What is the law for probability having Schrodinger equation?

[naima]

I think that all is in the title. If the amplitude $\phi$ obeys a Schrodinger equation, what is the law for $p = \phi^* \phi$?

 

[Orodruin]

You can find it by simply differentiating with respect to time and using the Schrödinger equation. However, you will not obtain a differential equation involving p only as you are losing the phase information when taking the absolute value squared.

 

[DrClaude]

Have a look at the Madelung equations.

 

[naima]

Density matrices are a generalization of states. Is ther a generalization of the law from states to density matrices?
Can we say that we have an equation for not measurable amplitudes and no equation for measurable probabilities?

 

[DrClaude]

Density matrices are a generalization of states. Is ther a generalization of the law from states to density matrices?
Can we say that we have an equation for not measurable amplitudes and no equation for measurable probabilities?

I have no idea what you mean. Can you rephrase that?

 

[naima]

In QM probabilities are a byproduct of amplitudes. We have a Schroedinger equation for them. With density matrices it is straightfoward to get the probabilities.
I found the law for them it is $\rho ^{'} = [H,\rho]$
Is it possible to avoid the amplitudes in the calculations but to use density matrices.
it is the way

 

[DrClaude]

With density matrices it is straightfoward to get the probabilities.

I don't see why you say that it is more straightforward to use the density matrix instead of the wave function. When dealing with pure states, it is often easier to work with the wave function.

I found the law for them it is $\rho ^{'} = [H,\rho]$

I guess you are talking about the Schrödinger equation written in terms of the density matrix, which is
$$
i \hbar \frac{d \rho}{dt} = [H,\rho]
$$

Is it possible to avoid the amplitudes in the calculations but to use density matrices.
it is the way

The density matrix formalism is simply an extension of the wave function that allows to treat mixed states. When it is not necessary to deal with mixed states, physicists will, in the majority of cases, work with the wave function (or state vectors). Density matrices are not "the way."

 

[naima]

I don't see why you say that it is more straightforward to use the density matrix instead of the wave function.

Density matrices have many intellectual advantages:
Any child can understand what a probability is.
The values of the probabilities can be found on the diagonal. No need to square mysterious things.
Probabilities can be measured just like density matrices. Amplitudes cannot.
There are only two things that interest a physicist: the possible values and their frequencies. They all come from operators not from states. we are so used to the usual presentation that we no longer see it.

 

[bhobba]

Density matrices have many intellectual advantages:

And disadvantages.

See Chapter 3 Ballentine where Schroedinger's equation etc is derived. As he explains at the start (page 63) mixed states present no new novelty and for simplicity is generally not included in time evolution etc and pure states are used. Its very simple to generalise because a mixed state is simply the convex sum of pure states ie of the form Σ pi |ui><ui| where observationally the pi give the probability of the system being in pure state |ui><ui|. We know how the pure states evolve hence we know how the mixed state evolves. It's the same for a lot of other areas - mixed states add nothing new and simply complicate the actual physics..

Thanks
Bill

 

[naima]

I can add another advantage to operators over states: You are less tempted to think that it is something which collapses.
@bhobba
I do not say that it adds something new. I only say that i prefer concrete things like probabilities. read #8
I am very close to your minimal interpretation.
Happy new year.

 

[Orodruin]

But density matrices do add something new, the possibility of taking effects of decoherence into account in the evolution of the states. For a pure state without decoherence, it is of course the same - as it should be - and a matter of preference only.

 

[bhobba]

But density matrices do add something new, the possibility of taking effects of decoherence into account in the evolution of the states. For a pure state without decoherence, it is of course the same - as it should be - and a matter of preference only.

Good point.

Thanks
Bill

 

[vanhees71]

Well, I'd say it has some advantages to use the density matrix (I prefer to call it the statistical operator) to define what a quantum-theoretical state is, because it's general and in my opinion simpler than the special case of pure states which are represented by rays (not vectors!) in Hilbert space.

A statistical operator is a positive semi-definite self-adjoint operator $\hat{\rho}$ with $\mathrm{Tr} \hat{\rho}=1$. It obeys the (picture-independent!) equation of motion
$$\frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}]+\left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{explicit}}=0.$$
It's meaning is the following: If $A$ is an observable, represented by the self-adjoint operator $\hat{A}$ and $|a,\alpha \rangle$ are the eigenvectors of $\hat{A}$ with eigenvalue $a$ then the probability to find the value $a$ when $A$ is measured on a system that is prepared in a state reprsented by $\hat{\rho}$ is given by
$$P_A(a|\hat{\rho})=\sum_{\alpha} \langle a,\alpha|\hat{|rho} a,\alpha \rangle,$$
where, of course, the sum could also be an integral or both a sum and an integral depending on whether there is a continuous and/or discrete set of labels $\alpha$ counting the eigenstates of $a$.

The expectation value of the observable is given by
$$\langle A \rangle_{\hat{\rho}} = \mathrm{Tr} (\hat{\rho} \hat{A}).$$
A pure state is a special case. Any pure state can be defined as being represented by a projection operator, i.e., in this case $\hat{\rho}^2=\hat{\rho}$. This equation implies that the eigenvalues of $\hat{\rho}$ can only be 0 and 1. Since $\mathrm{Tr} \hat{\rho}=1$ it can have only precisely one eigenstate $|\psi \rangle$ with the eigenvalue $1$, and this implies that
$$\hat{\rho}=|\psi \rangle \langle \psi|,$$
where $|\psi \rangle$ is any normalized eigenvector of $\hat{\rho}$, and by the arguments just given it's defined up to a phase factor, which is irrelevant for physics. That's the great advantage of the statistical operator also in the case pure states: It is uniquely defined, because the arbitrary phase factor cancels in the projector $|\psi \rangle \langle \psi|$. Of course, the statistical-operator formalism is for pure states equivalent to the usual definition of pure states as being represented by rays in Hilbert space, i.e., normalized Hilbert-space vectors modulo an arbitrary phase.

 

[naima]

How can we get from density matrices the surfaces with same phase?