- #1

- 53

- 8

It started with the wave function $$\nabla^2\psi=k^2\psi$$

I am a bit lost at this point. Where does the right side of the equation come from? What should I review to fix that part of my knowledge?

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- Thread starter BearY
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- #1

- 53

- 8

It started with the wave function $$\nabla^2\psi=k^2\psi$$

I am a bit lost at this point. Where does the right side of the equation come from? What should I review to fix that part of my knowledge?

- #2

- 350

- 81

- #3

- 53

- 8

The reasoning says this is the spatial part of the wave equation, which to my understanding, is $$\nabla^2\psi$$.

After this, we have $$k=\frac{2\pi mv}{h}$$

and $$mv^2 = 2(E−V)$$

and thus $$\nabla^2\psi = \frac{8\pi^2m}{h^2}(E-V)\psi$$

and rearrange to get the $$E_k + V = E$$

To my understanding, this reasoning just to show that Schrödinger Equation is indeed a wave function that was talked about earlier in the book.

- #4

- 350

- 81

Read in between the lines and try to answer my question this equation is wave equation for what?The reasoning says this is the spatial part of the wave equation, which to my understanding, is $$\nabla^2\psi$$.

After this, we have $$k=\frac{2\pi mv}{h}$$

and $$mv^2 = 2(E−V)$$

and thus $$\nabla^2\psi = \frac{8\pi^2m}{h^2}(E-V)\psi$$

and rearrange to get the $$E_k + V = E$$

To my understanding, this reasoning just to show that Schrödinger Equation is indeed a wave function that was talked about earlier in the book.

Also make a comparison with the differential equation for a classical plane wave.

- #5

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What textbook? Please give a specific reference.When reading a textbook

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