# Confused about some of the basic principles

1. Dec 5, 2006

### Aero

First of all, am I right in saying the following basic postulates?

1. Each observable has an associated Hermitian operator
2. States of systems are represented by complex functions on space-time
3. When an observable is to be measured, one can calculate the probability distribution by expanding the state of the system in eigenvectors of the observable's operator, and then the probability of measuring each eigenvalue is proportional to the component of the state vector in the eigenstate, squared.

Next, the operators themselves.
Is it correct that x^ = x
p^ = ih d/dx
t^ = t
E^ = ih d/dt ?

Schrodinger's equation: all that this says to me, in light of the above, is that
E^ = ((p^) ^ 2) / 2m + V

This seems perfectly reasonable on the one hand, because it suggests that familiar results about values of observables should be applied to operators of observables. However, on the other hand, it seems completely unreasonable to use a classical definition of energy in a quantum formulation.

One thing I do not understand at all, however, is the time-independent Schrodinger equation. What is its meaning, use or relevance?

Thanks

PS How can I use MathType to create LaTeX in this forum?

2. Dec 5, 2006

### Parlyne

yep

Well, not really. States of a system are vectors in a Hilbert space. If we express them in the basis of position states, their representation is as complex functions of position and time. (Note, not necessarily scalar functions of position and time.) But, there are plenty of other ways to express them. For example, the state functions can be Fourier transformed to get momentum space wave functions.

Two problems. First, momentum is $$\hat{p} = \frac{\hbar}{i}\frac{\partial}{\partial x}$$. Second, time is not an operator; it's a parameter.

Really, this isn't a definition of energy. It's a dispersion relation between energy and momentum. And, to be totally correct, we should use a wave equation the incorporates the relativistic dispersion relation ($$E^2 = p^2c^2 + m^2c^4$$) instead of the classical one. It turns out, however, that no matter what you do, quantizing the relativistic dispersion relation always leads to an infinite number of available negative energy states; which, in the real world would be catastrophic. And, the only way to solve this (rigorously) is to step a level up to quantum field theory.

All this said, the Schroedinger equation, for all its faults, actually does a very good job of describing the dynamics of most quantum systems. And, in most cases where it begins to fail, we can introduce small correction terms to account for the strongest relativistic effects.

In systems where the potential energy does not depend on time, the eigenstates can be written as a time dependent piece multiplied by a time independent, position dependent piece. The time independent SE is essentially the time dependent SE divided by the time dependent part of the wave function. (This assumes the understanding that the time derivative of the time dependent part is $${\frac{-iE}{\hbar}$$.)

As far as I know, you can't use MathType. You actually have to type the LaTeX by hand. It's actually not that hard to learn.

Last edited: Dec 5, 2006
3. Dec 5, 2006

### Staff: Mentor

Actually, you can use MathType. Go to the Preferences menu, choose Translators, and then select LaTeX as the format for equations in the clipboard. Then just copy the equation from MathType, paste it into the message-editing window, and replace the stuff before and after the equation with ["tex"] and ["/tex"] tags (Remove the quote marks! I had to put something extra inside the tags so the forum software wouldn't interpret them as attempts to insert LaTeX.)

That's how I produced the equation below:

$$- \frac{{\hbar ^2 }}{{2m}}\nabla ^2 \Psi + V\Psi = i\hbar \frac{{\partial \Psi }}{{\partial t}}$$

4. Dec 6, 2006

### dextercioby

Nope, the operator needs to be self-adjoint.

Nope, the states are points in the projective Hilbert space of the system.

Correct, this time.

That's true iff the Hilbert space chosen is

$$L^{2}\left(\mathbb{R}, dx\right)$$

The issue of the time operator in nonrelativistic QM has only been recently solved (topics?). Search for "E. Galapon" on <<arxiv>>.

As for "E^ = ih d/dt", this is nonsense.

The SE is much more than that.

And what is then the classical definition of energy ? And what is the quantum one, if there is at all ?

There's no such thing as "time-independent SE"

Daniel.

5. Dec 6, 2006

### Parlyne

Hermitian is the same thing as self-adjoint.

How so? The SE state that that operator's action on the state gives the same result as the action of the Hamiltonian operator.

Not really. On a fundamental level, that is what it says. In fact, writing $$\hat{E}\Psi = \left (\frac{\hat{p}^2}{2m} + \hat{V} \right )\Psi$$ is more general that the standard representation, since it's basis independent.

There most certainly is a time-independent SE. It's generally writen: $$E\psi(\vec{x}) = \frac{-\hbar^2}{2m} \nabla^2 \psi(\vec{x}) + V(\vec{x})\psi(\vec{x})$$

This is the diff. eq. for the spacial wave function in cases where the time dependence in the eigenfunction is just a complex phase.

6. Dec 6, 2006

### dextercioby

True only for bounded linear operators.

Yes, but there's no such thing as $\hat{E}$

See above.

The SE is

$$\frac{d|\psi (t)\rangle}{dt}=\frac{1}{i\hbar} \hat{H}(t)|\psi (t)\rangle$$

it is not time-dependent, nor time-independent, it's a first order differential equation for the state vector.

Daniel.

7. Dec 6, 2006

8. Dec 6, 2006

### Parlyne

It may not be a standard notation to write $$\hat{E}$$. However, $$i\hbar \frac{\partial}{\partial t}$$ is certainly an operator; and, it absolutely serves the role of an energy operator (consider, for example, the role it plays in the Dirac equation and the transformation properties between it and the momentum operator).

Time-dependent vs. time-independent is a statement about the solutions of the equation. The time-independent SE is what you get in cases where the state vector can be written as $$|\Psi(t)\rangle = e^{\frac{-iEt}{\hbar}}|\psi\rangle$$ by dividing through the "normal" SE by the temporal part of the wave function. In the position basis, it's what I wrote above. In general, it states $$\hat{H}|\psi\rangle = E|\psi\rangle$$.

Importantly, the time-independent SE is the equation solved to find the wave functions in such problems as the square well, the harmonic oscillator and the hydrogen atom (or, in other words, any example you've ever seen in an introductory class, and many others).

9. Dec 6, 2006

### Parlyne

10. Dec 6, 2006

### dextercioby

First of all, there's something to say:

$$A\psi = B\psi$$ doesn't automatically mean that A=B.

Next, the Hamilton operator can be unbounded and, since required to be self-adjoint, by Hellinger-Toeplitz theorem, it cannot be defined on all Hilbert space $\mathcal{H}$. Therefore:

$$D(H) \varsubsetneq \mathcal{H}$$.

On the other hand you need to show that $D\left(i\hbar \frac{d}{dt}\right) = D(H)$. Then you can make your claim.

What do you mean by coefficient of $\hat{p}$ ?

Daniel.

11. Dec 6, 2006

### dextercioby

It's pretty obvious that there's no such thing as an "energy operator". QM acknowledges the "energy observable" and to it the operator called "Hamiltonian". End of story.

I also resent calling $$\hat{H}|\psi\rangle = E|\psi\rangle$$ "the time-dependent SE". It should be called by its name: the spectral equation for the Hamilton operator.

Daniel.

12. Dec 6, 2006

### vanesch

Staff Emeritus
I would even go much further: it is only in the case of a quantum system composed of one single point particle that the wavefunction is a function over "space and time". If we have a quantum system corresponding to a classical system (which doesn't even need to be the case!), then it is a function over configuration space and time. It is only in the case of a single point particle that the configuration space is also an Euclidean space with 3 dimensions.
2 particles have already a configuration space of 6 dimensions etc... So the wavefunction of a 2-particle system is already not defined in "space and time".

13. Dec 6, 2006

### vanesch

Staff Emeritus
I think the discussion between dexter and parlyne is one about language: dexter is such a sophisticated physicist that he has forgotten the language of elementary QM texts I will try to provide for a translation...

In elementary treatments, one associates the hamiltonian with the non-relativistic expression of energy (the non-relativistic hamiltonian), and one calls the equation $$H \psi = i \hbar \frac{d\psi}{dt}$$ the time-dependent Schroedinger equation, and the equation $$H \psi = E \psi$$ the time-independent Schroedinger equation.

But when one gets further into quantum theory, one will actually understand that what is essential, is that there is a unitary time evolution over hilbert space, which is usually written as $$U(t_1,t_0)$$. That is, there is a set of unitary operators which transport the state (element of Hilbert space) at time t_0 to the state at time t_1. This is understood as well in quantum field theory as in non-relativistic QM. Moreover, this set of operators U is a representation of the set of time translations (the real axis). It has hence a Hermitean generator, and that is nothing else but the Hamiltonian: the hamiltonian is the generator of time translations, which is then what dexter wrote about.

What is sometimes called the "time independent Schroedinger equation" is in fact nothing else but the spectral equation of the hamiltonian. Its solutions are the stationary states of the system (those states that do not change under time evolution, and for which their hilbert representation takes a time-dependent phase factor).

d/dt is not strictly an operator over hilbert space, given that the elements of hilbert space itself are not time-dependent.
(in the same way that the points in configuration space in classical physics are not time-dependent).

14. Dec 6, 2006

### Severian

Lol! No he isn't. It is quite clear that he is a mathematician, not a physicist. Complaining that there is no such thing as $$\hat E$$ just because he is used to writing $$\hat H$$ demonstrates that nicely.

15. Dec 6, 2006

### Aero

When you try to "solve" a problem, you are trying to find the wavefunction after a certain time t, given the initial wavefunction, right? Why does finding the energy eigenstates help? In other words, why do we only look at "steady state" problems?

Suppose I wanted to model a particle moving under a constant force, starting at rest. The potential would be V = -Fx, and so the SE would be:

$$i\hbar \frac{{\partial \psi }}{{\partial t}} = - \frac{{\hbar ^2 }}{{2m}}\frac{{\partial ^2 \psi }}{{\partial x^2 }} - Fx\psi$$

How do you solve this? I tried separation of variables, and got a second order non-linear differential equation that looks soluble but that I don't know how to solve. But why does separation of variables find all solutions anyway?

16. Dec 6, 2006

### vanesch

Staff Emeritus
Because if you know how to write your initial wavefunction as a linear combination of energy eigenstates at a certain time, then you know that after a later time, it will be the same linear combination of the evolved energy eigenstates. And the "evolved energy eigenstates" are the same eigenstates, but simply with a complex phase factor in front of it.

17. Dec 7, 2006

### dextercioby

Yes.

For simple problems, with the Hamiltonian being a constant of motion, i.e. explicitely time-independent in the Schroedinger picture, the answer has already been given by Vanesch. For more complicated problems, with the Hamiltonian being explicitely time-dependent, the logics is basically the same. It's either one tries to write the time-dependent part as a perturbation and do perturbation theory, or simply compute the time-evolved state vector using Dyson formula. But anyhow, the (possibly generalized) eigenstates of a time-independent Hamiltonian are useful. The expansion of an arbitrary $|\psi \left(t_{0}\right)\rangle$ in terms of the complete set of (possibly generalized) Hamiltonian eigenstates is possible due to the spectral theorem.

Why nonlinear ?

Daniel.

18. Dec 7, 2006

### masudr

I share your resentment. I also hate the fact that the notation implies that this equation holds for any state vector. Of course this is completely untrue, it only holds for eigenstates of the Hamiltonian; which implies the equation anyway.

Last edited: Dec 7, 2006
19. Dec 7, 2006

### masudr

Assume that F acts on the space-part of the wavefunction only, this may help.

We assume that the wavefunction can be written as a product of space and time parts understanding that if it can't then the diff. equation will throw out some inconsistency. Instead, we find that the equation doesn't, and so our inital assumption was indeed correct.

P.S. The quote function appears to be attributing this post by Aero to dextercioby; I had to manually modify this post for that reason.

Last edited: Dec 7, 2006
20. Dec 7, 2006

### dextercioby

I assumed F was constant, and would simply be the classical "force". So the integration of the ODE is simpler.

Daniel.