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The one-dimension Schrodinger's equation(spatial component of the S.E.) is

$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ E-U\left ( x \right ) \right ]\psi=0$$

where E is the sum of the kinetic and potential energy of the system

If the potential energy U(x)=0, the equation reduces to

$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ E^{*} \right ]\psi=0 $$

where

$$E^{*}=\frac{1}{2}mv^{2}$$

But in the case where the potential energy of the system is non-zero

$$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ E-U_{0}\left ( x\right ) \right ]\psi=0 $$

What can I say about E,specifically?

Is $$E=\frac{1}{2}mv^{2}+U_{k}\left ( x \right )$$ such that $$U_{k}=U_{0}$$ necessarily?

Asked because if $$U_{k}=U_{0}$$ then $$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ \left ( \frac{mv^{2}}{2}+U_{k} \right )-U_{0}\left ( x \right ) \right ]\psi=0$$ reduces to $$\frac{d^{2}\psi}{dx^{2}}+\frac{8 \pi^{2}m}{h^{2}}\left [ \left ( \frac{mv^{2}}{2} \right ) \right ]\psi=0$$

which says that potential energies can never exists in the spatial component of the wave function

Clarification would help

Edit: $$\psi=\psi\left ( x\right )$$ to be clear

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  • $\begingroup$ Why is this down voted? $\endgroup$
    – Physkid
    Dec 11, 2015 at 14:26
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    $\begingroup$ Let me guess about the down vote. This site was originally designed for professionals to ask research-related questions. There already were several forums for education-related questions. But a forum is open by definition, and questions that do not meet the original purpose are asked. Regular contributors don't agree on how to deal with that. The down vote may have come from someone who feels that the question is too far out of bounds for us. The expression $E*=\frac{1}{2}mv^2$ indicates a shaky understanding of quantum mechanics. The down voter may think you need to do more homework. $\endgroup$
    – garyp
    Dec 11, 2015 at 14:42
  • $\begingroup$ @garyp Even in the non-relativistic case? Because this is what my book uses for a non-relativistic case. $\endgroup$
    – Physkid
    Dec 11, 2015 at 15:04
  • $\begingroup$ Your book is not a quantum mechanics book! $\endgroup$
    – garyp
    Dec 11, 2015 at 16:55
  • $\begingroup$ @garyp Do you recommend QM by David Griffin? $\endgroup$
    – Physkid
    Dec 12, 2015 at 14:55

2 Answers 2

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Forget about $E=\frac{mv^2}{2}$ for a minute because in QM there's no Classical concept of speed ($v$). Potential and kinetic energy are nonetheless used in QM too.

In QM, the values of $E$ are the eigenvalues of the time-independent Schrödinger equation and $\psi$ a set of wave functions that are the eigenfunctions of that equation.

The simplest example of a quantum system with bound states is the particle in a 1D infinite well and you find its solutions of the SE here. If you study that system (follow the links!) you'll start to understand how that SE works, before attempting to explore more complicated Q systems.

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  • $\begingroup$ Helpful that you mentioned eigenvalues and eigenfunction. Have dealt with diffusion problems so I might try to see what this eigenvalues and eigenfunction tells me in a quantum system. $\endgroup$
    – Physkid
    Dec 11, 2015 at 16:04
  • $\begingroup$ @Physkid: the math of the SE is remarkably similar to that of other physical systems like Classic waves, vibrating strings, diffusion and even the use of Fourier's Law (in some cases). Just try to abandon Classical concepts of position and speed for now. Things will then become clearer as you go on. $\endgroup$
    – Gert
    Dec 11, 2015 at 16:10
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You seem to be confusing different things. There is no such thing as "$E = \frac{1}{2} m v^2 + E_p$" in quantum mechanics, this is a classical interpretation of the energy. In the Schrödinger representation, $E$ represents some eigenvalue of the Hamiltonian operator, corresponding to an eigenstate of the system of energy $E$. Kinetic and potential energy are not numbers but operators, in fact the equation you wrote is just the eigenvalue-equation of the Hamiltonian poerator in 1-D cartesian coordinates : $(\hat{E_c} + \hat{E_p}) |\Psi> = E |\Psi> $, with $\hat{E_c} = \frac{(\frac{\hbar}{i} \vec{\nabla})^2}{2m}$ and $\hat{E_p} = U_0(x)$.

To find the energy, you need to solve this equation with a given form for $U_0(x)$, and boundary conditions will give you the possible values of the energy $E$.

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  • $\begingroup$ Is E the linear energy operator? My book uses E instead of E-hat so it confuses me. Currently doing advance studies in preparation for my final year. The stuffs you posted does looked rather alien. $\endgroup$
    – Physkid
    Dec 11, 2015 at 14:51
  • $\begingroup$ $E$ is a number, it's not an operator. It is one of the eigenvalues of the Hamiltonian operators. Usually in undergrad quantum mechanics textbooks, operators are written with a hat. $\endgroup$
    – Dimitri
    Dec 11, 2015 at 14:54
  • $\begingroup$ My text mentioned also that E is the mechanical energy(combination of the potential plus kinetic energy) in the non-relativistic case and in fact uses $$E=K=(1/2)mv^{2}$$ I would agree with you that in the relativistic case, there is no half mv squared. The text is "fundamental of physics" authored by Jearl walker. $\endgroup$
    – Physkid
    Dec 11, 2015 at 15:01
  • $\begingroup$ The context there is classical mechanics. It does not apply to quantum mechanics, at least not without a lot of work. $\endgroup$
    – garyp
    Dec 11, 2015 at 16:54

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