# Prove that the exchange operator is Hermitian

• Sofie RK
In summary, the exchange operator can be proved to be hermitian by showing that the integral expressions <φ|Pψ> and <Pφ|ψ> are equivalent. This is because the specific labels used for the integration variables do not affect the numerical value of the integral. Therefore, the exchange operator is hermitian if the condition <φ|Pψ> = <Pφ|ψ> is satisfied.
Sofie RK

## Homework Statement

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Let P be the exchange operator:

Pψ(1,2) = ψ(2,1)

How can I prove that the exchange operator is hermitian?

I want to prove that <φ|Pψ> = <Pφ|ψ>

## Homework Equations

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<φ|Pψ> = <Pφ|ψ> must be true if the operator is hermitian.

## The Attempt at a Solution

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<φ(1,2) | P ψ(1,2) > = ∫φ*(1,2) P ψ(1,2) dτ = ∫φ*(1,2) ψ(2,1) dτ

<P φ(1,2) | ψ(1,2)> = ∫P* φ*(1,2) ψ(1,2) dτ = ∫φ*(2,1) ψ(1,2) dτ

DrClaude said:
You're almost there. Just remember that inside the integrals, the labels 1 and 2 are simply dummy integration arguments.

If labels 1 and 2 are integration arguments, can I write it as ##x_1## and ##x_2##? Which means that ##x_1## and ##x_2## are integration variables? And I can write the integrals as

$$\int{\phi*(x_1,x_2)\psi(x_2,x_1) dx_1dx_2}$$

and

$$\int{\phi*(x_2,x_1)\psi(x_1,x_2) dx_1dx_2}$$

But, I still don't understand why these integrals are the same. How can I use these integration arguments to prove that?

Sofie RK said:
If labels 1 and 2 are integration arguments, can I write it as ##x_1## and ##x_2##? Which means that ##x_1## and ##x_2## are integration variables? And I can write the integrals as

$$\int{\phi*(x_1,x_2)\psi(x_2,x_1) dx_1dx_2}$$

and

$$\int{\phi*(x_2,x_1)\psi(x_1,x_2) dx_1dx_2}$$

But, I still don't understand why these integrals are the same. How can I use these integration arguments to prove that?

Note that these integrals are numbers, not functions, and that ##x_1, x_2## are dummy variables that can be replaced with any other dummy variables. For example:

In your first integral, set ##x_1 = X## and ##x_2 = Y##.

In your second integral, set ##x_1 = Y## and ##x_2 = X##.

DrClaude
PeroK said:
Note that these integrals are numbers, not functions, and that ##x_1, x_2## are dummy variables that can be replaced with any other dummy variables. For example:

In your first integral, set ##x_1 = X## and ##x_2 = Y##.

In your second integral, set ##x_1 = Y## and ##x_2 = X##.

Okey, then the integrals will be the same. But if ##x_1## refers to particle 1 and ##x_2## refers to particle 2, why can I choose that ##x_1 = X## in the first integral, ##x_1 = Y## in the second integral, and then compare the integrals when the variables don't refer to the same particle anymore?

Sofie RK said:
Okey, then the integrals will be the same. But if ##x_1## refers to particle 1 and ##x_2## refers to particle 2, why can I choose that ##x_1 = X## in the first integral, ##x_1 = Y## in the second integral, and then compare the integrals when the variables don't refer to the same particle anymore?

The definite integrals are just complex numbers. Whatever the origin of the dummy variables, you are looking for numerical equality of two complex numbers.

If you had two particles, say, and some number associated with each of them which you expressed as a definite integral. And you got, perhaps after some calculations:

##n_1 = \int_0^1 f(x_1)dx_1##

and

##n_2 = \int_0^1 f(x_2)dx_2##

Then, you might ponder over whether ##n_1 = n_2##? But, we clearly have:

##\int_0^1 f(x_1)dx_1 = \int_0^1 f(x_2)dx_2 = \int_0^1 f(X)dX##

The value of the definite integral depends only on the function, not on the integration variable.

If you are still not convinced, you might imagine that you know ##f##. Let's say ##f(x) = x##. Then:

##n_1 = \int_0^1 x_1 dx_1 = 1/2##

and

##n_2 = \int_0^1 x_2 dx_2 = 1/2##

And, ##1/2## is ##1/2##.

@Sofie RK

Alternatively, try to find two functions ##f## and ##g## where:

##\int_0^1 \int_0^1 f(x_1, x_2)g(x_2,x_1) dx_1 dx_2 \ne \int_0^1 \int_0^1 f(x_2, x_1)g(x_1,x_2) dx_1 dx_2##

If you go through the process of evaluating the integrals, you'll see that the choice of variables themselves can't make any difference to the number that comes out.

PeroK said:
The definite integrals are just complex numbers. Whatever the origin of the dummy variables, you are looking for numerical equality of two complex numbers.

If you had two particles, say, and some number associated with each of them which you expressed as a definite integral. And you got, perhaps after some calculations:

##n_1 = \int_0^1 f(x_1)dx_1##

and

##n_2 = \int_0^1 f(x_2)dx_2##

Then, you might ponder over whether ##n_1 = n_2##? But, we clearly have:

##\int_0^1 f(x_1)dx_1 = \int_0^1 f(x_2)dx_2 = \int_0^1 f(X)dX##

The value of the definite integral depends only on the function, not on the integration variable.

If you are still not convinced, you might imagine that you know ##f##. Let's say ##f(x) = x##. Then:

##n_1 = \int_0^1 x_1 dx_1 = 1/2##

and

##n_2 = \int_0^1 x_2 dx_2 = 1/2##

And, ##1/2## is ##1/2##.

Thank you! This really helped!

## What is the exchange operator?

The exchange operator is a mathematical operator that switches the positions of two particles in a quantum system. It is commonly denoted as "P" and is used in quantum mechanics to describe the behavior of identical particles.

## What does it mean for an operator to be Hermitian?

An operator is said to be Hermitian if it is equal to its own adjoint, or conjugate transpose. In other words, if the operator is applied to a vector and then multiplied by its complex conjugate, the result will be equal to the original vector.

## Why is it important to prove that the exchange operator is Hermitian?

Proving that the exchange operator is Hermitian is important because it ensures that the operator is a valid physical observable in quantum mechanics. Hermitian operators have real eigenvalues, which correspond to measurable quantities in experiments.

## How do you prove that the exchange operator is Hermitian?

To prove that the exchange operator is Hermitian, we must show that it is equal to its own adjoint. This can be done by applying the operator to a general state vector and then taking its complex conjugate. If the result is equal to the original vector, the operator is Hermitian.

## What are some applications of the exchange operator in quantum mechanics?

The exchange operator is used in many different areas of quantum mechanics, such as in the study of identical particles, quantum statistics, and quantum chemistry. It is also an important concept in the development of quantum algorithms for quantum computing. In addition, the exchange operator plays a key role in understanding the behavior of fermions and bosons in quantum systems.

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