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The Time Independent Schrodinger Equation is as follows: $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi(x)=E\cdot\psi(x)$$

Due to the fact that $E=E_{kinetic}+E_{potential}$ and the fact that $V$ is the potential energy, the equation could be simplified to $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=(E-V(x))\cdot\psi(x)=\frac{1}{2}mv^2\cdot\psi(x)$$ which can be rearranged to get $$-\frac{\hbar^2}{p^2}\frac{d^2\psi}{dx^2} = \psi(x)$$ After applying the De Brogile relationship, we have that $$-\left(\frac{\lambda}{2\pi}\right)^2\frac{d^2\psi}{dx^2}=\psi(x)$$ which can be simplified to $$\frac{d^2\psi}{dx^2} + k^2\psi(x) = 0$$

This is the differential equation for a system with zero potential energy, and it would be absurd to think that all systems behave as such. As such, what is the mistake in this simplification of the Schrodinger Equation, or if there is none, then is there any confirmation in the literature of this form?

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    $\begingroup$ The last equation is equivalent to TISE if $k:=\sqrt{2m(E-V)}/\hbar$ is a function of $x$. $\endgroup$ – Qmechanic Oct 1 '17 at 4:38
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    $\begingroup$ What is $v$ in your second line? It better not be velocity; wavefunctions don't have a well defined velocity. $\endgroup$ – Jahan Claes Oct 1 '17 at 4:43
  • $\begingroup$ @Jahan How about lets say that it is momentum divided by mass $\endgroup$ – Teoc Oct 1 '17 at 4:47
  • $\begingroup$ Then it should be written as $(-\frac{i\hbar}{m}\frac{\partial}{\partial x})$ instead of $v$, right? Because a wavefunction doesn't have a definite momentum either! $\endgroup$ – Jahan Claes Oct 1 '17 at 4:50
  • $\begingroup$ @Qmechanic $k=2\pi/\lambda = 2\pi\frac{\nu}{c}=2\pi \frac{E}{hc} = \frac{E}{\hbar c}$ so it is constant for constant-energy systems? $\endgroup$ – Teoc Oct 1 '17 at 4:54
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Thanks to @QMechanic and @JahanClaes for clearing up this question in the comments.

As $$k=\frac{\sqrt{2m(E-V)}}{\hbar} = \frac{\sqrt{\hat{p}^2}}{\hbar}=-i\frac{\partial}{\partial x},$$ we have that $$k^2\psi=-\frac{\partial^2 \psi}{\partial x^2}$$ and thus when substituted into the derived equation we obtain an identity, and thus the derived equation is equivalent to the Schrodinger Equation.

The confusion arose by considering momentum and kinetic energy as artifacts of classical mechanics, but it is resolved by considering momentum in the context of quantum mechanics, where $\hat{p}=-i\hbar \frac{\partial}{\partial x}.$

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