## Clarrification of EΨ

$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

Anthony_3C
Posts: 30
Joined: Wed Sep 30, 2020 10:00 pm

### Clarrification of EΨ

In the Schrodinger equation, I understand the left part (Hamiltonian), but I'm confused about what EΨ actually means. Is it just "one" symbol, so that EΨ means the total energy of an electron, or is it that E means the energy and Ψ just represents the wave function itself, so that EΨ equals E (energy) times Ψ (wavefunction). Basically I'm unsure if there's an omitted multiplication symbol between E and Ψ.

Farah Abumeri 1B
Posts: 39
Joined: Wed Sep 30, 2020 9:51 pm
Been upvoted: 1 time

### Re: Clarrification of EΨ

The EΨ means the energy of the wave function. This value also corresponds to the orbitals (since the second derivative of the wave function = the probability of finding an electron). In other words, the Hamiltonian (the double or second derivative) of the wave function = the energy of the wave function (the orbitals). We can also compare the energy with the atomic spectra to verify the energy value.

I think that we just have to understand the conceptual basis of the Schrodinger equation, rather than actually use it for calculations. I hope this makes some sense!

Chinyere Okeke 2A
Posts: 37
Joined: Wed Sep 30, 2020 10:05 pm
Been upvoted: 1 time

### Re: Clarrification of EΨ

Farah Abumeri 1B wrote:The EΨ means the energy of the wave function. This value also corresponds to the orbitals (since the second derivative of the wave function = the probability of finding an electron). In other words, the Hamiltonian (the double or second derivative) of the wave function = the energy of the wave function (the orbitals). We can also compare the energy with the atomic spectra to verify the energy value.

I think that we just have to understand the conceptual basis of the Schrodinger equation, rather than actually use it for calculations. I hope this makes some sense!

I was also a little confused with the definition, but you're statement helped clarify for me. But would that mean that the Hamiltonian of the wave function = the wave function squared; since the wave function squared also represents the probability of finding an e-?

reyvalui_2g
Posts: 35
Joined: Wed Sep 30, 2020 9:52 pm

### Re: Clarrification of EΨ

If I remember correctly, the second derivative of the wave function doesn't necessarily equate to the probability of finding an electron, as that would be the wave function squared. All the second derivative would do is tell you the curvature of the wave function at a certain point.

As for the Hamiltonian, it would be the sum of the potential energy of the wave function plus the kinetic energy of the wave function.

Manseej Khatri 2B
Posts: 35
Joined: Wed Sep 30, 2020 9:42 pm

### Re: Clarrification of EΨ

Hi. Schrodinger's equation essentially states that the Hamiltonian multiplied by Ψ (wave function) is equivalent to some energy value E multiplied by Ψ . It's essentially saying that there is some value E (energy) when multiplied by the wave function is the same as the Hamiltonian function(s) multiplied by Ψ.

ColmConnolly1E
Posts: 35
Joined: Wed Sep 30, 2020 10:06 pm
Been upvoted: 1 time

### Re: Clarrification of EΨ

Are the Hamiltonian functions something that would be unique to each atom, or will all atoms share a set of wave functions?