# A few conceptual questions on time evolution of quantum states

1. Aug 8, 2014

### "Don't panic!"

Hi guys,

Sorry if this isn't quite the right place to post this, but I have a few conceptual questions that I'd like to clear up about time evolution of a quantum state.

Firstly, what is the exact argument for the evolution operator $\hat{U}\left(t,t_{0}\right)$ being independent of the initial state $\lvert\psi\rangle$?
Is it that we assume that all operators in quantum mechanics are linear and thus implying that operators are independent of the states that they act upon?
Or is it that we make a physical assumption that the laws of physics are universal and therefore act uniformly on each given state. As such, if a given quantum state evolves according to a set of physical laws, then any other state subject to the same conditions should evolve in the same way and thus requiring operators to be independent of the state that they act upon (implying linearity)?

My second question pertains to the composition of evolution operators, i.e. $$\hat{U}\left(t_{2},t_{0}\right)= \hat{U}\left(t_{2},t_{1}\right)\hat{U}\left(t_{1},t_{0}\right)$$ for $t_{0}\leq t_{1}\leq t_{2}$.

Does this follow from considering time as homogeneous (as the same laws of physics that apply today should apply at any subsequent time) and thus the state of a given system at $t=t_{0}$ and the it's evolved state at $t=t_{2}$ should therefore be independent of the path taken through time between the two states?

2. Aug 8, 2014

### WannabeNewton

1. $U(t,t_0) = e^{-i \int_{t_0}^t H dt / \hbar}$ so it's independent of the initial state $|\psi,t_0 \rangle$ simply by definition. There's really nothing else to it. The Hamiltonian $H$ certainly doesn't care about the initial state and it's clear from the definition of the propagator that neither does it. It's just a formal fundamental solution of the Schrodinger equation $i\hbar \partial_t |\psi \rangle = H |\psi \rangle$ just like Green's functions from classical field theory.

2. The product of the map taking us from $t_0$ to $t_1$ and the map taking us from $t_1$ to $t_2$ is the same as the single map taking us from $t_0$ to $t_2$ by the group composition law so the product of the associated unitary operators have to be equal to the single unitary operator.

3. Aug 8, 2014

### "Don't panic!"

What about in the case where the Hamiltonian is time-dependent, as subsequent Hamiltonian's don't necessarily commute, right?

For the second question, thanks for the information. I was hoping to gain an insight into the physical reasoning behind it all as well. Is what I said correct?

4. Aug 8, 2014

### Oxvillian

I'm not sure this really means anything

The time evolution operator is just list of rules telling you what final state any given initial state is mapped onto. Does this list of rules depend on the initial state? I guess not.

5. Aug 8, 2014

### "Don't panic!"

If the evolution operator were dependent on the state, then we would never be able to consider a superposition of states (which we know already satisfy the Schrodinger equation) as each sub-state would have its own evolution operator, an incomprehensible task to try and resolve. That was part of my thoughts behind it anyway.

6. Aug 8, 2014

### Oxvillian

But again, that doesn't make sense. The evolution operator, by definition, already contains information on how any state evolves.

Suppose I have a shopping list of items that tells me whether or not I should buy them:

Eggs: yes
Bacon: yes
Milk: no
Orange juice: yes
Apples: no

Should I take five shopping lists to the supermarket, one for each item? I could have five copies of the same list, each with one of the items highlighted. I guess I could even change the yes/no entries on some of the non-highlighted items. Would that make sense?

Making a shopping list for each item makes exactly as much sense as making a time-evolution operator for each state.

7. Aug 8, 2014

### atyy

Maybe something like this? The evolution operator should be unitary, so as to preserve probability. According to Wikipedia, requiring unitarity is enough to prove linearity: http://en.wikipedia.org/wiki/Unitary_operator.

But I don't know if there is really a fundamental need for unitarity, since the normalization can be put into the Born rule. For example, collapse is not unitary, and it is an acceptable time evolution. A very wide class of acceptable time evolutions are the completely positive trace-non increasing maps, which include collapse.

One way I've seen of justifying this is that one should be able to put the Heisenberg cut in several places, so that a quantum system can always be "purified". This axiom plus several others can lead to a derivation of finite dimensional quantum mechanics http://arxiv.org/abs/1011.6451.

Another famous derivation of quantum theory from other axioms is Hardy's http://arxiv.org/abs/1303.1538. (Actually, his famous one is http://arxiv.org/abs/quant-ph/0101012, but I like the tweaks he's done in his latest version.) Depending on one's aesthetics, one may consider the restriction of these derivations to finite dimensional Hilbert spaces as problematic.

Last edited: Aug 8, 2014
8. Aug 8, 2014

### "Don't panic!"

Yeah, I've kind of got myself into a spiral trying to justify it in my mind!
Anyway, thank you guys for your help so far.

9. Aug 8, 2014

### "Don't panic!"

Yes, that's kind of what I was eluding to in my comment about state dependence (although I didn't word it very well, I know).

Would it be correct to analogise the evolution operator with the differential operator in the sense that the domain of the differential operator is the set of all differentiable functions and its codomain the set of derivatives of those functions. It is a linear mapping from those functions to their derivatives.
In the case of the evolution operator, its domain is the set of all pure quantum states and it's codomain the set of time evolved states of those quantum states. It is a linear mapping from those (initial) states to their corresponding evolved states after a given time interval. Is this correct? (sorry, I understand that's a bit of a convoluted statement, but hopefully you catch my drift?!)

Last edited: Aug 8, 2014
10. Aug 8, 2014

### WannabeNewton

Yes that's fine. But I wouldn't use the term "pure state" to mean the initial states because "pure state" has a precise meaning in QM.

Have you seen Green's functions in electrodynamics? $U(t,t_0)$ plays the same role for Schrodinger's equation as the retarded propagator does for Maxwell's equations.

11. Aug 8, 2014

### bhobba

Its tied up with the very important Wigners theorem:
http://arxiv.org/abs/0808.0779to the symmetry view

Basically it says, for continuous transformations between vector spaces, then such a unitary transformation always exists.

This is very fundamental to the view of QM dynamics because, exactly the same as it is for classical dynamics, it is determined by symmetry considerations.

You can only really appreciate it by studying it.

For Classical Mechanics see Landau:
https://www.amazon.com/Mechanics-Third-Edition-Theoretical-Physics/dp/0750628960

I cant help myself - every time I give that link I must mention what a reviewer said (I concur of course which is why I give it):
'If physicists could weep, they would weep over this book. The book is devastingly brief whilst deriving, in its few pages, all the great results of classical mechanics' 'The reason for the brevity is that, as pointed out by previous reviewers, Landau derives mechanics from symmetry. Historically, it was long after the main bulk of mechanics was developed that Emmy Noether proved that symmetries underly every important quantity in physics. So instead of starting from concrete mechanical case-studies and generalising to the formal machinery of the Hamilton equations, Landau starts out from the most generic symmetry and dervies the mechanics.'

For the Quantum case see Ballentine Chapter 3.

Logically the situation is this.

Classically the Principle of Least Action (PLA)and the resulting symmetries of the Lagrangian determines classical mechanics via Noether's theorem.

In Quantum Mechanics symmetries applied to the probabilities (such that they are frame independent via the Principle Of Relativity) imply Schroedinger's equation etc.

But the PLA follows from QM anyway so really mechanics is simply the working out of the symmetries of the underlying formalism.

And in Quantum Field Theory (QFT) its even more important as Noether's beautiful theorem really comes into its own.

Thanks
Bill

Last edited by a moderator: May 6, 2017
12. Aug 8, 2014

### atyy

But there are non-unitary evolutions (collapse) consistent with the principle of relativity.

Last edited by a moderator: May 6, 2017
13. Aug 8, 2014

### bhobba

That, as I know you are aware, is very interpretation dependant.

In some interpretations, collapse only applies to filtering type observations which is viewed as a state preparation procedure. Indeed a state is viewed as its preparation procedure. Unitary evolution is simply that - how the state evolves - its viewed as logically distinct from the state itself. It simply changes it from one equivalence class of preparation procedures to another.

I am not going to discuss which is the correct view - merely to point out different takes exist.

And I have said it before, and will say it again, those features you do not like about QM can be removed by simply going to a different interpretation. But what we don't have is one that removes them all. That to me is what's weird about QM - not any particular worry such as non-unitary evolution.

Thanks
Bill

14. Aug 8, 2014

### atyy

But it does mean that PLA is not sufficient to prove that unitary evolution is the only acceptable evolution. It has to be PLA plus some other postulate. Hardy's axioms, for example, permit collapse.

Edit: Whoops, sorry - that should be POR throughout, not PLA.

Last edited: Aug 8, 2014
15. Aug 9, 2014

### bhobba

No worries. :rofl:

Yea - I would agree with that.

Thanks
Bill

Last edited: Aug 9, 2014
16. Aug 9, 2014

### Barry911

I believe one related problem is the space-time dependence of QM. Further the SU1 formalism for the
psi function seem unrealistic.

17. Aug 9, 2014

### Barry911

How much of Q.M. interpretation is the fault of language? There are no "particles" in the quantum realm nor
are their "waves". A more meaningful description would recognize Laplacian operators on the grid, high nable
values can appear particle like. Further "waves" should be understood as correlations, nothing more.

Last edited: Aug 9, 2014
18. Aug 9, 2014

### bhobba

A LOT IMHO eg confusing observation with an organic observer.

Thanks
bILL

19. Aug 9, 2014

### "Don't panic!"

Thanks for the help guys, really appreciate the input on the subject.