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# Time-independent Schrodinger equation

| 🗖 nosplit | ↑ parent "Schrödinger equation" | words: 416
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As shown at https://quantummechanics.ucsd.edu/ph130a/130_notes/node124.html from quantum physics by Jim Branson (2003), using separation of variables we can break up the general Schrödinger equation into:
• a time-only part that does not depend on space and does not depend on the potential
• a space-only part that does not depend on time:
$$Equation 30. Time-independent Schrodinger equation H^[ψ(x∣)]=Eψx∣ (30)$$
Since this is the only non-trivial part, unlike the time part which is trivial, this spacial part is just called "the time-independent Schrodinger equation".
Note that the here is not the same as the in the time-dependent Schrodinger equation of course, as that psi is the result of the multiplication of the time and space parts. This is a bit of imprecise terminology, but hey, physics.
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It is interesting to note that the time-independent Schrodinger equation can also be seen exactly as an eigenvalue equation, since the value of the energy E can be any fixed value that can satisfy the boundary conditions conditions imposed. The only difference from usual matrix eigenvectors is that we are now dealing with an infinite dimensional vector space.
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Then, just like in solving partial differential equations with the Fourier series of the heat equation, the boundary conditions may make it so that only certain discrete values of E are possible solutions.
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Then, the time part o the equation can be solved explicitly in the general case, as it does not depend on the potential , and it is just an exponential.
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Therefore, all we need to do to solve the general Schrodinger equation is to solve the time-independent version, and then decompose the initial condition in terms of it like as done in Section "Solving partial differential equations with the Fourier series".
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Furthermore:
• we immediately see from the equation that the time-independent solutions are states of deterministic energy because the energy is an eigenvalue of the Hamiltonian operator
• the probability density of such a state does not change with time because the exponential time part cancels out on the conjugate
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