Schrodinger equation

[swansont]

Probablity waves don't occur in nature. They are mental constructs.

 

And yet natures behaves as if they do...

[Johnny5]

And yet natures behaves as if they do...

 

Actually swansont, De Broglie and Born were totally different minds.

 

It was De Broglie who postulated that matter has wave properties. He didn't say in what sense. Schrodinger thought the energy density was waving, and Born said probability is waving.

 

But it was De Broglie who started it all, with the following postulate:

 

[math] p = \frac{\hbar}{\lambda} [/math]

 

So how do you go from what De Broglie thought, to what Born thought?

[□h=-16πT]

Probability of what?

 

I knew you or someone would ask, I couldn't remember at the time. I believe (although my lack of education in QM is great) that it gives the probability of finding a particle in some state [math]\Psi[/math].

[swansont]

I knew you or someone would ask, I couldn't remember at the time. I believe (although my lack of education in QM is great) that it gives the probability of finding a particle in some state [math]\Psi[/math'].

 

But ^{[math]\psi[/math]} can be a superposition of eigenstates. ^{[math]\psi[/math]}*^{[math]\psi[/math]} gives the probability density of finding a system in a particular state.

[Johnny5]

But ^{[math]\psi[/math]} can be a superposition of eigenstates. ^{[math]\psi[/math]}*^{[math]\psi[/math][/sup'] gives the probability density of finding a system in a particular state.}

 

^{What is just psi? Before psi star psi?}

 

^{If [math] \psi^* \psi [/math] is a probability, what is just psi?}

 

^{Oh nevermind.}

[ed84c]

Wave Function isnt it?

[Johnny5]

Wave Function isnt it?

 

 

No, I know what it is guys, but WHAT IS IT?

 

What in nature is complex valued?

[ed84c]

Hmm now, is this not the function thats dead hard to explain? I read somewhere (I thikn its wave function) that only a few scientists understand what it is, and as i say I THINK its this, so if you dont get a responce, thats why!

[swansont]

No' date=' I know what it is guys, but WHAT IS IT?

[/quote']

 

It's the wave function. The function that, when multiplied by its complex conjugate, gives you the probability density.

[Severian]

The Schrodinger Equation is just conservation of energy in a quantum mechanical form.

 

Classical (non-relativistic) conservation of energy is: Kinetic energy + potential energy = total energy

 

Kinetic energy = 1/2 m v² = [math]\frac{p^2}{2m}[/math] where p=mv is the momentum.

 

Potential energy depends on the form of the potential V(x)

 

So classically [math]\frac{p^2}{2m}+ V(x) = E[/math]

 

In qunatum mechanics, the momentum of a field [math]\psi(x)[/math] is given by [math]-i \hbar \frac{\partial}{\partial x} \psi(x)[/math] or in other words the momentum operator is [math]\hat p = -i \hbar \frac{\partial}{\partial x}[/math].

 

Also the energy operator is [math]\hat E = i \hbar \frac{\partial}{\partial t}[/math].

 

So the Schrodinger equation is just [math](\frac{\hat p^2}{2m}+V(x))\psi(x) = \hat E \psi(x)[/math] or [math]-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+V(x)\psi(x) = i \hbar \frac{\partial}{\partial t} \psi(x)[/math].

[Johnny5]

It's the wave function. The function that, when multiplied by its complex conjugate, gives you the probability density.

 

That is the Born interpretation right?

[Johnny5]

The Schrodinger Equation is just conservation of energy in a quantum mechanical form.

 

Classical (non-relativistic) conservation of energy is: Kinetic energy + potential energy = total energy

 

Kinetic energy = 1/2 m v² = [math]\frac{p^2}{2m}[/math] where p=mv is the momentum.

 

Potential energy depends on the form of the potential V(x)

 

So classically [math]\frac{p^2}{2m}+ V(x) = E[/math]

 

In qunatum mechanics' date=' the momentum of a field [math']\psi(x)[/math] is given by [math]-i \hbar \frac{\partial}{\partial x} \psi(x)[/math] or in other words the momentum operator is [math]\hat p = -i \hbar \frac{\partial}{\partial x}[/math].

 

Also the energy operator is [math]\hat E = i \hbar \frac{\partial}{\partial t}[/math].

 

So the Schrodinger equation is just [math](\frac{\hat p^2}{2m}+V(x))\psi(x) = \hat E \psi(x)[/math] or [math]-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+V(x)\psi(x) = i \hbar \frac{\partial}{\partial t} \psi(x)[/math].

 

Severian, if I may, i would like to ask you an intelligent question, which is related to the original poster's question.

 

Why does [math] \sqrt -1 [/math] show up in Schrodinger's formulation? I seem to recall it has something to do with making the velocity of the wave be equivalent to the velocity of the object. In other words, do you know how the square root of negative one ended up in the wave equation?

 

 

Thank you

[Guest MikeM]

The original poster wanted a complete explanation that's a very tall order. The three dimensional form should be used' date=' and the spherical harmonics for a hydrogen atom should be derived. Can anyone here do that for him?

 

[b']To Muralisankar: [/b] Do you know spherical coordinates?

 

Start off with the 3-D Schrodinger, with the H atom potential, transformed into spherical coordinates...

 

Separate the variables, leading to 3 distinct ORDINARY differential equations..

 

(radial, angular and time)...

 

The radial equation will need solving to give the energy levels. This, if I remember correctly, should lead to Legendre polynomials, with eigenvalues corresponding to the 'energy levels'...

 

However, to do this requires some knowledge of Partial Differential Equations, and if the original poster has yet to study them in depth, then it's a tall order to get very far...

 

As to the 'derivation' of the Schrodinger Equation, you can start from First Principles with Calculus of Variations.. The equation will 'fall out' with some manipulation.. This is not 'water-tight', as there is some prestidigitation required....

 

Hope this helps...

[Johnny5]

Start off with the 3-D Schrodinger' date=' with the H atom potential, transformed into spherical coordinates...

 

Separate the variables, leading to 3 distinct ORDINARY differential equations..

 

(radial, angular and time)...

 

The radial equation will need solving to give the energy levels. This, if I remember correctly, should lead to Legendre polynomials, with eigenvalues corresponding to the 'energy levels'...

 

However, to do this requires some knowledge of Partial Differential Equations, and if the original poster has yet to study them in depth, then it's a tall order to get very far...

 

As to the 'derivation' of the Schrodinger Equation, you can start from First Principles with Calculus of Variations.. The equation will 'fall out' with some manipulation.. This is not 'water-tight', as there is some prestidigitation required....

 

Hope this helps...[/quote']

 

How about the fourth quantum number. How do you get that to just "fall out" given that there are only three ordinary differential equations to solve, after you have used the method of separation of variables.

 

spin quantum number is still missing.

 

Do you know?

 

Regards

[Bob182]

google on the Dirac equation

[Severian]

Severian' date=' if I may, i would like to ask you an intelligent question, which is related to the original poster's question.

 

Why does [math'] \sqrt -1 [/math] show up in Schrodinger's formulation? I seem to recall it has something to do with making the velocity of the wave be equivalent to the velocity of the object. In other words, do you know how the square root of negative one ended up in the wave equation?

 

 

Thank you

 

Well, the 'i' arises because the wave is decribed by an expenential of a complex number.

 

So a field may look like [math]\psi(x,t) = \psi_0 exp(i(kx-\omega t))[/math] for example. Remember that [math]e^{ix} = \cos x + i \sin x[/math] so this form is really just saying that the field is 'waving'.

 

Now the momentum of this wave is [math]\hbar k[/math] and we can extract the k from the field only by differentiating with respect to x:

 

[math] \frac{\partial}{\partial x} \psi (x,t) = \psi_0 exp(i(kx-\omega t)) ik = \psi (x,t) ik[/math]

 

So to get the momentum [math]\hbar k[/math] the operation we must perform on the field is the differentiation and then multiply by [math]\hbar[/math] and finally multiply by -i to get rid of the extra i.

 

Therefore the momentum operator (which extracts the mometum from the field) is [math]\hat p \equiv -i \hbar \frac{\partial}{\partial x}[/math] and [math]\hat p \psi(x,t) = \hbar k \psi(x,t)[/math] as desired.

[Johnny5]

Well' date=' the 'i' arises because the wave is decribed by an expenential of a complex number.

[/quote']

 

How did the square root of negative 1 get in the formula historically. Who put it in, and why?

 

Thank you

 

The concept that the probability is waving is useless btw.

 

Regards

[swansont]

The concept that the probability is waving is useless btw.

 

I think a large number of physicists would disagree with that, given the success of QM.

[Johnny5]

I think a large number of physicists would disagree with that, given the success of QM.

 

 

They can all disagree with me, that will change nothing.

 

Kind regards Dr Swanson

[swansont]

They can all disagree with me' date=' that will change nothing.

 

Kind regards Dr Swanson[/quote']

 

Ironic that you used a computer to convey that sentiment.

[Guest MikeM]

How about the fourth quantum number. How do you get that to just "fall out" given that there are only three ordinary differential equations to solve' date=' after you have used the method of separation of variables.

 

spin quantum number is still missing.

 

Do you know?

 

Regards[/quote']

 

It's been a LONG LONG time since I've looked at this, (I no longer use mathematics), but I have a feeling it's 'defined' from the Legendre polynomial. If anybody knows better, please correct me...

[Guest MikeM]

How did the square root of negative 1 get in the formula historically. Who put it in' date=' and why?

 

Thank you

 

The concept that the probability is waving is useless btw.

 

Regards[/quote']

 

Obviously, I'm new here, so please don't misinterpret the following...

 

You are asking about quantum spin, you are saying that probability functions have no real application in QM, yet you don't know how 'i' comes into the equation...

 

Can I ask what your background is?

[Tom Mattson]

The spin quantum number doesn't come from the Schrodinger formalism. In fact it cannot, because spin is not a function of coordinates. But that doesn't mean that spin cannot be entered into the Schrodinger equation. Feynman pointed out that the operator for kinetic energy (p²/2m) can actually be written as follows:

 

[math]

(\sigma \cdot p)^2/2m

[/math]

 

where [math]\mathbf{\sigma}[/math] stands for a vector operator whose components are the Pauli spin matrices.

[Johnny5]

Obviously' date=' I'm new here, so please don't misinterpret the following...

 

You are asking about quantum spin, you are saying that probability functions have no real application in QM, yet you don't know how 'i' comes into the equation...

 

Can I ask what your background is?[/quote']

 

I think that i comes into the equation in order to get the speed of the wave to be equivalent to the speed of the objects center of mass, but I haven't recieved a genuine answer on this yet. Years ago I was told that, but I never actually checked it, because at the time I was told it, I already knew that Schrodinger put the equation together piecemeal, and furthermore that he designed it to predict E=1/2mv^2 for a free particle, but that later Dirac changed it to be E^2 = (pc)^2+(mc^2)^2. So by the time I began studying the equation, I already was a bit umm whats the word... well I felt the usage of i was bogus. So here I am asking the question, yet again, and still no answer. Not to mention I don't like anything with 'i' in it, because you cannot order the complex numbers.

 

As for my background, sure you can ask... I read alot.

 

Regards

[Tom Mattson]

The Schrodinger equation is complex because the time evolution operator is complex. And the time evolution operator is complex because it must be unitary, which guarantees probability conservation. In other words, if the i is missing from the Schrodinger equation, its solutions will decay instead of wave.