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We can take the complex conjugate of schrodinger equation, and obtain $$ -\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = i \hbar \frac{\partial \psi}{\partial t} $$

$$ -\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* = -i \hbar \frac{\partial \psi^*}{\partial t} $$

this seems natural to me, however, does it indicate also the following matrix form is valid?(replace $E$ with $i\hbar d/dt$, the second being a little uncomfortable)

$$ H\phi=E\phi $$ $$ H\phi^*=-E\phi^* $$

Suppose we have a Hamiltonian $H$ in matrix form, and solve for the eigenvalue problem, then how are we supposed to know which is "$E$", which is "$-E$".

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  • $\begingroup$ If the energy eigenvalue is real, then why should there be a minus sign from complex conjugating the TISE? (eqn 4) $\endgroup$ – theage Dec 10 '14 at 10:29
  • $\begingroup$ @theage it's obtained from TDSE(eq.2) $\endgroup$ – Lorniper Dec 10 '14 at 10:37
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The origin of the eigenvalue equation $H\phi=E\phi$ is the separation ansatz $$\psi(x,t)=\exp{\left(-i\frac{E}{\hbar}t\right)}\phi(x)$$ If you conjugate this, this will obviously change the sign of the exponent and therefore you will the same eigenvalue.

What you are trying to state would be something like "if $\lambda$ is an eigenvalue of $H$, so is $-\lambda$, which is obviously not true. Suppose you have an eigenfunction $\phi$, then $\phi^*$ is an eigenfunction to the same eigenvalue (not the negative) due to self-adjointness of the Hamiltonian.

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  • $\begingroup$ I thought E can be replaced by $ ihd/dt$ $\endgroup$ – Lorniper Dec 10 '14 at 10:58
  • $\begingroup$ @Lorniper not when you have just reversed the time direction (by applying complex conjugation). $\endgroup$ – Ruslan Dec 10 '14 at 11:03
  • $\begingroup$ @Lorniper This trivially only holds if you use an ansatz of the above kind. If I say $\psi(x,t)=\exp{\left(+i\frac{E}{\hbar}t\right)}\phi(x)$, which you would say if you conjugate your stuff, this obviously does not hold. $\endgroup$ – Daniel Dec 10 '14 at 11:03

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