# What makes matrix mechanics matrix mechanics?

1. Aug 8, 2012

### Einstein Mcfly

Hello all.

I haven't worked hard enough yet (reading the original papers etc) to really get at the differences here, but I thought I'd ask the local experts if there was more to it than I already know. From what I've gathered so far, Heisenberg's formulation dealt only with matrix representations of operators expressed in some basis and derived observables from them (I assume by diagonalizing them). There were no "orbitals" in the sense of functions defined in space with particular "shapes", because all he cared about was finding a way to get the spectra correct. If I'm not right up to this point, please correct me....

So, now when I'm, say, doing quantum chemistry and I have the fock operator or some such thing and I'm diagonalizing it to get orbitals and energies and doing the rinse and repeat needed to converge, and I doing "matrix mechanics"? Is phrasing the problem in terms of matricies rather than a differential equation all it takes, or am I just using the algebra of matricies to do Schrodinger wave mechanics? I know that they're necessarily equivalent, but its the differences in perspective that I'm interested in.

2. Aug 9, 2012

### ardie

in the Heisenberg picture, the states are constant in time, and the evolution is carried on by the operators, thereby allowing you to put the equations in tensorial form (matrix), as the gradient of a state tensor is always of the second rank, this simplifies the calculations provided you know the state tensor and the initial conditions.
In the Schrödinger picture, one assumes the operators of the system are constant's in time. For example a bound particle that may not escape in the course of a measurement. So long as the Hamiltonian remains the same, the states (the solutions) will remain the same. But in this case the Hamiltonian is time changing, which means the states will also change accordingly. Now you may combine these 2 tensors to form the tonsorial (matrix) object that you need to solve, and it is a well known theory in linear algebra, it's slightly less intuitive and uncommon to use this method though.
So everyone has stuck to the way Heisenberg solved things because it was already very common among mathematicians to work in vector algebra.

3. Aug 9, 2012

### Amok

What? Heisenberg picture is barely mentioned in many standard QM books. In fact, when Heisenberg came up with his theory no one in the physics community knew what matrices were (not even him). The Schroedinger picture, on the other hand, involed tons differential equations
which everyone knew well. That's why the Schroedinger picture stuck.

4. Aug 9, 2012

### Einstein Mcfly

I learned these pictures in class, but it was always just a matter of whether you folded the time evolution operators into the states (SP) or the operators (HP). My question was whether or not solving the problem using matrix algebra for a time-independent case (variationally minimizing the energy by diagonalizing the fock matrix in a self-consistent procedure, for example) is what is meant by Matrix Mechanics or am I just using matricies to do Schrodinger wave mechanics.

Is this a proper question? Is it clear what I'm getting at? There have been a lot of reads on this thread but only a few replies....

5. Aug 9, 2012

### Amok

I don't see why you'd be using matrix mechanics. All of quantum chemistry is based on the Schroedinger equation, and in fact HF theory is about of finding an approximate solution to the (time-independent)Schroedinger equation for a many-electron atom (or molecule). Just because you writing everything down using a finite basis-set so you can solve it on a computer doesn't mean you're doing matrix mechanics (I think), everything is still derived from the SE.

6. Aug 9, 2012

### Einstein Mcfly

I think you're seeing where my confusion comes in. Obviously solving these problems using matrix mechanics is 100% equivalent to solving the SE, so what is the real difference IN THE PROCEDURE that makes it one or the other?

7. Aug 9, 2012

### ardie

The Heisenberg picture is the one that is physically realisable in real experiments. In reality the particle is moving through some potential which leads to changes in it's basis and we can do experiments to find out what that potential looks like. we can make some guesses about what the state looks like at the beginning and thereby manage to solve the problems at hand for the final states. The schrodinger picture is almost never used, unless you can manage to lock the system into a standstill (time-independent cases)

8. Aug 9, 2012

### Amok

Almost never used by whom?

9. Aug 9, 2012

### Naty1

Einstein: Seems like you have the background to be able to read Wikipedia's Heisenberg MATRIX MECHANICS and then SCHRODINGER EQUATION and maybe answer

for example, the former has this description:

If this example is helpful, check out the articles. [I'd answer
you question if I could!!]

10. Aug 9, 2012

### genericusrnme

The schrodinger picture and the heisenberg picture are entirely equivelant...

11. Aug 9, 2012

### Einstein Mcfly

Yes indeed, I already read through those. They didn't really clarify anything vis a vis what is DONE functionally in a matrix mechanics calculation.

From the responses in this thread, it seems like I'm not the only one who's confused about this.

12. Aug 9, 2012

### Jorriss

I wouldn't say never, but practicing chemists definitely prefer matrix mechanics.

13. Aug 9, 2012

### Einstein Mcfly

It's great that you're responding so confidently because I'm a practicing quantum chemist and haven't know if what I'm doing was MM or SWM. So what I do in QC calculations is choose some basis functions, construct linear combinations of them, store the coefficients in a matrix, diagonalize that matrix to get molecular orbitals, rinse and repeat until the energy reaches a minimum (hopefully) and stops changing.

You're saying that what I'm doing here is matrix mechanics. Now here's the big question: If I wanted to do the same thing using the Schrodinger wave picture, what would I do instead?

14. Aug 9, 2012

### Mark M

This certainly isn't true, the two are physically equivalent. The following is from page 3 of Zettlli's textbook:

15. Aug 10, 2012

### ardie

diagonalising a matrix is equivalent to rearranging its eigenbasis such that they are all linearly independent, and thereby equivalent to solving a set of differential equations coefficients to find unique solutions for each one.
for more information you may refer to a mathematical methods for physics and engineering, I recommend the Cambridge version which has a whole section on eigenfunction methods for differential equations. A differential equation of rank n can always be solved provided n coefficients are given, that should be a clue. in your methods you first choose a guess eigenbasis and insert in the matrix, if the matrix can be diagonalised then you have found the unique solution, else you try a different combination. So you skip over having to solve each differential equation to find the exact coefficient of the eigenbasis, because you simply do not have all the information about the system, correct?

Last edited: Aug 10, 2012
16. Aug 10, 2012

### Naty1

I've been out of school WAAAAAAAY too long to be able to dive into such details....so they remain 'above my paygrade'....

But I like your question and have wondered about the equivalency of the formulations beyond the general for some time. The extraction of physicality from mathematical formulations provides a lifetime of interesting issues.

17. Aug 10, 2012

### kith

Maybe this helps a bit. If I have a system with initial state vector |ψ(0)> and want to calculate expecatation values for another time t, I can do this in two different ways:
1) Apply the time evolution operator to the state and get a new vector |ψ(t)> = U(t)|ψ(0)>. From there I can calculate expectation values by <ψ(t)|A|ψ(t)>.
2) Apply the time evolution operator to the observable/matrix and get a new observable/matrix A(t) = U+(t)AU(t). From there I can calculate expectation values by <ψ(0)|A(t)|ψ(0)>.

In terms of equations, (1) corresponds to solving the Schrödinger equation, which is an equation for state vectors. (2) corresponds to solving the Heisenberg equation, which is an equation for observables/matrices.

18. Aug 10, 2012

### vanhees71

The important point is that all pictures of time evolution are unitary equivalent by definition. This ensures that the physical meaning of the quantum theoretical formalism is independent of the choice of the picture. Neither state-reprsenting vectors nor eigenvectors of operators, representing observables, are measurable quantities but only the probabilities
$$P_{\psi}(a)=|\langle a| \psi \rangle|^2.$$
The time dependence of the mathematical quantities $|\psi \rangle$, $\hat{A}$ (and following from this $| a \rangle$ can be quite arbitrarily chosen. Usually it depends on the problem you want to solve, which choice is best. For the final result it doesn't matter at all.

The wave function in the "a basis", $\psi(a,t)=\langle a|\psi \rangle$ is, by the way, a picture independent quantity.

19. Aug 10, 2012

### Fredrik

Staff Emeritus
The Heisenberg picture and the Schrödinger picture have very little to do with the difference between matrix mechanics and wave mechanics. (Edit: "Wave mechanics" is what some people call the quantum theory of a single spin-0 particle influenced by a potential, i.e. the theory that's taught in introductory classes. I think it's essentially the same as Schrödinger's theory).

My understanding* is that Schrödinger was working with the semi-inner product space of square-integrable functions, while Heisenberg was working with an inner product space of "column vectors with infinitely many rows". (Yes, this seems rather ill-defined, but Heisenberg wasn't trying to do rigorous mathematics).

*) I could be wrong, since I haven't read any of the original sources. I'm really just guessing based on what I know about Hilbert spaces and the stuff that's being taught in introductory QM classes.

If you define two members of Schrödinger's inner product space to be equivalent if they differ only on a set of measure zero, and then consider the set of equivalence classes, you can easily give it the structure of an inner product space. This inner product space turns out to be a separable Hilbert space.

The rows of Heisenberg's "∞×1 matrices" can be thought of as the components in some specific orthonormal basis, of a vector in a separable Hilbert space.

This is the sense in which von Neumann's Hilbert space approach "unifies" the two original approaches by Heisenberg and Schrödinger.

The two "pictures" on the other hand are only different opinions about whether the time dependent factors in an expression like $\langle\alpha|e^{iHt}A e^{-iHt}|\alpha\rangle$ should be considered part of the state vectors or part of the operators, i.e. should we say that $|\alpha\rangle$ is the state and $e^{iHt}A e^{-iHt}$ is the observable (Heisenberg picture) or should we say that $e^{-iHt}|\alpha\rangle$ is the state and $A$ the observable (Schrödinger)?

Last edited: Aug 10, 2012
20. Aug 10, 2012

### Einstein Mcfly

This is of course true. This is the method of solving systems of differential equations that I learned in my "differential equations with linear algebra" class my sophomore year of college. This (the use of matrix algebra to solve differential equations where each eigenvector of the fock operator is an "orbital" with a particular energy) I had always thought was just the simplest way of doing "Schrodinger wave mechanics" because I was THINKING of the orbitals in the way that Schrodinger/Born specified. My question was, is this actually "Matrix Mechanics" because I'm using matrix algebra and, if not, how WOULD I solve these problems (or any other) with matrix mechanics.