# Is matrix mechanics better suited for relativity than wave mechanics?

1. Mar 8, 2013

### snoopies622

"The Heisenberg picture does not distinguish time from space, so it is better for relativistic theories than the Schrödinger equation", says Wikipedia's entry on Matrix Mechanics.

Since the Heisenberg equation of motion

$$\frac {dA}{dt} = \frac {i}{\hbar} [H,A] + \frac {\partial A}{\partial t}$$
contains time derivatives but no space derivatives, how is this not distinguishing time from space?

2. Mar 8, 2013

### dextercioby

You realize <matrix mechanics> is not the same thing with the Heisenberg picture, right ?

Next, the time-dependence of operators in the H. picture in the absence of special relativity is transfered to the time (together with space) dependence of quantum fields (which are operator valued distributions) in the presence of special relativity in which time and space are interchangeable.

3. Mar 8, 2013

### Bill_K

Quoting Bjorken and Drell,

4. Mar 8, 2013

### Darwin123

"Matrix mechanics" is not equivalent to the "Heisenberg picture". "Matrix mechanics" refers to a mathematical technique. The "Heisenberg picture" refers to a convention in quantum physics.

Matrix math can be used in any branch of mechanics. You can use matrices in classical mechanics, general relativity and quantum mechanics.

The "Schrodinger picture" can be analyzed with matrices as easily as the "Heisenberg picture". In fact, the Schrodinger equation can be expressed in matrix form as easily as it can be expressed in differential form. The expression for the Schrodinger equation usually given to sophomore physics students is in the form of a differential equation. However, it could also be expressed in the form of a matrix equation.

Both the "Schrodinger picture" and the "Heisenberg picture" have equations where an operator acts on a vector. There is a choice of different mathematical ways to represent the operators and vectors can be written. The operator can be matrices and the vector can be n-tuples. The operator can be differentials and the vector can be functions.

The "Schrodinger picture" and the "Heisenberg picture" differ as to where one places the variation in time. They do not really differ in terms of the mathematical formalism used.

The difference between the "Schrodinger picture" and the "Heisenberg picture" regards where to place the time variation. The equations in either picture can be expressed in terms of either matrices or differential equations.

The Schrodinger picture places the variation in time in the vector. The operators are invariant to time. The vectors can vary in time. In the Schrodinger picture, the equation for energy is often written as a differential equation. This is a convention, not a strict rule. If you look carefully at this expression, you will see that the coefficients of the differential equation do not vary in time. The function often varies in time. However, this same equation could have been written in matrix form. The Hamiltonian in matrix form would also be invariant in time. The n-tuple would have varied with time.

The "Heisenberg picture" places the the variation in time in the operator. The vector does not vary in time. In the Heisenberg picture, the equation for energy is often written in matrix form. This is a convention, not a rule. The same equation could have been written as a differential equation where the coefficients vary with time. The n-tuple in this case would not have varied in time.

I never worked on both general relativity and quantum mechanics at the same time. However, I will provide a well educated guess as to why someone would prefer the Heisenberg picture to the Schrodinger picture. I have solved a lot of problems using both matrix equations and differential equations together. What I generally find is a problem where the initial conditions and boundary conditions are given. The problem is to find the solution at times later than the initial time and points that are not at the boundary.

A scientist working with general relativity would also like to separate both boundary and initial conditions from the solution to the problem. Here, the Heisenberg picture may be more convenient than the Schrodinger picture.

The scientist working with general relativity usually wants the vector to be invariant with both space and time. The scientist working with general relativity does not want the operator to vary with space and the vector to vary with time. Therefore, this scientist would prefer the Heisenberg picture over the Schrodinger picture. He would want all properties that vary with the observer in the vector and all properties that do not vary with the observer in the operator.

Time in the Heisenberg picture is being placed on the same footing as space in that it is in the operator. He would not really care whether the equation is in matrix or differential form, as the problem may be as difficult either way. However, he wants to separate the "given" parts of the problem from the "unknown" parts of the problem. He wants the given parts of the problem

Suppose one wanted to write the equation of energy in differential form but in the Heisenberg picture. One would write a "differential equation" where the coefficients vary in both space and time. The function can incorporate both the initial conditions and the boundary conditions. The function itself would not have to incorporate forces or interactions. The effect of forces and interactions would be taken into account in the coefficients of the differential equation.

The equivalent statements can be made about a matrix form of the equations in the Heisenberg picture. The solution would be in the matrix. The n-tuple would contain both initial conditions and boundary conditions.

5. Mar 8, 2013

### The_Duck

In the Heisenberg picture of QFT, we have operators that depend on space and time, $\phi(x)$, where $x$ is a spacetime point. Then the equation of motion can be generalized to include space as well as time. In special relativity we have an energy-momentum 4-vector that contains both the energy operator (the Hamiltonian) and the spatial momentum operators: $P^\mu = (H, P^x, P^y, P^z)$. We can use this to write the operator equation of motion in a way that puts time and space on equal footing:

$[P_\mu, \phi(x)] = \frac{d \phi}{d x^\mu}$

The time component of this equation is the operator time evolution equation:

$[H, \phi(x)] = \frac{d \phi}{d t}$

and the space components give a relation between commutators of the momentum operator and spatial derivatives, e.g.:

$[P_z, \phi(x)] = \frac{d \phi}{d z}$

6. Mar 8, 2013

### snoopies622

Thanks all, looks like much to think about here when I finally have some free time later tonight.

7. Mar 11, 2013

### A. Neumaier

The covariant form of the Heisenberg equation of motion is
$$\frac {dA}{dx_i} = \frac {i}{\hbar} [P_i,A] + \frac {\partial A}{\partial x_i},$$
where $$P_i$$ is the i-th component of the momentum operator. Since $$H=P_0$$, time is not distinguished in any special way.

I do not share the other responent's reservations about euqating matrix mechanics and the Heisenberg picture. The lattter is just the modern continuation of the former; the term ''matrix mechnaics'' is simply no longer used. Once you discretize the Hilbert space (as in lattice gauge theory) you are back to old matrix mechanics, though the modern terminology and conceptual context makes it not so easy to recognize.

8. Mar 11, 2013

### snoopies622

So . . in the Heisenberg picture the vectors are constant and the operators change, and in the Schrödinger picture the operators are constant and the vectors change.

Aren't the components of both Heisenberg's matrices and Schrödinger's vectors functions of space and time?

9. Mar 11, 2013

### A. Neumaier

It depends on the way the Hilbert space is represented. But a common situation is the following:

In the relativistic Heisenberg picture, the state typically is written in terms of multi-momentum coordinates with values in a mass shell, whereas the operators are functions of space-time.

In a comparable nonrelativistic Schroedinger picture, the state would be a function of
time and multi-momentum coordinates with values in R^3, whereas the operators are functions of space only.

10. Mar 11, 2013

### snoopies622

Not perfectly clear yet, but almost. I guess the essence of my question is, in what way does the non-relativistic Heisenberg picture have a space / time equivalence that the non-relativistic Schrödinger picture does not?

Is it that in the Schrödinger picture the vector varies with time and the operators are functions of space, while in the Heisenberg picture the operators are functions of both time and space and the vectors are simply lists of invariant numbers?

11. Mar 11, 2013

### Darwin123

I think that is it.

12. Mar 11, 2013

### Darwin123

You can discretize the Hilbert space just as easily in the Schrodinger picture as with the Heisenberg picture. The set of equations in each picture will be matrix equations.

I think that you are mixing up historical contingency with mathematical necessity.

A
1) Heisenberg formulated his version of quantum mechanics in terms of matrix equations and what is now called the Heisenberg picture.

2) Schroedinger formulated his version of quantum mechanics in terms of differential equations and what is now called the Schroedinger picture.

3) Both formulations are sufficient for solving quantum mechanical problems.

It took about 20 years before scientists realized that that the two formulations are mathematically equivalent. However:

B
1) It is not necessary to use matrices with the Heisenberg picture.

2) It is not necessary to use differential equations with the Schrodinger picture.

3) One can formulate a quantum mechanical problem using both the Heisenberg picture and differential equations.

4) One can formulate a quantum mechanical problem using the Schrodinger picture and matrix mechanics.

As an example of B4, think about how the Schrodinger equation is usually modified to include spin. Usually, the Schrodinger equation is written as an equation that is a hybrid of matrix and differential forms. The space-time parts are in differential form while the spin parts are in matrix form.

Another example of B4 is the way introductory textbooks write the Dirac equation. The spin/parity parts are written up in matrix form while the space-time parts are in differential form.

Both the Schrodinger equation and the Dirac equation are usually written in the Schrodinger picture. However, they don't have to be. Both can be written in the Heisenberg picture.

I just realized this. The Schrodinger equation can be written in the Heisenberg picture!

13. Mar 12, 2013

### snoopies622

Yes, I believe you. My question is a reaction to a couple things:

1. I was looking in Dirac's Lectures On Quantum Field Theory and almost right away came across the quote,

" . . the Heisenberg picture is a good picture and the Schrödinger picture is a bad picture and the two pictures are not equivalent."

and then

2. that quotation from the Wikipedia essay,

"The Heisenberg picture does not distinguish time from space, so it is better for relativistic theories than the Schrödinger equation."

So I was wondering how the two formulations were different, specifically this alleged space/time equivalence that exists in the non-relativistic Heisenberg picture but not the non-relativistic Schrödinger picture.

If vectors in the Heisenberg picture really are just lists of numbers and not functions of either space or time, than I think I understand.

14. Mar 12, 2013

### atyy

My understanding is along the lines of Bill K's quote from Bjorken and Drell. In quantum field theory, the wave function becomes a wave functional - it is not a function of space, but a function of a field configuration in space.

Last edited: Mar 12, 2013
15. Mar 12, 2013

### Darwin123

If that is true, then I don't understand the difference. I may have been wrong all along.

I thought that was it. According to what I was taught, the vectors in the Heisenberg picture don't vary in time.

However, that quote by Dirac makes me wonder if I should even have replied to this thread. I will stop before I get dinged again.

16. Mar 12, 2013

### ftr

17. Mar 14, 2013

### A. Neumaier

They don't vary in time, but they are still functions of (multiple) positions or momenta.

The Heisenberg and the Schroedinger picture are fully equivalent in nonrelativistic N-particle quantum mechnaics. But in relativistic quantum field theory there are subtle issues that make the Heisenberg picture much better behaved. The reason is that unlike the Schroedinger picture, the Heisenberg picture does not require a fixed representation, hence allows for things such as superselection sectors.

- Comment to an earlier post: One cannot avoid differential equations in the Schroedinger picture.
And the use of matrices in the Schroedinger picture doesn't make the latter matrix mechanics in the sense this now essentially obsolete notion had at the time it was stille frequently used.

18. Mar 14, 2013

### strangerep

Oh, c'mon on. Chin up! Sometimes this must be the way we find things out. The subtleties involved in this point are very interesting, and (probably) quite important if humans are ever to achieve a rigorous QED.

There's a paper by Dirac: "QED without Dead Wood", which is (very) cutdown version of what's in his QFT lecture notes. If you have trouble obtaining the lecture notes, try googling

That's google, not google scholar. Send me a PM if you have difficulty...

19. Mar 16, 2013

### snoopies622

Last edited by a moderator: May 6, 2017
20. Mar 17, 2013

### snoopies622

If in the Heisenberg picture the elements of a vector are functions of position and the elements of the matrix are functions of time, then I am led back to the question I posed in entry #10: In what way does the non-relativistic Heisenberg picture have a space/time equivalence that the non-relativistic Schrödinger picture does not?

21. Mar 17, 2013

### A. Neumaier

After a Fourier transform of the basis, the N-particle states in the Heisenberg picture are functions of N momenta, while the operators are functions of a single position and time. This is the way things are usually described in field theory.

In the nonrelativistic case, the momenta are taken to be 3-momenta, and in the relativistic case, these 3-momenta become curved 3D mass shells in a 4D space of momenta. Only then one can see the space-time equivalence (which is of course absent in the nonrelativistic case).