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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!

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