derive hψ=Eψ where h is hameltonian and E is energy and psi is function
Answers
Explanation:
The energy operator is easily derived from using the free particle wave function (plane wave solution to Schrödinger's equation).[2] Starting in one dimension the wave function is
{\displaystyle \Psi =e^{i(kx-\omega t)}\,\!} \Psi =e^{{i(kx-\omega t)}}\,\!
The time derivative of Ψ is
{\displaystyle {\frac {\partial \Psi }{\partial t}}=-i\omega e^{i(kx-\omega t)}=-i\omega \Psi \,\!} {\frac {\partial \Psi }{\partial t}}=-i\omega e^{{i(kx-\omega t)}}=-i\omega \Psi \,\!.
By the De Broglie relation:
{\displaystyle E=\hbar \omega \,\!} E=\hbar \omega \,\!,
we have
{\displaystyle {\frac {\partial \Psi }{\partial t}}=-i{\frac {E}{\hbar }}\Psi \,\!} {\frac {\partial \Psi }{\partial t}}=-i{\frac {E}{\hbar }}\Psi \,\!.
Re-arranging the equation leads to
{\displaystyle E\Psi =i\hbar {\frac {\partial \Psi }{\partial t}}\,\!} E\Psi =i\hbar {\frac {\partial \Psi }{\partial t}}\,\!,
where the energy factor E is a scalar value, the energy the particle has and the value that is measured. The partial derivative is a linear operator so this expression is the operator for energy:
{\displaystyle {\hat {E}}=i\hbar {\frac {\partial }{\partial t}}\,\!} {\hat {E}}=i\hbar {\frac {\partial }{\partial t}}\,\!.
It can be concluded that the scalar E is the eigenvalue of the operator, while {\displaystyle {\hat {E}}\,\!} {\hat {E}}\,\! is the operator. Summarizing these results:
{\displaystyle {\hat {E}}\Psi =i\hbar {\frac {\partial }{\partial t}}\Psi =E\Psi \,\!} {\hat {E}}\Psi =i\hbar {\frac {\partial }{\partial t}}\Psi =E\Psi \,\!
For a 3-d plane wave
{\displaystyle \Psi =e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\,\!} {\displaystyle \Psi =e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\,\!}
the derivation is exactly identical, as no change is made to the term including time and therefore the time derivative. Since the operator is linear, they are valid for any linear combination of plane waves, and so they can act on any wave function without affecting the properties of the wave function or operators. Hence this must be true for any wave function. It turns out to work even in relativistic quantum mechanics, such as the Klein–Gordon equation above.