What is the Conservation of Information in Quantum Mechanics?

[Samo_phy]

Hi everyone,
I was watching Susskinsd's lectures posted on youtube about Quantum mechanics and he started explaining his -1st basic law of physics saying: the conservation of information means basically that if you start with 2 quantum states (vectors) which are observably different, or mathematically speaking, the both vectors are orthogonal, and if you let both of them evolve then they will stay orthogonal. Meaning that if two states that are observably different will not evolve into states which are not observably different.
I only need a few physical examples to understand it better. Anyone?

 

[Tio Barnabe]

For examples,

1 - for a particle having two independent spin states there must be two independent state vectors, because the spin state is represented by a state vector. The two independent state vectors must remain orthogonal all later times, because otherwise the particle would loose its property of having two independent spin states.

2 - the same story as above, but with spin substituted by any other physical property of a particle that can be described by a state vector in our analysis (it's important), which is to be preserved all the times.

What does it has to do with information? We name a particle by its properties. We say an electron is a particle having "bla bla bla". If we loose some property assigned to the electron, it will no longer be an electron. But in the case we are considering, we still want to analyse an electron. So that would be a contradiction.

 

[Tio Barnabe]

If some moderator feels that my responses in some threads are not appropriate, please let me know and I'll no longer be responding to members.

 

[Leo1233783]

Could you point on the exact Susskind reference ? TY

 

[PeterDonis]

if you start with 2 quantum states (vectors) which are observably different, or mathematically speaking, the both vectors are orthogonal

These aren't the same thing. "Orthogonal" is a much stronger condition than "observably different".

if you let both of them evolve then they will stay orthogonal

This is true for "orthogonal" (in fact it's one way of defining what "orthogonal" means), but not for "observably different", which, as above, is a much weaker condition.

To use the example @Tio Barnabe gave, if we have two orthogonal spin states--for example, the "spin up" and "spin down" states for a spin-1/2 particle and a particular direction of the spin axis--then they will stay orthogonal under time evolution (in this case, time evolution is very simple, the states just stay the same).

However, suppose we consider the two spin states "spin-z up" and "spin-x up", i.e., the spin "up" states for spin axis in the z and x directions, respectively. These two spin states are not orthogonal, but they are still observably different. In this particular case, the states will stay observably different under time evolution, since, as above, spin states just stay the same under time evolution.

But now consider the ground state of a hydrogen atom, and one of its excited states, say the 2p state. These states are not orthogonal, but they are observably different. However, they will not stay observably different under time evolution, since the excited state time evolves into the ground state (by emitting a photon), while the ground state stays the same.

 

[tom.stoer]

But now consider the ground state of a hydrogen atom, and one of its excited states, say the 2p state. These states are not orthogonal, but they are observably different. However, they will not stay observably different under time evolution, since the excited state time evolves into the ground state (by emitting a photon), while the ground state stays the same.

The ground state $|1s\rangle$ and the excited state $|2p\rangle$ are orthogonal, i.e. $\langle 1s|2p\rangle = 0$.

In non-rel. qm they stay different for any finite time. But this scenario is a little bit obscure from a fundamental perspective b/c the quantum system is not isolated, one has to mess around with an external classical field and a time-dependent Hamiltonian.

Strictly speaking the state vectors stay different for all times b/c one must include the photon. The ground state $|1s,0\rangle$ remains the same, whereas the excited state evolves from $|2p,0\rangle$ into a superposition of this excited state and $|1s,\gamma\rangle$.

So they stay "observably" different b/c one state will contain a photon $\gamma$.

And they stay different in principle b/c
$$\langle 1s,0|\,(c_{2p} |2p,0\rangle + c_{1s} |1s,\gamma\rangle) = 0$$
b/c
$$\langle 1s,0|1s,\gamma\rangle = 0$$

 

[tom.stoer]

the conservation of information means basically that if you start with 2 quantum states (vectors) which are observably different, or mathematically speaking, the both vectors are orthogonal, and if you let both of them evolve then they will stay orthogonal.

This is basically the unitary time evolution according to the Schrödinger equation.

The cosine of the angle between two state vectors $|a\rangle$ and $|b\rangle$ is given by the scalar product
$$\cos\phi_{ab} = \langle a|b\rangle$$
The Schrödinger equation says that a state vector $|a\rangle$ given for t = 0 evolves unitarily according to
$$|a\rangle \to |a(t)\rangle = e^{-iHt} |a\rangle$$
Calculating the angle at later times means
$$\cos\phi_{ab} (t)= \langle a(t)|b(t) \rangle = \langle a|e^{+iHt}\,e^{-iHt}|b\rangle = \langle a|b \rangle = \cos\phi_{ab}$$
simply b/c
$$ e^{+iHt}\,e^{-iHt} = 1$$
So this whole stuff of conservation of information is just a new buzzword for the universal validity of unitary time evolution of the Schrödinger equation.

Up to now we haven't talked about the meaning of "observably" different. That is clearly a new story b/c we must discuss an experiment and an observable, represented by a self-adjoint operator.

 

[PeterDonis]

The ground state $|1s\rangle$ and the excited state ##|2p\rangle are orthogonal

Yes, you're right, I should have said the state the atom is in when it is coupled with the field and has a nonzero probability of emitting a photon and dropping from the 2p state to the 1s state. This state is really a time-dependent superposition of 2p and 1s. But, as you say, even this isn't complete because it omits the state of the EM field.

In non-rel. qm they stay different for any finite time. But this scenario is a little bit obscure from a fundamental perspective b/c the quantum system is not isolated, one has to mess around with an external classical field and a time-dependent Hamiltonian.

Yes, agreed. Your statement of the general issue in post #7 is much better than my attempt was.

 

[vanhees71]

But in this case you have either $|2p \rangle$ (i.e., a "one-particle state", describing an H atom in the intrinsic state 2p) or $|1s,\gamma \rangle$ (i.e., a "two-particle state" describing an H atom in the intrinsic state 1s and a photon). These vectors are again orthogonal to each other. Also in QED the time evolution of a closed system (here consisting of an H atom or an electron, a proton, and the electromagnetic field).

 

[tom.stoer]

yes, of course, that's what I wrote

 

[PeterDonis]

in this case you have either $|2p \rangle$ (i.e., a "one-particle state", describing an H atom in the intrinsic state 2p) or $|1s,\gamma \rangle$ (i.e., a "two-particle state" describing an H atom in the intrinsic state 1s and a photon)

I think this is a bit misleading, since it could be taken to imply that the EM field is not "part of the state" in the 2p case, which is not correct. The full quantum system consists of the atom and the field; in the "2p" state the full atom-field state is $\vert 2p, 0 \rangle$, i.e., the atom is in the 2p state and the field is in the vacuum state.

 

[Samo_phy]

If some moderator feels that my responses in some threads are not appropriate, please let me know and I'll no longer be responding to members.

No not at all you were helpful, thanks a lot!

 

[Samo_phy]

Could you point on the exact Susskind reference ? TY

It was in his Advanced Quantum Mechanics lecture # 5 as far as I remember.

 

[Samo_phy]

This is basically the unitary time evolution according to the Schrödinger equation.
$$\cos\phi_{ab} (t)= \langle a(t)|b(t) \rangle = \langle a|e^{+iHt}\,e^{-iHt}|b\rangle = \langle a|b \rangle = \cos\phi_{ab}$$
simply b/c
$$ e^{+iHt}\,e^{-iHt} = 1$$
So this whole stuff of conservation of information is just a new buzzword for the universal validity of unitary time evolution of the Schrödinger equation.

Wow that was really illuminating, now I have a new perspective on that. However, I want to point out something that I have in common with Leonard Susskind.
He approaches the big questions by constructing a formal system just like in mathematics. He then understands the physical reality in terms of a small set of intuitive axioms, the principle of least action, the conservation principles, and the consistency of nature. I think that is what Einstein had in mind as he was building his general theory of relativity, he started from a basic set of principles he found self-evident. We see this idea clearly through the work of Dirac who tried to reduce everything to a formal system that all of quantum mechanics quantities come from. However, Susskind keeps referring to this buzzword as a basic principle in physics because it made him win against Hawking in the 20 years long war ^^

 

[vanhees71]

Yes, Susskinds "Theoretical Minimum" books are marvels. I don't know the 1st volume, but the 2nd, and I start reading the 3rd. I guess, I've to buy finally also vol. 1 :-)).

 

[tom.stoer]

Up to now we haven't talked about the meaning of "observably" different. That is clearly a new story b/c we must discuss an experiment and an observable, represented by a self-adjoint operator.

So let's discuss observables.

Two states are observably different if there is at least one observable represented by a self-adjoint operator $\mathcal{O}$ for which we expect different measurement results for the two states $|a\rangle$ and $|b\rangle$.

 

[vanhees71]

Within the formalism, two states are different, if their representing Statistical Ops. are different.

Operationally that's measurable by, e.g., measuring complete sets of compatible observables on both states (preparing the corresponding ensembles of quantum system). If the resulting probability distributions within the significance of the statistics (assuming that systematical errors are under control and negligible or controllable) are different, you have different states.

 

[tom.stoer]

Operationally that's measurable by, e.g., measuring complete sets of compatible observables on both states ... If the resulting probability distributions are different, you have different states.

Agreed.

I can reformulate my statement

Two states are observably different if there is at least one observable represented by a self-adjoint operator $\mathcal{O}$ for which we expect different measurement results for the two states $|a\rangle$ and $|b\rangle$.

that two states are observably identical if there is no observable for which we expect different measurement results for the two states $|a\rangle$ and $|b\rangle$.

Theoretically we can chose the projectors $E_\lambda$ to one-dim. eigenspaces of $\mathcal{O}$ to be our set of observables. Having identical probabilities
$$p_a(\lambda) = |\langle\lambda|a\rangle|^2 = |a_\lambda|^2$$
and
$$p_b(\lambda) = |\langle\lambda|b\rangle|^2 = |b_\lambda|^2$$
for all $\lambda$ means that all coefficients $a_\lambda$ and $b_\lambda$ in
$$|a\rangle =\sum_\lambda a_\lambda\,|\lambda\rangle$$
and
$$|b\rangle =\sum_\lambda b_\lambda\,|\lambda\rangle$$
must be identical, i.e. $a_\lambda = b_\lambda$ for all $\lambda$. But that essentially means that both states $|a\rangle$ and $|b\rangle$ are identical.

However - this seems to blur the difference between "identical" and "observably identical". In practice the projectors are not our observables but we have a small set of observables from which it is not automatically clear that we can distinguish between two states. So the question is not whether two states are different, but which properties a set of physically measurable observables must have in order to be able to distinguish between any two different states.

 

[vanhees71]

Again it is of utmost importance for QT that you DO NOT define a state vector as the representant of the state for two reasons: (a) it's limiting the notion of state to pure states, which is a pretty unusual case given the fact how complicated it is to prepare them and (b) for pure states it's not the vector in Hilbert space that represents it but the corresponding Statistical Operator $|\psi \rangle \langle \psi|$ or, equivalently but unnecessarily complicated for beginners, a unit ray in Hilbert space. Particularly when discussing the uniqueness of states it's important to be precise on point (b)!

For a more detailed discussion about state preparation and state determination, see Ballentine, Quantum Mechanics.

 

[tom.stoer]

Again it is of utmost importance for QT that you DO NOT define a state vector as the representant of the state ... for pure states it's not the vector in Hilbert space that represents it but the corresponding Statistical Operator $|\psi \rangle \langle \psi|$ or ... a unit ray in Hilbert space.

I agree, sorry for being imprecise (using rays the problem of equivalent vectors differing by a phase vanishes)