# Time in QM

1. Dec 8, 2009

### surena1980

The Time is not a ket nor an operator, so what role does the Time play in Physics?

2. Dec 8, 2009

3. Dec 8, 2009

### Demystifier

According to
http://xxx.lanl.gov/abs/0811.1905 [Int. J. Quantum Inf. 7 (2009) 595]
both operator and hidden variable ;)

4. Dec 8, 2009

### Landau

Sakurai calles it a "parameter" (without a proper definition, but that's Sakurai). Also see https://www.physicsforums.com/showthread.php?t=136126 [Broken].

Last edited by a moderator: May 4, 2017
5. Dec 8, 2009

### Demystifier

Dmtr, thanks for the links. It seems very interesting.

6. Dec 8, 2009

7. Dec 8, 2009

### meopemuk

There can be no "operator of time" in quantum mechanics. Suppose there is such an operator $$T$$. Then, there it should possess mutually orthogonal eigenvectors $$|t \rangle$$ with eigenvalues $$t$$. However, the state represented by the eigenvector $$|t \rangle$$ is physically impossible. In this state particle exists only in a short time interval around $$t$$. Such a state violates all existing conservation laws.

So, time is not an observable. Time is not a property or attribute of a physical system. Time is rather an attribute of the measuring device or observer. When we measure certain true observable (position, spin, etc) we attach a label (time) to this measurement in accordance with what the laboratory clock showed at the instant of measurement. So, mathematically it is OK to treat time as a numerical parameter, as it is done in ordinary quantum mechanics.

Also I think that it is not useful to demand that time and space should be treated on "equal footing". Space-time unification is not an essential feature of relativistic physics. The true postulates of relativity are: (1) all inertial reference frames are equivalent (2) the group of transformations between inertial reference frames is the Poincare group. In QM this implies that the Hilbert space of any isolated physical system carries an unitary representation of the Poincare group. This is the true and complete mathematical manifestation of the principle of relativity. The "symmetry" between space and time is an additional unjustified assumption.

Edit: for more details see

E. V. Stefanovich, "Is Minkowski space-time compatible with quantum mechanics?", Found. Phys., 32 (2002), 673.

Eugene.

Last edited: Dec 8, 2009
8. Dec 8, 2009

### statespace101

Time is linked to energy on QM though, but that is if you assign any credibility to the "weak" uncertainty of time and energy and since the relation is non-commutative and yields a constant it must be an operator. How else could information of correlated spins between two “particles” be transferred instantaneously if time isn’t an operator?

Just looked at a john Baez site and he gives a compelling reason that time is not an operator. I still think it is and by not coming to terms with it may be missing some great discoveries in QM.

I may be wrong? I took a turn out of math and physics and went to the horrible world of economics, econophysics and finance.

Last edited: Dec 8, 2009
9. Dec 8, 2009

### meopemuk

Yes, there is the time-energy uncertainty relationship. The primary example is an unstable particle, whose lifetime is inversely proportional to the energy (mass) uncertainty. However, the nature of this relationship is completely different from the position-momentum uncertainty relationship (where two non-commuting operators are involved). Particle decays with all their features can be described pretty well in ordinary quantum mechanics, where time is not an operator.

Eugene.

10. Dec 8, 2009

### statespace101

The link dmtr posted by just reading the beginning made me realize that time must be an operator for the delyaed choice quantum eraser to be possible. I looked at John von's memory space and that just clicked.

Or how about Wigner's friend, Eugene? Just kidding Wigner Eugene Wigner irony :lol:

Last edited: Dec 8, 2009
11. Dec 8, 2009

### Neo_Anderson

We may never make satasfactory progress in QM until we respect Special Relativity. This means, of course, that we must regard the notion of 'time' as 'spacetime,' whether in cosmology or QM.

12. Dec 8, 2009

### statespace101

Another good example I just posted in another thread is the Aharonov-Bohm effect. That time is an operator, or you belive it's a hidden variable. Ballentine goes over this in his book, Quantum Mechanics: A Modern Development by Leslie E. Ballentine.

Last edited: Dec 8, 2009
13. Dec 8, 2009

### meopemuk

Let us suppose that you are right and that space and time are unified in a 4-dimensional continuum. I would like to show that this assumption leads to a contradiction with other important postulates of relativity and quantum mechanics.

First, the space-time idea is equivalent to the assumption that Lorentz transformations

$$x' = \frac{x-vt} {\sqrt{1-v^2/c^2}}$$.......................(1)
$$t' = \frac{t -vx/c^2} {\sqrt{1-v^2/c^2}}$$................(2)

are exact and universal. For example, if $$x, x'$$ are particle positions in different frames, then the above transformations are true independent on whether this particle is free or it is a part of an interacting multiparticle system. The transformations do not depend on the strength of this interaction as well.

Now, let me look at these transformations from another perspective. In relativistic quantum mechanics a multiparticle system is described in the Hilbert space, in which a unitary representation of the Poincare group is defined. If the system is interacting then, as it is well-known, the generator of time translations (the Hamiltonian $$H$$) is interaction-dependent (i.e., it is different from the free-particle form). It is less-known that the vector of boost generators $$\mathbf{K}$$ must be interaction-dependent too (you can read more about that in S. Weinberg's "The quantum theory of fields", vol. 1). The boost generator is exactly what we need to find out how particle position operators $$X,X'$$ are related to each other in relatively moving reference frames. If the movement is along the x-axis with rapidity $$\theta$$, then the QM formula is

$$X' =\exp(-iK_x \theta)X\exp(iK_x \theta)$$..................(3)

Since operator $$K_x$$ is interaction-dependent, the transformation (3) must have different forms depending on the nature of the multiparticle system and on interactions acting there. This conclusion is in direct contradiction with the exactness and universality of (1) assumed earlier.

Eugene.

14. Dec 11, 2009

### Demystifier

15. Dec 11, 2009

### dmtr

Are you assuming a single history line? In the MWI you are going to have contributions from all possible futures, I guess they will destructively interfere / cancel out.

16. Dec 11, 2009

### Demystifier

Well, I have written a comment on a particular paper which does not use MWI.
In the spirit of this comment, I would object to you that you don't explain why only futures cancel out and not the pasts.

In a more general (not related to the criticized paper) MWI context, I object to you that different MACROSCOPIC histories do not interfere due to decoherence.

17. Dec 11, 2009

### RUTA

18. Dec 13, 2009

### dmtr

Paper: http://arxiv.org/abs/0802.0438
Criticism: http://xxx.lanl.gov/abs/0912.1947

IMHO a good theory (or criticism) should hold under any QM interpretation (as QM interpretations do not change the underlying equations, they are essentially the ways to visualize different aspects of the equations). The original article didn't use the term 'single, well defined future', you did. That's why I've tried to apply MWI and show that your criticism about the traces from the future is not a very sound one.

As far as the original goes about the 'traces from the future' - the author says that we will consider these traces to be from the past:

Last edited: Dec 13, 2009
19. Dec 13, 2009

### dmtr

20. Dec 14, 2009

### Demystifier

Perhaps the original article didn't use the term "single, well defined future", but it used "single, well defined past" and "symmetry between the two time directions", which, taken together, imply "single, well defined future". That is my point. It does not depend on the interpretation of QM.