Schroedinger's equation

Oct 2010
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I am trying to understand, why did Schroedinger choose to write the total energy, and then multiply with psi and then insert the wave function?



What does the total energy which is potential energy plus kinetic energy has to do with subatomic particles and waves?
 

topsquark

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Apr 2008
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This is a little long winded. Sorry for that.

When "converting" a Classical equation to a QM equation we use the
substitution \(\displaystyle p \to \dfrac{\hbar}{i} \dfrac{d}{dx}\). (This had been known for around the same time as the Bohr model of the Hydrogen atom.)

Note then, that the kinetic energy \(\displaystyle T = \dfrac{p^2}{2m} \to
- \dfrac{\hbar ^2 }{2m} \dfrac{d^2}{dx^2}\)

We've got that derivative but it has to have something there to operate on. The Physics at this level is pretty much done. Now we hit Math.

The form of the kinetic energy here is what's know as a (linear) operator. The S- equation itself is an eigenvalue equation. They are of the form:
\(\displaystyle L \psi = \lambda \psi\). The L is a linear operator acting on the state \(\displaystyle \psi\) (an eigenstate) and the \(\displaystyle \lambda\) is called an eigenvalue and is simply a constant, not an operator.

So we have a conservation of energy statement as an eigenvalue equation:
\(\displaystyle \left ( \dfrac{p^2}{2m} + V \right ) \psi = E \psi\)

The energy operator is kinetic plus potential energies so we have an operator \(\displaystyle T + V = -\dfrac{\hbar ^2}{2m} + V\) is the energy operator and E is the eigenvalue of that operator. Thus we can write
\(\displaystyle -\left ( \dfrac{\hbar ^2}{2m} \dfrac{d^2}{dx^2}+ V \right ) \psi = E \psi\) and you get the S-equation

\(\displaystyle -\dfrac{\hbar ^2}{2m} \dfrac{d^2}{dx^2} \psi + V \psi = E \psi\)

-Dan
 
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Oct 2010
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Thanks, but can you explain with words what is the intuitive reason why we take the total energy in classical mechanics to derive what is important as a Newton's law in quantum mechanics.

I don't understand what was Schroedinger thinking, when deriving this. Why shouldn't we use Newton's laws to derive an equation which has some significance in quantum mechanics?
 
Oct 2017
536
256
Glasgow
Thanks, but can you explain with words what is the intuitive reason why we take the total energy in classical mechanics to derive what is important as a Newton's law in quantum mechanics.
We don't.

I don't understand what was Schroedinger thinking, when deriving this. Why shouldn't we use Newton's laws to derive an equation which has some significance in quantum mechanics?
QM doesn't depend on Newton's laws or any other classical physics models. You can make comparisons between QM and classical mechanics (like your first post) as a learning exercise, so it might appear that some equations can be derived as such, but QM is the more fundamental theory and we need only expect QM to provide results in agreement with Newton's laws at macroscopic scales (which seems to be the case).

Also, Newton's laws do not reproduce the observed behaviour of particles or collections of them at atomic and sub-atomic scales. It doesn't work.
QM, love it or hate it, does reproduce the observed behaviour.
 
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Oct 2017
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Is this because we don't understand QM very well?
Nope, it's just not required. Newton's laws don't work at atomic and sub-atomic scales. When dealing with atoms and sub-atomic phenomena, just forget Newton's laws; they're unlikely to predict anything useful.

QM is a notoriously difficult theory to understand (does anyone truly understand it?), hence why it has been introduced to you using comparisons to classical theories.

By the way, not all equations are just derived from others. There are all sorts of hypotheses that compete with each other and there are all sorts of reasons for choosing one over another. Perhaps one hypothesis is more consistent with existing physics, or one matches experimental evidence better than the other, or one covers a greater range of scope than the other, or one is based on first-principles models rather than empirical or semi-empirical models, or one has fewer free parameters/fudge factors, etc....

Specifically for QM, it is one of the best examples of a "paradigm shift". QM is so completely different from all other preceding physics theories that it has had a massive impact about how scientists view the world. To this day, many philosophers are still debating the ramifications. Furthermore, QM is probably the biggest reason why we have the concept of "classical" physics versus modern, "non-classical" physics, along with general relativity.
 
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topsquark

Forum Staff
Apr 2008
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On the dance floor, baby!
Is this because we don't understand QM very well? So we don't know what was Schroedinger thinking when deriving this equation.
I have been told (by Eugene Hect himself. Name drop!) that the Schrodinger equation was first derived using optics methods. I haven't looked up this derivation for myself.

Also note that Schrodinger first derived what is now known as the Klein-Gordon equation, which is a relativistic equation. He had to abandon it because there was a problem with the state space. He published a low velocity approximation which is now known as the Schrodinger equation.

-Dan
 
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Oct 2010
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How did they get this operator:

\(\displaystyle p \to \dfrac{\hbar}{i} \dfrac{d}{dx}\)?
 
Oct 2017
536
256
Glasgow
How did they get this operator:

\(\displaystyle p \to \dfrac{\hbar}{i} \dfrac{d}{dx}\)?
I don't know who came up with it or how, but Wikipedia seems to suggest that the momentum operator can be inferred by considering a 1D plane wave solution to Schrodinger's equation and then taking the first-order derivative. Perhaps someone determined the momentum operator when studying de Broglie's wave theory? After all, it's not unusual for scientists to discover things by inventing the solution they want to see and then processing the answer backwards through the formulae they have to see what initial conditions or relationships would be required to provide that solution.

Also, as a minor point, I tend to see the momentum operator expressed in the 1D form:

\(\displaystyle \hat{p} = -i \hbar \frac{d}{dx}\)

which is equivalent the other form expressed above, or

\(\displaystyle \hat{p} = -i \hbar \nabla\)

as the general form.
 
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