Prove that the exchange operator is Hermitian

[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]

<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

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

See also https://www.physicsforums.com/threads/hermiticity-of-permutation-operator.520787/

 

[Sofie RK]

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

See also https://www.physicsforums.com/threads/hermiticity-of-permutation-operator.520787/

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?

 

[PeroK]

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 $x_2 = X$.

Now, compare your integrals.

 

[Sofie RK]

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 $x_2 = X$.

Now, compare your integrals.

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?

 

[PeroK]

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 $1/2$.

 

[PeroK]

@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.

 

[Sofie RK]

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!