- #1
vact
- 2
- 0
Hi, there. It should be yes, but I'm very confused now.
Consider a simple one-dimensional system with only one particle with mass of m. Let the potential field be 0, that's V(r) = 0. So the Hamiltonian operator of this system is:
H = -hbar^2/(2m) * d^2/dx^2
[tex]
\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}
[/tex]
To obtain the eigenvalue and eigenfunction of H, one should solve the following equation:
H f = E * f
[tex]
\hat{H}f = Ef
[/tex]
That leads:
d^2f/dx^2 + 2mE/hbar^2 * f = 0
[tex]
\frac{d^2f}{dx^2} + \frac{2mE}{\hbar^2}f = 0
[/tex]
However, there is always a solution f for any value of E, even imaginary number, such as E = i, E = 1+i... But a Hermitian operator can never have an imaginary eigenvalue.
I don't know what's wrong. Can you help me? Thank you!
Consider a simple one-dimensional system with only one particle with mass of m. Let the potential field be 0, that's V(r) = 0. So the Hamiltonian operator of this system is:
H = -hbar^2/(2m) * d^2/dx^2
[tex]
\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}
[/tex]
To obtain the eigenvalue and eigenfunction of H, one should solve the following equation:
H f = E * f
[tex]
\hat{H}f = Ef
[/tex]
That leads:
d^2f/dx^2 + 2mE/hbar^2 * f = 0
[tex]
\frac{d^2f}{dx^2} + \frac{2mE}{\hbar^2}f = 0
[/tex]
However, there is always a solution f for any value of E, even imaginary number, such as E = i, E = 1+i... But a Hermitian operator can never have an imaginary eigenvalue.
I don't know what's wrong. Can you help me? Thank you!
Last edited: