Parity of function of multiple variables

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 6K views
wuhtzu
Messages
9
Reaction score
0
Hi everyone

I was wondering how to determine the parity of a function of multiple variables.

Say the function is:

[tex]f(x,z) = xz[/tex]

How would I determine its parity?

For the above function it is true that #1 f(x,z) = f(-x,-z) but its also true that #2 [tex]\int_{-a}^{a}\int_{-a}^{a}xz \dx \dz = 0[/tex].

In one variable functions #1 would indicate an even function and #2 would indicate an odd function. So I guess you cannot directly extend the concept of odd and even function from one variable to multiple variables?

The ultimate need to answer such question is to determine the parity of potentials in quantum mechanics :)

Best regards
Wuhtzu
 
Mathematics news on Phys.org
That is correct, but it doesn't remove the need to determine the parity of potentials suchs as

V(x) = E*x
V(x,z) = c*x*z
V(z) = B*L

in quantum mechanics :( Most of them is easy because they are only dependent on one direction ie. one variable, but some, like v(x,z) = c*x*z, is dependent on more...
 
Parity (math term) is well defined for functions of one variable. I have never seen a definition for functions of more than one variable, other than for each variable separately.
 
I will post what ever answer we get when our quantum mechanics professor gets back from vacation.

Thank you for looking
 
So for this problem it turned out to be sufficient to just change sign of all variables, making it an even function.


The original problem if anyone is interested: Determine which matrix elements of the form

<n m l | V | n' m' l'> , n = n' = 2, m-m' = ? and l-l' = ?

was zero for a spherically symmetric quantum mechanical system by using selection rules.

The potential V was of the from V = x*z. The eigenstates of the spherically symmetric system has alternating even and odd parity with increasing l quantum number and the potential V = x*z is a rank k = 2 tensor.

This imposes the restriction [tex]l-l' = \Delta \le 2 \le 2[/tex].

Getting restrictions on m-m' is another story.

Thanks again.

Furthermore for [tex]<n m l | V | n' m' l'> \ne 0[/tex] the product of the wavefunctions <n m l | and |n' l' m'>, and the potential V has to be an even function - if it is odd it integrates to zero. Since we know that V is even both of the two wavefunctions (<n m l | or |n' m' l'>) have to have the same parity. Imposes futher restrictions on l-l' : [tex]\Delat l=0 , 2[/tex] since placing the l states 1 apart will cause one of them to be even and one of them to be odd (remember the alternating parity). But choosing l-'l = 0 or l-l'=2 causes both to be of the same parity which makes the total product even and hence the intergral non-zero.