Proof that (p + ix) operator is non-hermitian (easy)

In summary, to get from the first expression to the second, we can use the fact that i = (-i)* and that p and x are Hermitian operators. By rearranging and taking the complex conjugate, we can see that the operator on the right-hand side is not Hermitian, which explains why the inequality does not hold.
  • #1
halfoflessthan5
16
0
theres one line that keeps coming up in proofs that I don't get. How do i get from

[tex]\int[/tex] ([tex]\hat{p}[/tex][tex]\Psi[/tex]1)*[tex]\Psi[/tex]2 + i [tex]\int[/tex] ([tex]\hat{x}[/tex][tex]\Psi[/tex]1)[tex]\Psi[/tex]2

to

[tex]\int[/tex] ( ([tex]\hat{p}[/tex]-i[tex]\hat{x}[/tex]) [tex]\Psi[/tex]1)*[tex]\Psi[/tex]2

using the fact that p and x are Hermitian.

im sure its painfully simple but i can't see it.
 
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  • #2
halfoflessthan5 said:
theres one line that keeps coming up in proofs that I don't get. How do i get from

[tex]\int[/tex] ([tex]\hat{p}[/tex][tex]\Psi[/tex]1)*[tex]\Psi[/tex]2 + i [tex]\int[/tex] ([tex]\hat{x}[/tex][tex]\Psi[/tex]1)[tex]\Psi[/tex]2
I guess you mean that there is a complex conjugate on (x Psi_1) in the second term)
to

[tex]\int[/tex] ( ([tex]\hat{p}[/tex]-i[tex]\hat{x}[/tex]) [tex]\Psi[/tex]1)*[tex]\Psi[/tex]2

using the fact that p and x are Hermitian.

im sure its painfully simple but i can't see it.

Well, for that step you actually do not need at all to use the fact that x and p are hermitian.

All you need to use is that i = (-i)*
 
  • #3
Why the * in the first integral, but not the second? The second integral only makes sense in QM if the wave function is real, when, in fact, most wave functions are complex.

Anyway, both p and x are Hermitian. Thus (P+iX)* =(P-iX), virtually by definition, as i is anti-Hertmitian.
Regards,
Reilly Atkinson
 
  • #4
okay got it.

I put the operator into the RHS of the hermitian condition, took the complex conjugate and re-arranged it so that it was in the same form as the LHS of the hermitian condition. the inequality obviously doesn't hold because the 'i' put a minus in one, so the operator wasn't hermitian. thankyou so much guys. and sorry for the typo
 

1. What is a non-hermitian operator?

A non-hermitian operator is an operator that does not satisfy the conditions of a Hermitian operator, which include being self-adjoint and having real eigenvalues. This means that the operator does not have the property of symmetry and its eigenvalues can be complex numbers.

2. How do you prove that (p + ix) operator is non-hermitian?

To prove that (p + ix) operator is non-hermitian, we can use the definition of a Hermitian operator, which states that the operator must equal its adjoint. If we take the adjoint of (p + ix) operator, we get (p - ix), which is not equal to the original operator. This difference in the operator and its adjoint proves that (p + ix) is non-hermitian.

3. What is the significance of the (p + ix) operator being non-hermitian?

The (p + ix) operator being non-hermitian has several implications in quantum mechanics and other areas of physics. It means that the operator does not have real eigenvalues, which can affect the solutions to quantum mechanical equations and the physical interpretations of the results. It also has implications in the study of non-conservative systems and time evolution.

4. Can a non-hermitian operator still have real eigenvalues?

No, a non-hermitian operator cannot have real eigenvalues. This is because the definition of a Hermitian operator requires the operator to have real eigenvalues. If the operator does not satisfy this condition, it is considered non-hermitian.

5. How does the (p + ix) operator relate to complex numbers?

The (p + ix) operator relates to complex numbers in that it is a combination of a real number (p) and an imaginary number (ix). This means that the operator does not have real eigenvalues and has implications for the solutions to equations involving complex numbers. It also has connections to the study of non-conservative systems and time evolution in physics.

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