• pellman
In summary, the conversation discusses a specific step in a problem involving a fermionic oscillator. The problem involves resolving an apparent contradiction and finding the eigenvalues of the number operator N. The conversation explores the effect of the operator a on the eigenstates of N and concludes that, although the number eigenvalue of a|0\rangle may appear to be 1, it is also 0 and all other numbers, making it a trivial case.
pellman
This is a question with regard to a specific step in a problem. I don't think it is necessary to elaborate the whole problem.

## Homework Statement

Resolve this apparent contradiction. I get two different answers for $$Na|n\rangle$$

## Homework Equations

For a fermionic oscillator, we have raising and lowering operators $$a^\dag$$ and $$a$$ that obey

$$\{a,a^\dag\}=aa^\dag+a^\dag a=1$$
$$\{a,a\}=0$$
$$\{a^\dag,a^\dag\}=0$$.

The last two amount to $$a^2=0$$ and $$(a^\dag)^2=0$$.

$$N=a^\dag a$$ is the number operator. It can be shown to have only two eigenvalues: 0 and +1.

## The Attempt at a Solution

The effect of $$a$$ on an eigenstate of N is

* if the eigenvalue is 1, a lowers it to 0, or
* if the eigenvalue is 0, a annihilates it.

Proof: Suppose $$N|n\rangle=n|n\rangle$$.

$$Na|n\rangle = a^\dag a a|n\rangle = a^\dag(a^2|n\rangle)=0$$ since $$a^2=0$$. So either the eigenvalue of $$a|n\rangle$$ is 0 or $$a|n\rangle=0$$.

$$Na|n\rangle=(a^\dag a) a|n\rangle$$
$$=(\{a,a^\dag\}-aa^\dag)a|n\rangle$$
$$=(1\cdot a - a(a^\dag a))|n\rangle$$
$$=a(1-N)|n\rangle$$
$$=(1-n)a|n\rangle$$

So if $$n=0$$, the number eigenvalue of $$a|n\rangle$$ is 1? That can't be right. We get the same relation for the eigenvalue $$a^\dag|n\rangle$$ (which is what we actually expect).

Can anyone see where I went wrong?

Well, if $n=0$, $a\left| n\rangle$ isn't even a state, so you can't talk about its eigenvalue with any sense. This is like an operator version of those "paradoxes" that try to show that 1=2 or something by hiding a division by zero under a bunch of algebra.

Thanks for the quick response.

diazona said:
Well, if $n=0$, $a\left| n\rangle$ isn't even a state, so you can't talk about its eigenvalue with any sense. This is like an operator version of those "paradoxes" that try to show that 1=2 or something by hiding a division by zero under a bunch of algebra.

Isn't the Hilbert space of states necessarily a vector space and have a zero (identity vector). When $$a$$ annihilates $$|0\rangle$$, the result is the zero (state) vector of the Hilbert space, isn't it? We write it as $$a|0\rangle=0$$ as if it were scalar, but it is a state vector, though a trivial one.

Well... OK, that makes sense. But wouldn't the zero state have all numbers as eigenvalues, since
$$A\vert z\rangle = \vert z\rangle$$
for all operators $A$? And (as a special case)
$$n\vert z\rangle = \vert z\rangle$$
for all numbers $n$? Then
$$A\vert z\rangle = n\vert z\rangle$$
for any choice of $A$ and $n$. Which would imply that yes, the number eigenvalue of $a\vert 0\rangle$ is 1, but it is also 0, and 7.5, and $\pi$... so it's kind of meaningless.

I guess the important thing to think about is this: in that last calculation in your original post, none of the individual expressions is inconsistent with the statement that $Na\vert 0\rangle = 0$.

diazona said:
Well... OK, that makes sense. But wouldn't the zero state have all numbers as eigenvalues,
. Yep. Just like a N-vector with all zero components is a eigenvector of any NxN matrix, with eigenvalues consisting of all numbers. It's the trivial case.

I guess the important thing to think about is this: in that last calculation in your original post, none of the individual expressions is inconsistent with the statement that $Na\vert 0\rangle = 0$.

Bingo! I couldn't see the forest for the trees. Thanks!

## 1. What is a Fermionic oscillator?

A Fermionic oscillator is a theoretical model used in quantum mechanics to describe the behavior of fermions, which are particles with half-integer spin. It is similar to a classical harmonic oscillator, but takes into account the special properties of fermions such as the Pauli exclusion principle.

## 2. What is the contradiction associated with Fermionic oscillators?

The contradiction arises when trying to apply the rules of quantum mechanics to models of fermions, as these rules lead to predictions that are inconsistent with experimental observations of fermionic systems.

## 3. How does the Fermionic oscillator contradiction affect our understanding of quantum mechanics?

The contradiction challenges our current understanding of quantum mechanics and highlights the need for further research and refinement of our theories. It also raises questions about the limitations of our current mathematical and conceptual frameworks for describing the behavior of particles at the quantum level.

## 4. Can the Fermionic oscillator contradiction be resolved?

There is ongoing research and debate among scientists to try and resolve the contradiction. Some proposed solutions involve modifying the existing theories of quantum mechanics, while others suggest the need for a completely new framework. However, a definitive resolution has not yet been reached.

## 5. How does the Fermionic oscillator contradiction impact practical applications of quantum mechanics?

The contradiction does not have a direct impact on practical applications of quantum mechanics, as these applications are based on well-established principles and equations that have been extensively tested. However, a better understanding of the contradiction could lead to new insights and advancements in quantum technologies and potential applications in fields such as computing and communication.

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