# Schrodinger equation

1. May 16, 2005

### zcapa14

what is the physical significance of each of the terms in the 1-D time indipendant schrodinger equation?

2. May 16, 2005

### dextercioby

Why are u asking us...?What are your ideas...?What u've read,i presume it's not a curiosity,but some sort of homework.

Daniel.

3. May 16, 2005

### zcapa14

The question is from last years exam that i am doing for revision. however i have no answers, my notes are a little sketchy when they get round to schrodinger equation and QMT...

i am aware of the definition of each of the terms, but the question wants more than this, namely the 'physical significance' of each of these terms. That is what i am puzzled at.

4. May 16, 2005

### dextercioby

I see.There are two ways of looking at it.Traditional way,in which the equation is postulated and everything is deduces from there,or the symmetry way,i'd like to call it J.J.Sakurai way.

So the trick is simple.The axiomatical approach asserts that the speed of variation in time of the state vector is proportional to the hamiltonian applied to that state vector.The other way,is to derive this equation by stating that the Hamiltonian of the system is the self-adjoint generator of the abelian group of time translations...

So,for further reference for this interesting symmetry-based approach,i invite you to read the second chapter (i think the 2-nd or the 3-rd section) from [1].

Daniel.

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[1]J.J.Sakurai,"Modern Quantum Mechanics",Addison-Wesley,any of the 2 editions.

5. May 16, 2005

### masudr

I think he said time-independent, as opposed to time-dependent. Furthermore, the two approaches you have mentioned look exactly the same to me. One of them says that the rate of change of the state vector is given by the Hamiltonian acting on the state vector, the other says the Hamiltonian (is self-adjoint, but that's obvious anyway) generates infinitesimal translations in time, which is the rate of change.

6. May 16, 2005

### werty

The equation reads

$$\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi = E\Psi$$

If you divide out $$\Psi$$ then the first term could be interpreted as the kinetic energy, the second potential energy and the right side as the total energy. So basically its a statement about energy conservation.

Remember that $$\frac{p^2}{2m}$$ is replaced by $$\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$ in QM.

Offcourse its not legal to divide out the $$\Psi$$, so the first term is some kinetic energy times a probability density function, and so on.

7. May 25, 2005

### smoslemi

Could you explan what is V (the potential energy)?

8. May 25, 2005

### James Jackson

Well, that's in 1D cartesian. Being more rigorous, the first term becomes

$$\frac{-\hbar^2\nabla^2}{2m}\Psi$$

as the momentum operator is:

$$\hat p=-i\hbar\nabla$$

Last edited: May 25, 2005
9. May 26, 2005

### James R

V is the potential energy related to some kind of conservative force acting on the system to which you are applying the Schrodinger equation.

For example, if you're solving the equation for the hydrogen atom, V is the electrostatic potential energy of the proton and the electron.

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