# I The symmetry argument and expectation value

#### i_hate_math

In 1D QM:
I understand that if a given potential well, U(x), is symmetric about x = L, then the expectation value for operator [x] would be <x> = L. (I am not even entirely sure why this is, guessing that the region where x<L and x>L are equally probable)

Is it possible to draw conclusion about expectation values of other operators?
Heres one example:

Given ψ(x) = sin(nπx/2L) for a infinite potential well with barriers at x=0 and x=2L,
Using symmetry arguments or otherwise, explain why or show that the expectation value of the particle momentum in this infinite well is <p> = 0

I got <p> = 0 by applying the momentum operator [p]ψ and evaluated it at the point of symmetry x=L, which gave [p]ψ = 0 (at x=L). => <p> = ∫ψ*[p]ψ = 0

Is that correct? Could someone explain to me what really is the "symmetry argument"?

Cheers

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#### blue_leaf77

Homework Helper
I understand that if a given potential well, U(x), is symmetric about x = L, then the expectation value for operator [x] would be <x> = L.
Only if the expectation value is evaluated with respect to a wavefunction with definite parity.
I am not even entirely sure why this is, guessing that the region where x<L and x>L are equally probable
It's easiest to assume a symmetric potential centered around x=0. The action of symmetry (or parity) operator $\Pi$ on the operator x is defined by $\Pi^\dagger x \Pi = -x$. If the potential is symmetric then the Hamiltonian commutes with $\Pi$ and thus the eigenfunctions of the Hamiltonian must also have a definite parity (since the momentum operator also obeys $\Pi^\dagger p \Pi = -p$), that is it must be either symmetric or antisymmetric. With this information you can get two results when trying to calculate the quantity $\langle u_n |\Pi^\dagger x \Pi | u_n\rangle$, where $|u_n\rangle$ is an eigenfunction of the Hamiltonian. One way using the definition of the action of parity operator on x you get
$$\langle u_n |\Pi^\dagger x \Pi | u_n\rangle = -\langle u_n | x | u_n\rangle$$
The other way using the fact that $\Pi |u_n\rangle = \pm |u_n\rangle$ you get
$$\langle u_n |\Pi^\dagger x \Pi | u_n\rangle = \langle u_n | x | u_n\rangle$$
So, in the end you find that $\langle u_n | x | u_n\rangle = -\langle u_n | x | u_n\rangle$ and there is only one solution namely $\langle u_n | x | u_n\rangle =0$.

#### mikeyork

It's a purely classical situation. A symmetric physical situation implies a symmetric distribution of $x$. Given a distribution $f(x)$, the expectation is $<x> = \int^\infty_{-\infty}x f(x)dx$. If $f(x)$ is symmetric about $x=L$, put $y = x-L$ and split the integral about $y=0$. Then $f(x) = g(y) = g(-y)$ and $<y> = \int_0^\infty y g(y)dy - \int_0^\infty (-y)g(-y)d(-y) = 0$ so $<x> = <y> + L = L$.

(The QM identification of $f(x)= |\psi(x)|^2$ adds nothing of significance.)

#### vanhees71

Gold Member
As I understood it, it's about the expectation value of the position for an energy eigenstate. It's clear that you can in the described case you can use a basis with states of good parity, and then $|u_n(x)|^2=u_n(-x)|^2$. So indeed, the assumption in the OP is only true, if you work with parity eigenstates (see #2).

#### i_hate_math

Only if the expectation value is evaluated with respect to a wavefunction with definite parity.

It's easiest to assume a symmetric potential centered around x=0. The action of symmetry (or parity) operator $\Pi$ on the operator x is defined by $\Pi^\dagger x \Pi = -x$. If the potential is symmetric then the Hamiltonian commutes with $\Pi$ and thus the eigenfunctions of the Hamiltonian must also have a definite parity (since the momentum operator also obeys $\Pi^\dagger p \Pi = -p$), that is it must be either symmetric or antisymmetric. With this information you can get two results when trying to calculate the quantity $\langle u_n |\Pi^\dagger x \Pi | u_n\rangle$, where $|u_n\rangle$ is an eigenfunction of the Hamiltonian. One way using the definition of the action of parity operator on x you get
$$\langle u_n |\Pi^\dagger x \Pi | u_n\rangle = -\langle u_n | x | u_n\rangle$$
The other way using the fact that $\Pi |u_n\rangle = \pm |u_n\rangle$ you get
$$\langle u_n |\Pi^\dagger x \Pi | u_n\rangle = \langle u_n | x | u_n\rangle$$
So, in the end you find that $\langle u_n | x | u_n\rangle = -\langle u_n | x | u_n\rangle$ and there is only one solution namely $\langle u_n | x | u_n\rangle =0$.
Thanks very much for your reply. I've not learnt in great details parity and eigenstates, but I get the main idea.

#### i_hate_math

It's a purely classical situation. A symmetric physical situation implies a symmetric distribution of $x$. Given a distribution $f(x)$, the expectation is $<x> = \int^\infty_{-\infty}x f(x)dx$. If $f(x)$ is symmetric about $x=L$, put $y = x-L$ and split the integral about $y=0$. Then $f(x) = g(y) = g(-y)$ and $<y> = \int_0^\infty y g(y)dy - \int_0^\infty (-y)g(-y)d(-y) = 0$ so $<x> = <y> + L = L$.

(The QM identification of $f(x)= |\psi(x)|^2$ adds nothing of significance.)
Thanks!

#### i_hate_math

As I understood it, it's about the expectation value of the position for an energy eigenstate. It's clear that you can in the described case you can use a basis with states of good parity, and then $|u_n(x)|^2=u_n(-x)|^2$. So indeed, the assumption in the OP is only true, if you work with parity eigenstates (see #2).
Thanks!!

"The symmetry argument and expectation value"

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