Question about Schrodinger equation, potential and energy

In summary, the 1D time-independent Schrodinger equation allows for the mixing of potentials and energies in the same equation, as long as the potential is converted to an energy first. This is due to the relationship between potential and potential energy for charged particles. However, some confusion may arise if the relationship between the quantum mechanical Hamiltonian and classical Hamiltonian dynamics is not well understood.
  • #1
Nylex
552
2
Ok, I know that the 1D time-independent Schrodinger equation is [tex]-\frac {\hbar^2} {2m} \frac {d^2 \psi(x)} {dx^2} + V(x) \psi(x) = E \psi(x)[/tex]. Why is it that you can mix potentials and energies in the same equation? For example, if you're saying that V(x) has a constant value, say, [tex]V(x) = V_{0}[/tex] and you're talking about a potential step, then you get something like [tex]k = \frac {\sqrt{2m(V_{0} - E)}} {\hbar}[/tex] for the wavenumber if the total energy is less than the step height. How can you subtract two different quantities? It doesn't make sense to me :(. I guess I didn't think about this much before now, when we've been set a problem where we're given values for E (in eV) and V (in V). Does the potential need to be converted to an energy first? It seems other people have been getting confused by this as well, so it isn't just me.

If anyone can explain, thanks.
 
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  • #2
Nylex said:
Ok, I know that the 1D time-independent Schrodinger equation is [tex]-\frac {\hbar^2} {2m} \frac {d^2 \psi(x)} {dx^2} + V(x) \psi(x) = E \psi(x)[/tex]. Why is it that you can mix potentials and energies in the same equation? For example, if you're saying that V(x) has a constant value, say, [tex]V(x) = V_{0}[/tex] and you're talking about a potential step, then you get something like [tex]k = \frac {\sqrt{2m(V_{0} - E)}} {\hbar}[/tex] for the wavenumber if the total energy is less than the step height. How can you subtract two different quantities? It doesn't make sense to me :(. I guess I didn't think about this much before now, when we've been set a problem where we're given values for E (in eV) and V (in V). Does the potential need to be converted to an energy first? It seems other people have been getting confused by this as well, so it isn't just me.

If anyone can explain, thanks.

Er... that "V" is potential ENERGY, not potential as in electrostatic potential. Y'know, as in gravitational potential energy is mgh, etc... Haven't you seen something like this when you do Lagrangian mechanics?

Zz.
 
  • #3
ZapperZ said:
Er... that "V" is potential ENERGY, not potential as in electrostatic potential. Y'know, as in gravitational potential energy is mgh, etc... Haven't you seen something like this when you do Lagrangian mechanics?

Zz.

I thought at first it was potential energy, but our lecturer refers to it as the step height.. and then he sets a problem with the step height in volts. I don't do Langrangian mechanics, it was an option module (although I didn't get any options this year and wouldn't have done it anyway.. looks too mathematical).
 
  • #4
Nylex said:
I thought at first it was potential energy, but our lecturer refers to it as the step height.. and then he sets a problem with the step height in volts. I don't do Langrangian mechanics, it was an option module (although I didn't get any options this year and wouldn't have done it anyway.. looks too mathematical).

But think about it. If you put a charge particle in a potential of V, it has the potential to gain an energy by an amount qV! If it is an electron, the energy it can gain is eV. So just simply having an electrostatic potential is equivalent to a potential energy for any charged particle.

However, your lecturer is obviously sloppy in setting up k with E-V, while still keeping V as the potential. What is meant here is that you have E - qV.

Zz.
 
  • #5
That's bad.Lagrange & Hamilton formalisms for CM are COMPULSORY PREREQUISITES for QM...

Daniel.
 
  • #6
Yes, it's very bad indeed. I wonder how they justify allowing students to take QM without classical Hamiltonian dynamics. How do they expect the students to know what the quantum mechanical Hamiltonian is? :confused:
 
  • #7
Tom Mattson said:
Yes, it's very bad indeed. I wonder how they justify allowing students to take QM without classical Hamiltonian dynamics. How do they expect the students to know what the quantum mechanical Hamiltonian is? :confused:
It's not really compulsory to have taken Hamiltonian mechanics.
Without it, you can certainly learn to do quantum mechanics.

Everyone can see that [itex]\frac{p^2}{2m}+V[/itex] is an expression for the total energy of the particle. The name 'Hamiltonian' may sound new, but wouldn't be a problem.
It's just that the 'derivation' of the SE and links to stuff that uses the action principle (which is generally considered somewhat 'advanced') remains ununderstood.
For an introductory course on QM. You can do without Hamiltonian and Langranian mechanics.
 
  • #8
Galileo said:
It's not really compulsory to have taken Hamiltonian mechanics.
Without it, you can certainly learn to do quantum mechanics.

I agree that you can learn to do it, but I think it would be more of a hollow algorithm than a real understanding.

Everyone can see that [itex]\frac{p^2}{2m}+V[/itex] is an expression for the total energy of the particle. The name 'Hamiltonian' may sound new, but wouldn't be a problem.

But you wouldn't know why H is a function of the momentum and the position, instead of the velocity and the position. In Physics I, you learn that that expression is [itex]\frac{1}{2}mv^2+V[/itex]. Somewhere, the break has to be made from this convention. If you show that expression you quote without explaining Hamiltonian dynamics, then the switch from (x,v) to (x,p) appears to be without motivation, does it not?

edit: Does anyone know why LaTeX puts all that whitespace in my expression?
 
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  • #9
In a QM class at this level (I had one) it goes like this:

There exist a wave function psi such that the partial of psi with respect to position is momentum, and the partial of psi with respect to time is the Energy. Now, energy and momentum are related thusly, so we have the shrodinger equation.

Q: "Why is partial of psi wrt to x the momentum?"

A: "Because that is what gives us the right answer".

To be honest, my class wasn't even this clear. Instead we started with:

psi = exp( i(kx-wt)) because the double slit particle suggested waves, and then noticed that the derivatives of psi were related according to the shrodinger equation... :cry: :zzz:
 
  • #10
ZapperZ said:
But think about it. If you put a charge particle in a potential of V, it has the potential to gain an energy by an amount qV! If it is an electron, the energy it can gain is eV. So just simply having an electrostatic potential is equivalent to a potential energy for any charged particle.

However, your lecturer is obviously sloppy in setting up k with E-V, while still keeping V as the potential. What is meant here is that you have E - qV.

Zz.

Thanks Zz :smile:. The recommended book for our course (Eisberg and Resnick) does the same) with V, unless they mean potential energy and not just potential.

Tom Mattson said:
Yes, it's very bad indeed. I wonder how they justify allowing students to take QM without classical Hamiltonian dynamics. How do they expect the students to know what the quantum mechanical Hamiltonian is? :confused:

Do you mean the Hpsi(x) = Epsi(x) thing? We did get taught that in our QM course.
 
  • #11
No. The whole QM formalism comes from the Hamiltonian by quantizing, and understanding the nature of state space, and dynamics. The Hamiltonian comes from the Lagrangian. Also the Lagrangian formalism carries across to QFT easily too.

I'm currently in the first year of a physics undergrad course, and feel that they ought to teach us the "Advanced CM" before QM, but they won't - full QM will be taught next year, much to my dismay. Lucky for me, I already know Lagrangian mechanics, and Hamiltonian will be something for me to read this summer.
 
  • #12
This is kind of sad - but I took 3 semesters of QM as an undergrad (1 semester undergrad, and 2 semesters grad) without taking a CM course.
 
  • #13
No say,then what the heck did they teach you there...?I think the formulations of Lagrange & Hamilton are called (at least in my country) "analytical mechanics" and they correspond to the first theoretical physics course which is taught in the 3-rd semester (of the academic programme).Newton's formulation is called "classical mechanics" and is taught in the second semester,while QM is taught in the fifth & the 6-th...

I don't know what system of teaching you got there people,but it's definitely not the one i like.

Daniel.
 
  • #14
juvenal said:
This is kind of sad - but I took 3 semesters of QM as an undergrad (1 semester undergrad, and 2 semesters grad) without taking a CM course.
Let me recommend Goldstein's book Classical Mechanics to you. There may better ones now, but I strongly doubt it. At JHU, where I got my Ph.D. in physics 40 years ago, it was the standard for the CM course.
The book is so good that there were no lectures. The graduate students met once per week and some were called upon to put their solutions to last week's assigned homework problems on the board. A new set of problems for home work was assigned and we went away when we wanted to (15 minutes or two hours later, as we thought best.)
It was one of the best courses I ever had. Work your way thru that book (It will take at least a hundred hours, unless you are very exceptional, if you do 1/3 of the problems.
Goldstein's introductory note tells it all. I can't quote it after fourty years, but the essense is "Be ye doers and not just readers." Do 1/3 of his chapter end problems, and you will really know CM. (Don't do just the first three, they get progressively tougher.) You don't need a formal course, but of course a good one, or someone to ask when you get stuck, sure helps. Organize your own informal course with a few friends, if need be.
 
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  • #15
In 40 years,there's no other book better than Goldstein in CM...(Landau & Lifschitz is a little shallow).

Daniel.
 
  • #16
I don't know where juvenal's from, but I'm from the UK at Oxford University, and we don't get advanced CM, unless we choose to do it in the 3rd year.
 
  • #18
Billy T said:
Let me recommend Goldstein's book Classical Mechanics to you. There may better ones now, but I strongly doubt it. At JHU, where I got my Ph.D. in physics 40 years ago, it was the standard for the CM course.
The book is so good that there were no lectures. The graduate students met once per week and some were called upon to put their solutions to last week's assigned homework problems on the board. A new set of problems for home work was assigned and we went away when we wanted to (15 minutes or two hours later, as we thought best.)

Thanks - but I took a Goldstein course my first term of grad school. That was torturous.
 
  • #19
dextercioby said:
That sounds really devious...

What do you mean?
 
  • #20
What do you get to choose it...?Isn't it COMPULSORY...?And when is the QM course...?

Daniel.
 
  • #21
dextercioby said:
In 40 years,there's no other book better than Goldstein in CM...(Landau & Lifschitz is a little shallow).

I'll be dammed if have to sit here and hear you complain about Landau's books!
(Just joking :wink: )
Now seriously, Goldstein's book is more complete (wich is obvious from the fact it's a LOT bigger), but I really prefer Landau's "axiomatic" approach.
Nevertheless, the classical mechanics book I like the most is Saletan & Jose's.

On the other point, I believe you don't have to know Lagrangian and Hamiltonian mechanics in order to learn or understand QM. Think about this, you have an elephant and it's picture, if you want to study the elephant, it's obviously easier if you've first studied its picture, but that's certainly not necessary.
In fact, when you say:
masudr said:
The whole QM formalism comes from the Hamiltonian by quantizing
I think you're pushing the correspondence way too far, remeber, classical mechanics should come as a limiting of quantum mechanics and not the other way around.
In other words, you can certainly give plausible classical arguments to justify some things in QM, but you can't (and shouldn't) "derive" QM from CM.
In the 3rd tome of Feynman Lectures, it's very clear he also thought that way about both: the fact that you don't need Lagrangian & Hamiltonian mechanics to learn QM and that you shouldn't push the correspondence too far.

Finally:
Crosson said:
Q: "Why is partial of psi wrt to x the momentum?"

A: "Because that is what gives us the right answer".
Ultimately, all axiomatic approaches to science, will give you that answer, or more correctly:
"Because that is what gives us the answer that agrees with experiments"
unless you can read God's mind, there's no escaping to that answer.
The reason to justify something in QM should NEVER be "because in classical mechanics is so and so".
At the end of the day, classical mechanics also answer the ultimate questions with that answer, for example, how would you answer:
¿Why does the action must be an extremum?
(Purely from the classical point of view, of course)
 
  • #22
BlackBaron said:
I'll be dammed if have to sit here and hear you complain about Landau's books!
(Just joking :wink: )


There are GOOD books,all 10 of them...However,the second part of classical field theory (the one dealing with GR) and CM are a bit "rushed",e.g.i was very annoyed to see the lack of Noether's theorem in the CM (particle) case...

Black Baron said:
On the other point, I believe you don't have to know Lagrangian and Hamiltonian mechanics in order to learn or understand QM. Think about this, you have an elephant and it's picture, if you want to study the elephant, it's obviously easier if you've first studied its picture, but that's certainly not necessary.

I'm sorry,i'll have to disagree...Much of the formalism of QM has roots in CM.

Black Baron said:
The fact that you don't need Lagrangian & Hamiltonian mechanics to learn QM and that you shouldn't push the correspondence too far.

I disagree.Nobody mentioned the correspondence principle.You're mising the point.


Daniel.
 
  • #23
Yeah; Hamiltonian mechanics was historically important in formulating QM and the the courageous men who led us into this brave new world used and modified the language to which they were happy with -- i.e. Hamiltonian.

btw, Daniel, in our course, QM is taught in the 2nd year. And whilst they hint at advanced CM, no formal treatment is given unless the student chooses it as one of his/her options for the 3rd year. Advanced CM is only regarded as absolutely necessary if you plan to do further study in physics; remember that many of us who graduate from here will probably end up working as a financial trader in the City of London.
 
  • #24
That figures it.Thank you for telling me that QM is a useful prerequisite for financial trading (maybe operators and eigenvalues) in the City of London...

Daniel.

P.S.For the record,i still think it's abnormal...Zeeman abnormal...
 
  • #25
dextercioby said:
I disagree.Nobody mentioned the correspondence principle.You're mising the point.
I didn't mention the correspondence principle either.
When I say "you shouldn't push the correspondence too far" I meant that you shouldn't try to build a quantum theory out of a classical one. Things should work the other way around (i.e. quantum theory -> classical theory).
I'm aware, of course, that that's really hard to do in practice, and that it's really helpful to study classical theories first, but my point is that, classical theories, outside their range of applications, should only be seen as a helpful instrument to build quantum theories, but not as a way to rigorously derive them.
 
  • #26
I (and no one else too) didn't mention "rigurously" deriving QM formalism from CM formalisms...I still believe in the Axiomatization of QM...

Daniel.

P.S.I think that settles it.No more reading too much into it,okay...?:wink:
 
  • #27
Nylex said:
Do you mean the Hpsi(x) = Epsi(x) thing? We did get taught that in our QM course.

No, I mean the fact that the Hamiltonian must be a function of (x,p) and not (x,v).
 
  • #28
Well,Tom,that's from the DEFINITION.If they don't do it in CM,and probably don't study Schwinger's view of QM dynamics,they cannot justify that...

Daniel.
 
  • #29
dextercioby said:
I (and no one else too) didn't mention "rigurously" deriving QM formalism from CM formalisms...
You got that right, but that's exactly my point, if you aren't using CM to "rigurously" derive QM, then CM it's not a "COMPULSORY PREREQUISITE".
It sure helps a lot, but it's certainly not essential.
 
  • #30
Oh,yes it is...You'd not be doing theoretical physics,then.And with that i'll give it a rest...I've said what i had to say.

Daniel.
 
  • #31
how to post picture?
 
  • #32
You can use the "manage attachement" tool...It should be less than 50KB and have one of the formats indicated there.

Daniel.
 
  • #33
Such as the formula of the Nylex post how does post come out?
 
<h2>1. What is the Schrodinger equation?</h2><p>The Schrodinger equation is a mathematical equation that describes how the quantum state of a physical system changes over time. It is a fundamental equation in quantum mechanics and is used to determine the probability of finding a particle in a certain location or state.</p><h2>2. What is the relationship between potential and energy in the Schrodinger equation?</h2><p>In the Schrodinger equation, potential and energy are related through the Hamiltonian operator. The potential energy represents the energy associated with the position of a particle in a given system, while the total energy is the sum of the potential and kinetic energy of the particle.</p><h2>3. How does the Schrodinger equation take into account the uncertainty principle?</h2><p>The Schrodinger equation takes into account the uncertainty principle by describing the quantum state of a particle in terms of probability amplitudes, rather than definite values. This means that the exact position and momentum of a particle cannot be simultaneously known, as described by the uncertainty principle.</p><h2>4. Can the Schrodinger equation be used to predict the behavior of all quantum systems?</h2><p>The Schrodinger equation is a powerful tool in predicting the behavior of quantum systems, but it is not applicable to all systems. In some cases, more complex equations or approximations may be needed to accurately describe the behavior of a system.</p><h2>5. How does the Schrodinger equation relate to wave-particle duality?</h2><p>The Schrodinger equation is a wave equation that describes the behavior of particles in quantum systems. It relates to wave-particle duality by showing that particles can exhibit both wave-like and particle-like behavior, depending on the context in which they are observed.</p>

1. What is the Schrodinger equation?

The Schrodinger equation is a mathematical equation that describes how the quantum state of a physical system changes over time. It is a fundamental equation in quantum mechanics and is used to determine the probability of finding a particle in a certain location or state.

2. What is the relationship between potential and energy in the Schrodinger equation?

In the Schrodinger equation, potential and energy are related through the Hamiltonian operator. The potential energy represents the energy associated with the position of a particle in a given system, while the total energy is the sum of the potential and kinetic energy of the particle.

3. How does the Schrodinger equation take into account the uncertainty principle?

The Schrodinger equation takes into account the uncertainty principle by describing the quantum state of a particle in terms of probability amplitudes, rather than definite values. This means that the exact position and momentum of a particle cannot be simultaneously known, as described by the uncertainty principle.

4. Can the Schrodinger equation be used to predict the behavior of all quantum systems?

The Schrodinger equation is a powerful tool in predicting the behavior of quantum systems, but it is not applicable to all systems. In some cases, more complex equations or approximations may be needed to accurately describe the behavior of a system.

5. How does the Schrodinger equation relate to wave-particle duality?

The Schrodinger equation is a wave equation that describes the behavior of particles in quantum systems. It relates to wave-particle duality by showing that particles can exhibit both wave-like and particle-like behavior, depending on the context in which they are observed.

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