvert 0 \rangle = 0$,

using the anticoummutation relation $[a,a^{\dagger}]_{+}=1$, so it does not look like my postulated $n-$th excited state is not correct.

 

[Orodruin]

Start with one single creation-annihilation pair. How many states does the space contain?

 

[Orodruin]

Also, you are not asked to order the states.

 

[spaghetti3451]

Start with one single creation-annihilation pair.

Do you mean I need to calculate $\langle 0 \lvert a a^{\dagger} \lvert 0 \rangle = \langle 0 \lvert (1-a^{\dagger}a) \lvert 0 \rangle = \langle 0 \lvert 0 \rangle$?

How many states does the space contain?

I'm not sure how to find that out.

 

[Orodruin]

Do you mean I need to calculate $\langle 0 \lvert a a^{\dagger} \lvert 0 \rangle = \langle 0 \lvert (1-a^{\dagger}a) \lvert 0 \rangle = \langle 0 \lvert 0 \rangle$?
I'm not sure how to find that out.

You start applying creation operators until you cannot apply any more without getting zero.

 

[my2cts]

Try to make a state with two or more identical quanta.

 

[spaghetti3451]

You start applying creation operators until you cannot apply any more without getting zero.

I need to see a few steps of calculation before I understand what you mean.

 

[Orodruin]

I need to see a few steps of calculation before I understand what you mean.

There is only one step of computation...

 

[spaghetti3451]

I give up.

 

[Orodruin]

What is the norm of $(a^\dagger)^2|0\rangle$?

 

[spaghetti3451]

What is the norm of $(a^\dagger)^2|0\rangle$?

Ah! I see!

$|(a^\dagger)^2|0\rangle | = \langle 0 | a (a a^\dagger) a^\dagger | 0 \rangle = \langle 0 | a (1 - a^\dagger a) a^\dagger | 0 \rangle = \langle 0 | a a^\dagger | 0 \rangle - \langle 0 | a a^\dagger (a a^\dagger) | 0 \rangle) = \langle 0 | a a^\dagger | 0 \rangle - \langle 0 | a a^\dagger (1 - a^\dagger a) | 0 \rangle) = \langle 0 | a a^\dagger | 0 \rangle - \langle 0 | a a^\dagger | 0 \rangle + \langle 0 | a a^\dagger a^\dagger a | 0 \rangle) = 0$.

Hmm ... I get the idea.

A similar calculation for the norm of $(a^\dagger)^3|0\rangle$ shows that it is 0.

Although a formal proof will use proof by induction, my intuition tells me all the higher order norms are zero as well.

Therefore, the state cannot have two or more quanta.

Let me try and write up the solution now.

(a) The generic normalised excited state is $a_{i}^{\dagger}a_{j}^{\dagger}\dots a_{r}^{\dagger}a_{s}^{\dagger}\lvert 0 \rangle,$ where only the species from $i$ to $s$ are in the excited state. This is under the assumption that the ground state $\lvert 0 \rangle$ is already normalised, i.e. $\langle 0 \lvert 0 \rangle = 1.$

(b) Take the $i$-th species. The argument can be retraced for all the other species because the creation or annihilation operator for one species commutes with the creation or annihilation operator for another species.

Consider the state with the $n-$th excitation of the $i$-th species, and ignore the excitations of all the other species (because of the commutation rule in the last sentence).

Write $(a_{i}^{\dagger})^{n}(a_{j}^{\dagger})^{m}\dots (a_{r}^{\dagger})^{k}(a_{s}^{\dagger})^{l}\lvert 0 \rangle$ as $(a^{\dagger})^{n}\lvert 0 \rangle$ to simplify the notation in the following calculation.

Now, $|(a^\dagger)^n|0\rangle| = \langle 0 | (a)^{n-1} (a a^\dagger) (a^\dagger)^{n-1} | 0 \rangle = \langle 0 | (a)^{n-1} (1 - a^\dagger a) (a^\dagger)^{n-1} | 0 \rangle = \langle 0 | (a)^{n-1} (a^\dagger)^{n-1} | 0 \rangle - \langle 0 | (a)^{n-1} a^\dagger a (a^\dagger)^{n-1}) | 0 \rangle).$

Therefore, the anticommutation relation allows us to switch the $a$ operator with the $a^{\dagger}$ operator. In this way, the $a$ operator shifts all the way to the right and annihilates the $\lvert 0 \rangle$ state. In the process, we obtain $n$ terms of the form $\langle 0 | (a)^{n-1} (a^\dagger)^{n-1} | 0 \rangle$ in an alternating $+\ -\ +\ -\ \dots$ series.

Now, when $n=1$, the norm is $\langle 0 | 0 \rangle$ = 1.

When $n=2$, the norm is 0 by cancellation of the two identical terms with each other.

When $n=3$, the norm is 0 because the first two terms are identical and cancel with each other and the third term is zero because the norm for $n=2$ is zero.

By induction, the norm for $n>1$ is zero.

N.B. The inductive step is necessary because the proof for odd values of $n$ greater than 1 (e.g. $n=3$) requires the use of the norm for the corresponding smaller even value of $n$ (e.g. $n=2$). For even values of $n$, cancellation alone shows that the norm if zero, but for odd values of $n$ greater than 1, cancellation of the identical terms leaves remianing one positive term which has to be zero because it is the norm for the smaller even value of $n$.Are my solutions correct?

 

[my2cts]

@failexam
Change your nick. Do yourself a favour !
a^†a^†=[a^†,a^†]₊ /2 = 0 .

 

[spaghetti3451]

@failexam
Change your nick. Do yourself a favour !

Why should I change my nick?

a^†a^†=[a^†,a^†]₊ /2 = 0 .

What does this mean?