# Linear combination of states with Pauli's principle

• I
• Salmone
In summary: Ok I understand you. But the state ##|\uparrow \downarrow\rangle## doesn't have a defined symmetry under exchange of the two particles, what I mean was: if the state ##|\psi\rangle## is a superposition of states with a well-defined symmetry (symmetric or antisymmetric), if we want ##|\psi\rangle## to be antisymmetric we need that all the states of the superposition are antisymmetric. I can't imagine a counter-example. To be clearer on what I mean: ##|\psi\rangle=|\alpha_1^s\rangle+|\alpha_2^s
Salmone
If I have two identical particles of ##1/2## spin, for Pauli's exclusion principle all physical states must be antysimmetrical under the exchange of the two particles, so ##\hat{\Pi}|\alpha\rangle=-|\alpha\rangle##. Now, let's say for example this state ##\alpha## is an Hamiltonian eigenfunction and we have a superposition of different eigenstates, what about the antysimmetric condition? I thought this applies: since the exchange operator ##\Pi## is linear and all the states must be antysimmetric, even a states which is a superposition of eigenstates need to be antysimmetric so it must be a superposition of antysimmetric eigenstates. Example: ##|\psi\rangle=\frac{1}{\sqrt{2}}|\alpha_1\rangle+\frac{1}{\sqrt{2}}|\alpha_2\rangle## I apply the exchange operator: ##\Pi|\psi\rangle=\Pi(\frac{1}{\sqrt{2}}|\alpha_1\rangle+\frac{1}{\sqrt{2}}|\alpha_2\rangle)=\Pi\frac{1}{\sqrt{2}}|\alpha_1\rangle+\Pi\frac{1}{\sqrt{2}}|\alpha_2\rangle=-\frac{1}{\sqrt{2}}|\alpha_1\rangle-\frac{1}{\sqrt{2}}|\alpha_2\rangle=-(\frac{1}{\sqrt{2}}|\alpha_1\rangle+\frac{1}{\sqrt{2}}|\alpha_2\rangle)=-|\psi\rangle## and so it is antysimmetric. Is this right?

More generally, if two states are eigenstates of a given operator with the same eigenvalue then any superposition of those states is also an eigenstate of the operator with the same eigenvalue. In other words, eigenspaces of a linear operator are sub-spaces (i.e. closed under linear vector operations).

PeroK said:
More generally, if two states are eigenstates of a given operator with the same eigenvalue then any superposition of those states is also an eigenstate of the operator with the same eigenvalue. In other words, eigenspaces of a linear operator are sub-spaces (i.e. closed under linear vector operations).
Sorry I don't understand, what do you mean with this? And in general, I was referring to superposition even of states that are not eigenstates with the same eigenvalue.

Salmone said:
Sorry I don't understand, what do you mean with this?
It's standard QM terminology. What don't you understand?
Salmone said:
And in general, I was referring to superposition even of states that are not eigenstates with the same eigenvalue.
All the states you use appear to be antisymmetric (i.e. eigenstates of the exchange operator with eigenvalue ##-1##).

Salmone
PeroK said:
It's standard QM terminology. What don't you understand?

All the states you use appear to be antisymmetric (i.e. eigenstates of the exchange operator with eigenvalue ##-1##).
Ok, so what I wrote is correct? And in general, what I am supposed to do when dealing with a superposition is to check that every state of the superposition is antisymmetric?

Salmone said:
Ok, so what I wrote is correct?
Yes.
Salmone said:
And in general, what I am supposed to do when dealing with a superposition is to check that every state of the superposition is antisymmetric?
That doesn't follow. In general, any state can be expressed in any basis. In particular, an antisymmetric state can be expressed in any basis and in a superposition of non-antisysmmetric states.

It's the whole state you need to check.

That said, if every state in the superposition is antisymmetric, then the state itself must be. That's the argument in posts #1 and #2.

Salmone
PeroK said:
Yes.

That doesn't follow. In general, any state can be expressed in any basis. In particular, an antisymmetric state can be expressed in any basis and in a superposition of non-antisysmmetric states.

It's the whole state you need to check.

That said, if every state in the superposition is antisymmetric, then the state itself must be. That's the argument in posts #1 and #2.
Ok I understand. I thought it was impossible to express an antisymmetric state as a superposition of non-antisymmetric states. Lastly, do you confirm that the exchange operator is linear?

Salmone said:
Ok I understand. I thought it was impossible to express an antisymmetric state as a superposition of non-antisymmetric states.
For example, we have the anti-symmetric singlet state:$$\ket {00} = \frac 1 {\sqrt 2}\big (\ket {\uparrow \downarrow} - \ket {\downarrow \uparrow} \big )$$In general, the properties of a vector do not apply to the basis vectors. They cannot, as the basis vectors span the space and cannot have every property themselves.
Salmone said:
Lastly, do you confirm that the exchange operator is linear?
You should be able to prove that.

In general, "operator" means "linear operator" in QM, unless otherwise stated!

Salmone
PeroK said:
For example, we have the anti-symmetric singlet state:$$\ket {00} = \frac 1 {\sqrt 2}\big (\ket {\uparrow \downarrow} - \ket {\downarrow \uparrow} \big )$$In general, the properties of a vector do not apply to the basis vectors. They cannot, as the basis vectors span the space and cannot have every property themselves.

You should be able to prove that.

In general, "operator" means "linear operator" in QM, unless otherwise stated!
Ok I understand you. But the state ##|\uparrow \downarrow\rangle## doesn't have a defined symmetry under exchange of the two particles, what I mean was: if the state ##|\psi\rangle## is a superposition of states with a well-defined symmetry (symmetric or antisymmetric), if we want ##|\psi\rangle## to be antisymmetric we need that all the states of the superposition are antisymmetric. I can't imagine a counter-example. To be clearer on what I mean: ##|\psi\rangle=|\alpha_1^s\rangle+|\alpha_2^s\rangle+|\alpha_3^a\rangle## with ##a,s## indicating respectively antisymmetric and symmetric states, ##|\psi\rangle## cannot be antisymmetric. I don't know if this is right.

Salmone said:
Ok I understand you. But the state ##|\uparrow \downarrow\rangle## doesn't have a defined symmetry under exchange of the two particles, what I mean was: if the state ##|\psi\rangle## is a superposition of states with a well-defined symmetry (symmetric or antisymmetric), if we want ##|\psi\rangle## to be antisymmetric we need that all the states of the superposition are antisymmetric. I can't imagine a counter-example. To be clearer on what I mean: ##|\psi\rangle=|\alpha_1^s\rangle+|\alpha_2^s\rangle+|\alpha_3^a\rangle## with ##a,s## indicating respectively antisymmetric and symmetric states, ##|\psi\rangle## cannot be antisymmetric. I don't know if this is right.
Well, a superposition of symmetric states must be symmetric. But, most states are neither symmetric nor antisymmetric.

You're leading yourself astray somewhat by these arguments. If we have:
$$\ket \psi = \ket {\psi_1} + \ket {\psi_2}$$Then, in general, the properties of ##\ket \psi## are not the same as the properties of ##\ket {\psi_1}## and ##\ket {\psi_2}##.

Insisting that if ##\ket \psi## is (anti-)symmetric, then both ##\ket{\psi_1}## and ##\ket{\psi_2}## are (anti-)symmetric is very wrong.

protonsarecool and Salmone

## 1. What is a linear combination of states?

A linear combination of states is a mathematical operation that combines multiple quantum states to create a new state. This operation is commonly used in quantum mechanics to describe the behavior of particles and their interactions.

## 2. What is Pauli's principle?

Pauli's principle, also known as the Pauli exclusion principle, states that no two identical fermions (particles with half-integer spin) can occupy the same quantum state simultaneously. This principle is a fundamental concept in quantum mechanics and explains many properties of matter.

## 3. How does Pauli's principle relate to linear combinations of states?

Pauli's principle is often used in conjunction with linear combinations of states to describe the behavior of multiple fermions in a system. The principle ensures that the resulting state is antisymmetric, meaning that it is unchanged when two particles are swapped. This is necessary to accurately describe fermions, which cannot occupy the same state.

## 4. Can linear combinations of states violate Pauli's principle?

No, linear combinations of states must adhere to Pauli's principle in order to accurately describe the behavior of fermions. If the resulting state does not satisfy the principle, it is not a valid solution and must be revised.

## 5. How are linear combinations of states with Pauli's principle used in practical applications?

Linear combinations of states with Pauli's principle are used in many areas of quantum mechanics, including quantum chemistry, solid-state physics, and nuclear physics. They are also used in the development of quantum algorithms and in quantum information processing. Understanding and utilizing these concepts is crucial for many modern technologies, such as transistors, lasers, and MRI machines.

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