# Very basic questions - Hamiltonian

1. Jun 26, 2008

### Xezlec

After several failures in the past (why does the universe have to be so complicated?!), I'm once again trying to learn to understand the basics of QM, out of sheer frustration with not knowing what the heck physicists are talking about all the time. I know, I still have a long way to go.

I'm led to understand that the rules governing the time evolution of the quantum state, together with the definitions of the observable quantities (in the form of their associated operators) take the place of Newton's laws of motion and the classical definitions of the quantities. Is that about right?

If so, why can't I find a list of the definitions of the operators of all the usual physics quantities written somewhere? I've found the operators for momentum, position, and spin, and they seem to make sense to me (with what little I know), but I can't find the definition of the Hamiltonian, which is needed for me to know how energy is defined in quantumland. More importantly, since the entire rules for how stuff happens are encoded in the SchrÃ¶dinger equation, which relies on the undefined Hamiltonian, I can't imagine (or compute) how anything happens at all.

Surely it has a definition somewhere? I mean, it can't just be that I make one up... thereby making up any laws of motion I want for my universe...

Thanks, and apologies for my utter n00bity.

2. Jun 26, 2008

### dx

The energy of a particle in quantum mechanics is the same as in classical mechanics :

$$H = \frac{p^2}{2m} + V(x)$$.

Just replace the classical quantities with the respective operators, and you get the hamiltonian operator.

3. Jun 26, 2008

### cmos

All classical dynamical variables may be defined in terms of position and momentum. Therefore you have already found what you need.

Regarding the Hamiltonian.....
dx has shown you the general form of the Hamiltonian, where the first term is simply the kinetic energy (again, define in terms of momentum as mentioned above) and the second term is the potential. In a sense, you do get to "make up" your Hamiltonian. The potential is determined by you, God, and/or may be inherent to your system.

For example, in the case of atomic hydrogen (one proton, one electron) the potential is simply the Coulomb potential. If you would like to set up your own crazy experiment with a charged particle placed off-center in an electric octupole, then you will have a more involved potential making for a more involved Hamiltonian.

4. Jun 26, 2008

### Xezlec

Thanks. But how/why can I replace quantities with operators? As I understood it, the relationship between a quantity and the operator is complicated and indirect: the quantity lives in the eigenvalues of the operator. I wouldn't expect to be able to add two operators in general and get the operator corresponding to a quantity that is the sum of the quantities represented by those two operators... would I? *brain smokes*

And I guess the operator for that potential V(x) is just the operator that multiplies V(x) by phi(x) for all x... that appears to work out right.

5. Jun 26, 2008

### cmos

Not exactly sure what you mean by "quantities." Physical observables are represented by operators. All of your classical observables may be defined by position and momentum, so that's all you need (spin is another story, but that doesn't have a classical counterpart).

What you might be getting tripped up on is the fact that operating the Hamiltonian on the wavefunction yields an eigenvalue equation. Well this is only true when you take out time dependence, hence the time-independent Schrodinger equation. Well, this is actually DERIVED from the more general time-dependent Schrodinger equation. From this derivation we see that the time-independent wavefunctions are actually energy eigenfunctions.

The same does not hold true for either momentum or position. The reason for this is that you may not have a determinant state of either momentum or position (a la Heisenberg). So your eigenvalue equations will not exactly behave in the simple manner as we have discussed above.

6. Jun 26, 2008

### pmb_phy

I believe that this is not something that can be proved through an arguement or derivation. I think that this is embedded in the postulates of quantum mechanics.
Only if there is a corresponding classical equivalent.
The x in V(x) is replaced with the operator X. This operator is defined such that X|$\Psi$> = x|$\Psi$>. V(X) is defined by its series expansion in the operator X.

Pete

7. Jun 26, 2008

### Xezlec

I was trying to put together another response, but I think my understanding of math is just not up to the task, which I think points to the fundamental problem. Maybe I still don't even understand the framework of the basic formulation of the theory. Can anyone recommend maybe what kinds of mathematics I should study to learn enough to approach basic QM? Any recommendations for URLs or, less desirably, books that might be a good first step?

8. Jun 26, 2008

### dx

You can study elementary QM using just the Schrodinger equation if you know some basic calculus and differential equations. If you want to study the full theory of QM you need to know at least advanced calculus and linear algebra (and also a good amount of physics to appreciate it fully).

9. Jun 26, 2008

### Xezlec

OK, well I think my vector cal is pretty good and my linear algebra is decent and improving, so I would think I should be able to understand this. But I guess I'm not getting the concept of an observable or an operator or something because so far it isn't making sense and I'm not even effectively communicating what the problem is.

10. Jun 26, 2008

### dx

If you're comfortable with vector calculus and linear algebra, you should be able to start learning quantum mechanics.

In classical mechanics, systems have states which are points in a set. For example, a particle on a line has a position which is an element of R. The observables in classical mechanics are basically just the states, or numbers representing the states. In quantum mechanics, these points in sets are replaced by vectors in a vector space. The observables of quantum mechanics are not the states, but operators on this vector space. It's normal to feel a little uncomfortable with this at the beginning, but I suggest you not worry about this and go ahead and learn quantum mechanics. You can worry about what exactly it means and how you interpret it after you learn it.

11. Jun 26, 2008

### Xezlec

Wait, you're saying the observable somehow *is* the operator? I don't even get wat that means... I need a better book.

12. Jun 26, 2008

### Xezlec

You know what, never mind. I'm chalking this up as another abortive attempt. I'll wait a few years and try again. Sorry about this.

13. Jun 26, 2008

### dx

No I'm sorry, I wasn't clear. Let's consider a particle moving on a line. In classical mechanics, it is assumed that at every instant of time t, the particle has a certain position on the line x. That is it's state at time t. In quantum mechanics, the thing that is known at every instant of time is not its position, but something else. This is the state vector. This is a complex function defined on the line: $$|\psi(x)>$$. That is the analog of the classical state.

Now I will give a certain interpretation, which is not accepted by everyone. You can think of this state as somehow representing your knowledge about the particle. This is loosely just a probability density on the line. Now there are certain states which correspond to a certain knowledge that the particle is at a particular position. These are the eigenstates of the position operator $$\hat{X}$$. In the case of a particle moving on the line, they are the Dirac delta functions. There are also momentum eigenstates, which are "states of knowledge" where the momentum is exactly known. These are the eigenvectors of the momentum operator which are

$$e^{i \frac{p}{h}x}$$.

If your "state of knowledge" is one of the momentum eigenstates, then it means that you know certainly what the momentum will be when you measure it.

14. Jun 26, 2008

### comote

Given a classical observable f(q,p) there is a self adjoint operator F(Q,P) acting on Hilbert space that represents the expected outcomes.

For simple cases
$$H = \frac{p^2}{2m} + V(q)$$.
this is easily understood. Unfortunately what to do for the
$$q^np^m$$
terms is a little more complicated and not fully understood. A common formula is to transform
$$qp\to\frac{1}{2}(QP+PQ)$$.

15. Jun 26, 2008

### dx

No! don't give up. Quantum mechanics is fascinating. Maybe you're just not reading the right books or something. I suggest the first few chapters of Feynman lectures vol 3 for an introduction. Also, Leonard Susskind (one of the founders of string theory) has a excellent set of video lectures on quantum theory you can find on youtube. Then try J. J. Sakurai's book if you are confident about your maths. Otherwise, try Shankar or Griffiths.

16. Jun 27, 2008

### Fredrik

Staff Emeritus
One of the postulates of QM is that states are represented by vectors in some vector space. A vector is often written using the "ket" notation: $|\alpha\rangle$. If you describe the system as being in state $|\alpha\rangle$, then an observer who's translated in time relative to you would describe the system as being in another state $|\alpha'\rangle$. There must be an operator U(t) that takes $|\alpha\rangle$ to $|\alpha'\rangle$, and it must be an exponential because it has to satisfy U(t+t')=U(t)U(t'), so we can write it as $U(t)=e^{At}$, where A is some operator. (Its existence is guaranteed by the condition U(t+t')=U(t)U(t')). It's convenient to choose U(t) to be unitary so that it doesn't change the norm of the state vectors it acts on. The requirement that U(t) be unitary implies that A must be anti-hermitian (i.e. $A^\dagger=-A$). We prefer to work with hermitian operators because their eigenvalues are real, so let's define H=iA. H is now hermitian, and $U(t)=e^{-iHt}$.

That's the definition of the Hamiltonian. It's the generator of translations in time. The momentum operators can be defined the same way as generators of translations in space, and the spin operators can be defined as generators of rotations.

17. Jun 27, 2008

### Fredrik

Staff Emeritus
Linear algebra is by far the most important kind, but you can probably just study those parts you need when you run into difficulties in your QM book. You obviously have to understand complex numbers too.

Yes, an observable is an operator.

18. Jun 27, 2008

### Xezlec

OK, let's keep going I guess. Most of what's being said here are things I already understand, or at least think I understand. It's the details that are confusing me.

Let me try stating the path I've followed and see if there are any objections. Maybe this will let me debug my brain glitch.

1. The state of a particle is described by a set of complex numbers called a wavefunction. From one point of view, there can be said to be N complex numbers for each point in space, where N is the number of legal spin states. In this representation (the position/spin eigenstate basis) the probability of observing the particle with a particular combination of spin and position equals the squared magnitude of the complex number for that particular combination. Spin and position are therefore two quantities that the wavefunction tells us something about.

2. The procedure for extracting information about other quantities (that is, quantities other than position and spin, which were very convenient in this basis) is more involved. To find out what the probability is of measuring the system to have a value p for a quantity (say, momentum), I have to solve the equation P Phi = p Phi, where Phi is an unknown vector of zillions of complex numbers and P is the operator corresponding to the quantity in question (momentum in this case). The result will be a wavefunction Phi that is the eigenstate of the system for the momentum value p. So now I take the inner product of Phi with the actual wavefunction and the result will be the probability of measuring the particle to have a momentum of p.

*catches breath*

3. Now that our wavefunction has some defined connection to measurable quantities, we have to find the rules for how wavefunctions behave. These rules are defined by saying that every wavefunction is a superposition of energy eigenstates (obviously). In each of those eigenstates, the components all keep the same magnitude with time, but their phases all spin around (counterclockwise!) at a frequency proportional to the value of energy for this eigenstate (that is, the eigenvalue).

4. But wait! This doesn't help us yet! In order for that to have any bearing on what happens to the observable quantities in time, we need to know what the energy eigenstates look like in terms of the basis we were using at the beginning (the position/spin eigenstate basis), and that can only be done if we know the energy operator so we can solve an equation as in #2. So apparently the total energy operator is related to the momentum operator the same way that the energy quantity/eigenvalue/<insert jargon here> is related to the momentum <whatever>. And that is true because...?

19. Jun 30, 2008

### Domenicaccio

I have vague memories but...

1) express the quantity/observable in terms of coordinates x, y, z and momentum $$p_x$$, $$p_y$$, $$p_z$$

2) the operator corresponding to a simple coordinate is the multiplication by such coordinate

3) the operator corresponding to a momentum is $$-i \frac{h}{2\pi} \frac{\delta}{\delta x}$$

Therefore for example, if you need the operator for kinetic energy

$$T=\frac{p_x^2+p_y^2+p_z^2}{2m}$$

then you substitute each p with the corresponding

$$-i \frac{h}{2\pi} \frac{\delta}{\delta x}$$,

square it to

$$-\frac{h^2}{4\pi^2} \frac{\delta^2}{\delta x^2}$$

and the final operator will be

$$-\frac{h^2}{8m\pi^2} \nabla^2$$

20. Jun 30, 2008

### Xezlec

Interesting. That derivation appears to implicitly assume that an operator applied twice corresponds to a quantity that is the square of the quantity represented by the original operator. Add that to the list of fascinating and unintuitive properties of quantum operators that have yet to be explained to me in any form I can understand. The other thing on the list right now is that adding two operators produces the operator corresponding to the quantity that is the sum of the quantites represented by the original two operators.

If there is a trend here, it sounds like it would be that whatever you do to the operators will always "look the same" notationally as whatever happens to the quantities. I wonder how this mysterious connection is achieved, and how far it goes. I'm still thinking about it. Maybe this new information will help me come up with the answer.

I'm trying to gather enough information that I can infer a full list of the basic postulates of QM. That way, I will be able to understand everything else based on how it is derived from those postulates. I don't know yet whether this particular question is related to some more postulates that I don't know, but it sounds interesting enough that it might be.

21. Jun 30, 2008

### Xezlec

OK, I can see why both of the above magical properties of operators are true if it is possible for the system to be in eigenstates of both operators at once. I do not, however, see any particular reason why they are true for operators for which this is not possible. So let me ask another question:

Do the postulates of QM include the postulates of Newtonian mechanics, or can the postulates of NM be viewed as derived from QM?

22. Jun 30, 2008

### nrqed

QM is a deep subject and there are many ways to introduce it. There are many layers of subtleties. Some people like to jump into the most general (and most abstract) formalism. Others prefer to introduce the subject more gently, not worrying about being as general as possible to start with, knowing that with time the student will eventually reach the point where he/she will be ready for the next layer of subtlety.

The most basic approach is what some have mentioned: take the classical expression for the Hamiltonian (NOTE: for some systems, there is NO classical hamiltonian to use as a starting point in which case you must use other principles to guide you. This is the case with spin interactions which have no classical analogue). But let's say you have a classical Hamiltonian to work with. Then simply leave x,y,z as the positions and replace $$p_x, p_y, p_z$$ by their operator expression $$- i \hbar \frac{\partial}{\partial_x}, \ldots$$ and so on. Usually this is straightforward (there could be so-called ordering ambiguities in some exceptional cases but no need to worry about this for now).

Then solve Schrodinger's equation to determine the time evolution of the wavefunction.
You must pick an initial state that satisfied the boundary conditions you have and which is square integrable. Once you have the wavefunction at any time you can answer any physical question like: if I measure the angular momentum at a specific time, what are the possible results I can get and with what probabilities?).

That's all there is to it.

When you include spin, things are a bit more subtle.

23. Jul 1, 2008

### Xezlec

I think it's the student that makes the difference between these approaches. I, for instance, am a detail thinker. I am not often capable of understanding or working with "high-level overviews" or partial descriptions of systems. This has been a big disadvantage for me in a world expressly designed for the other kind of person. I always had trouble reading block diagrams of circuits until I learned to visualize specific gate-level implementations of each block in the picture and mentally superimpose them on top of the blocks. What I am best at is instantly seeing every possible consequence of a system of a small number of absolutely precise, rigid, unambiguous rules.

I feel like the moment I comprehend that very last rule, I will immediately see everything. Thousands and thousands of true statements will quickly rush into mind. The moment before I understand that last rule, I will still see nothing.

My lack of a roaring blast of sudden realizations suggests to me that is not the case.

I don't intuitively see what the rules are for constructing an equation of operators based on an equation of classical variables. I have gathered from other posts that application of an operator is the equivalent of multiplication by a classical quantity, and addition is addition, but these two facts by no means suggest to me the general character of this weird transformation. What is the operator equivalent of integration of a classical quantity, for example? What about exponentiation? Sinusoids? More importantly, how would I figure out what one of these were if I didn't already know?

On top of that, I am not convinced that this procedure preserves the relationships I require to be able to think of this as a transformation of ordinary arithmetic. For example, I can't see how commutativity of multiplication is preserved when it turns into operator application.

I'm sorry, but, your valiant efforts notwithstanding, so far I still understand nothing. :-(

As I've mentioned already, spin is the only aspect of quantum physics I do understand.

24. Jul 1, 2008

### malawi_glenn

Xezlec: Just start with Sakurai modern quantum mechanics. Start on page 1, and go from page to page, and try to fill in the steps.

The first chapter starts with the postulates of QM, states and hilbert space, commutators and observables vs. operators and momentum and position.

The Second chapter deals with time evolution (dynamics), and it is here that the Hamiltonian will appear since it is, as Fredrik pointed out, the generator of time evolution.

The third chapter deals with angular momentum and spin in quantum mechanics.

My explanation to operator:
And operator is a general description of an 'action' you perform on a function (state in QM). An operator can be just a number, or a derivative, a second order derivative, or a combination of those.

The quantum mechanical operators relation to their classical identity dependend on many things, like what representation you have and so on. That are quite advanced topics, which you DON'T have to learn from the beginning in order to appreciate and study QM. Also, one needs quite solid knowledge of classical mechanics in order to 'obtain' their quantum mechanical counterparts.

http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/qm2.html#c1

http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/qmoper.html#c1

As a general example:

Consider state $$|\alpha > = |\vec{x} = (x',y',z'), S_z = + >$$

Now the primes are just labels, they don't symbolize anything with derivatives, it is a common thing, that one lets prime also denote other things than order of derivation.

consider an action of operator $$\hat{x} |\alpha > = x'|\alpha >$$

and

$$\hat{S_z}|\alpha > = +(1/2)\hbar|\alpha >$$

What do you mean by:
"What is the operator equivalent of integration of a classical quantity, for example?"

It is a pretty vauge question, can you clarify? You mean like integrating the kinetical energy of a system over time?

25. Jul 1, 2008

### nrqed

Commutativity of multiplication is in general not [/i] preserved when the classical quantities are replaced by operators. For example replacing XP by operetaors and replacing PX by operators does not give the same result even though the classical expressions are obviously equal (this is a so-called ordering ambiguity). This is because the operators X_op and P_op do not commute.

You know, I had the same frustrating feeling about superstring theory a few years ago. I kept reading and feeling more and more frustrated because I could not understand the BIG picture and see how everything felt together. It seemed as if I had to master supersymmetry, conformal field theory, quantum field theory, topology, differential geometry, etc all at once before I could understand string theory! I was getting very frustrated.

And then I realized that I should not worry about getting everything right away. I simply had to keep reading and to understand little bits by little bits. The trick was to keep reading different books and papers and sometimes to reread them again and again. At first I would understand only 5% of what a paper said. Then I would understand some other pieces from other papers and books, then I would come back and 15% of the same paper now made sense. And I would go back and read otherpapers and books. It was frustrating to read stuff and to not understand most of what was written, even papers who claimed to be pedagogical introduction to the topic. The key point was to not let myself stay stuck on papers that did not make sense to me. I would read them but not try to understand everything at first. It's only after a year of understanding tons of little details that suddenly I reached a critical mass of understanding that was enough to finally start to be able to learn from textbooks. Once you reach a critical mass, there is snowball effect, the more you understand, the easier it is to understand new aspects. But the key point is to reach the critical mass and during that initial period, the key point is not to worry if everything does not fit into place right away! Otherwise there is the danger of getting discouraged and frustrated and learning is no longer fun.

But the good thing about QM i sthat there are many excellent textbooks on it so I would suggest you pick a good one and just work your way throug it, aksing questions here if you are stuck. I would suggest Shankar.