# Hey,guys is ther a thing name after the eigenfunction of the time operator

1. Mar 11, 2004

### peter308

we know that there's position eigenket,P eigenket and energy eigenket
but is there something called the time operator,and the time eigenket
or this is only a terminology that doesnt even exist,it's just my illusion!
thanks gratefully

2. Mar 11, 2004

### outandbeyond2004

I believe in some QM formulations time is treated mathematically like position is (except for a different sign). Hence in these formulations the answer would be yes. I am not familiar with QM, though, so I will watch this topic and see what others say.

3. Mar 11, 2004

### jeff

There's no time operator in the sense of a position operator: time parametrizes the evolution of systems and is therefore just a number, which - unlike position - is constantly changing. But just as there's an operator which generates translations in position - the momentum operator - there's one that generates translations in time, the hamiltonian. It gives the energy of a system and can have well-defined energy eigenfunctions and eigenvalues.

This being the case, it shouldn't be too surprising that there's a relation satisfied by time and energy which looks like the uncertainty principle holding between position and momentum. However, as alluded to above, the relation between time and energy is different than with position and momentum: Unlike the latter, time and energy are not conjugate variables.

Thus this so-called time energy "uncertainty" principle $\tau \Delta E \geq \hslash$ isn't an uncertainty principle in the sense of the one for position and momentum and simply says that the length of time $\tau$ it takes for systems to evolve appreciably must be at least $\hslash/\Delta E$ in which $\Delta E$ is the root mean square of the spread in their possible energies.

4. Mar 11, 2004

### outandbeyond2004

But, Jeff, aren't there special relativity versions of QM? Minkowski put the time coefficient on the same mathematical footing (except for a different sign, as mentioned before) as the space coefficients. Moreover, the time coefficient does not necessarily have to be the parameter? Any affine parameter can be used?

5. Mar 11, 2004

### jeff

Just because in relativity spatial and temporal components can be handled covariantly doesn't mean that time and space have the same physical status. In particular, this is true in relativistic quantum theory, but the situation is more complicated than in quantum mechanics. For example, the position x&mu;(&tau;) and momentum p&mu;(&tau;) of a relativistic particle are parametrized in terms of &tau; which is not automatically the physical time. However, we require that the action associated with the system be invariant under changes in parametrization, and this unphysical degree of freedom must be removed. One way to do this is by making a specific choice of &tau;. In particular, we can choose it so that x0 = t, the physical time. Alternatively, we can keep everything covariant by requiring physical states satisfy a constraint related to the generator of time translations, the hamiltonian.

Last edited: Mar 11, 2004
6. Mar 11, 2004

### outandbeyond2004

"physical time"? I thought that an affine parameter can be the proper time of some physical object. You say that the action be invariant wrt the parameter; I thought that affine parameters are such parameters.

I would agree that we can choose x0 = the time parameter, if it happens to be a physically measurable and reasonable coordinate of the frame in which measurements that you are interested are being made. What about other frames? I guess you first do Lorentz transformations from frame to frame until you are in the frame whose measurements you want. Then do the QM calculations.

It does seem as though I need to understand relativistic QM better.

7. Mar 11, 2004

### jeff

By physical time I mean the parameter t which is actually newtonian time in the non-relativistic limit. Again using the example of a relativistic particle of mass m, it's action can be written

S = - m&int;d&tau; (&part;x&mu;/&part;&tau; &part;x&mu;/&part;&tau;)&frac12;.

Taking x0 = &tau; = t allows us to write this as

S = - m&int;dt (1 - &part;xi/&part;t &part;xi /&part;t)&frac12;,

which in the non-relativistic limit &part;xi/&part;t << 1 is

S ~ (1/2)m&int;dt &part;xi/&part;t &part;xi /&part;t.

On the subject of affine parameterizations, they by definition yield the geodesic equation Ta&nabla;aTb = 0 in which Ta is the tangent to the geodesic, and are unique up to linear transformations, that is, transformations &tau;&prime; = a&tau; + b which change the origin and units of time.

Last edited: Mar 11, 2004
8. Mar 11, 2004

### lethe

you probably want to square the derivative in there, n'est pas?

anyway, i guess i was asking a question along similar lines at some point in the Thiemann strings from LQG thread.

what does it mean to have relativistic quantum mechanics?

the example you showed above, where you break manifest Lorentz invariance by choosing a gauge x0=&tau; is perfectly clear: in this gauge, there no longer reparametrization invariance, no constraint, all the variables in the Lagrangian are independent.

time becomes a parameter, instead of a dynamical variable, and quantization is straightforward. there is no time operator, because time is not a dynamical variable.

but in the fully covariant theory, the dynamical variables are not really independent. if you want to quantize the covariant theory, then x0 is an operator, i think. as such, it may have eigenkets.

what i gathered from Urs is that in this theory, the Hamiltonian is not the conjugate variable to x0, but rather to the affine parameter &tau;

what would then be the conjugate variable to x0? p0, i would guess, but i thought that p0 was the Hamiltonian.

maybe, i am still confused...

9. Mar 11, 2004

### MathNerd

The marriage between quantum mechanics and special relativity gave birth to quantum field theory, in quantum field theory neither the position nor the time are treated as operators and therefore are on the same footing, making quantum field theory Lorentz invarient.

10. Mar 12, 2004

### outandbeyond2004

Let me see if I can sum it up correctly for Peter308, who may be thoroughly confounded by now. (Perhaps I am confused too, ha.) The answer is no, see Jeff's first answer, in the standard (non-relativistic) formulation of QM. However, in QFT, distance and time are on the same footing (neither are operators).

Maybe MathNerd can put the above much better, and at long last we all will attain to wisdom.

11. Mar 12, 2004

### MathNerd

Well you explain the essence of the answer. The answer could be explained in more detail but assuming the thread starter is not well versed in quantum mechanics I think it will just confuse.

12. Mar 13, 2004

### lethe

i think the question was about relativistic quantum mechanics, not quantum field theory

13. Mar 13, 2004

### MathNerd

Relatavistic quantum mechanics is quantum field theory! Quantum field theory solves non-locality issues in non-relatavistic quantum mechanics and enables particle creation and anhilation...

Relatavistic Quantum Mechanics = Quantum Field Theory

14. Mar 13, 2004

### lethe

no

true

not true.

Last edited: Mar 13, 2004
15. Mar 13, 2004

### MathNerd

I still disagree with you Lethe, but would you like to expand on your one word arguements against what I said to try and make what you're saying more viable.

16. Mar 14, 2004

### lethe

quantum mechanics is a theory where the configuration space contains the dynamical variables.

in field theory, functions on configuration space (fields) are the dynamical variables.

if you think relativistic quantum field theory and relativistic quantum mechanics are the same thing, then do you think nonrelativistic quantum field theory and nonrelativistic quantum mechanics are also the same thing?

quantum mechanics can describe the motion of a single quantum particle. quantum field theory can describe large (or indefinite) numbers of quantum particles.

quantum mechanics has a finite dimensional phase space, and so is subject to the Stone-von Neumann theorem. quantum field theory has an infinite dimensional phase space, and therefore is not.

quantum mechanics can be thought of as a quantum field theory on 0+1 dimensions. more familiar quantum field theories have more dimensions.

i don't see why you would think they are the same thing. they are very different.

17. Mar 14, 2004

### outandbeyond2004

Well, what term shall we use for the whole field of relativistic + quantum theory? QFT plus the other kinds of relativistic + quantum theories.

RQT, I guess!