Why not formulate QM in terms of |ψ| squared?

[Tosh5457]

I'm beginning to study QM, and as I understand all the information we can get out of system with 1 particle is the probability distribution function (which has position and time as variables). By knowing the wave function it's possible to know the probability distribution function, but that's an indirect way to find it. Why not formulate directly QM in terms of |ψ| squared?

 

[cattlecattle]

I'm beginning to study QM, and as I understand all the information we can get out of system with 1 particle is the probability distribution function (which has position and time as variables). By knowing the wave function it's possible to know the probability distribution function, but that's an indirect way to find it. Why not formulate directly QM in terms of |ψ| squared?

See this recent discussion:
https://www.physicsforums.com/showthread.php?t=655266

 

[bhobba]

That's because it only applies to what are called pure states. The general formula is given an observable R the expected value of R E(R) = Trace(pR) where p is the system state which is defined as a positive operator of unit trace. Pure states are a special case being the operator |u><u| and the squaring rule is a further special case when the pure state is expanded in terms of eigenvectors of the observable.

Thanks
Bill

 

[Demystifier]

I'm beginning to study QM, and as I understand all the information we can get out of system with 1 particle is the probability distribution function (which has position and time as variables). By knowing the wave function it's possible to know the probability distribution function, but that's an indirect way to find it. Why not formulate directly QM in terms of |ψ| squared?

QM is not only about the probability distribution of position. It is also about the probability distribution of other observables, such as momentum. From knowing the former you cannot determine the latter. But if you know the full wave function in the position space, by a Fourier transform you can also determine the wave function in the momentum space. The probability distribution of momenta is given by the absolute value squared of the wave function in the momentum space.

There is, however, a way to eliminate the unphysical total phase of the wave function. It can be done by reformulating QM in terms of density matrices, where each wave function is represented by a corresponding density matrix.

 

[DrDu]

Why not formulate directly QM in terms of |ψ| squared?

That's called density functional theory!

 

[the_pulp]

The answer is the 2 slit experiment. Suppose you have that the probability of a particle entering in slit 1 is 0,3 and the probability of the same particle entering in slit 2 is 0,5, then in classical probability the probability of entering in either slit 1 or 2 is 0,8. Nevertheless in QM the answer could be more or less than that depending on the distance of the slits, the velocity of the particles and such. So, they noticed that, in order to predict such behaviour, a particle should be described by 2 real numbers (real and complex part of the wave function).

There is a paper that states more precisely this point of view:

"Origin of Complex Quantum Amplitudes and Feynman's Rules"

Philip Goyal, Perimeter Institute, Waterloo, Canaday
Kevin H. Knuthz, University at Albany (SUNY), NY, USA
John Skillingx, Maximum Entropy Data Consultants Ltd, Kenmare, Ireland

Just to summarize, the AND & OR rules of probability theory does not describe the AND & OR rules of particle experiments, so they had to develope this "crazy new probability theory"

 

[wacki]

Hi the_pulp,

I was always wandering why a real valued wave function can't be used in QM.
The title "Origin of Complex Quantum Amplitudes and Feynman's Rules"
sounds very interesting. Unfortunately it costs $25 to download :-(
Could you give a "light weight" version of it in this forum?

 

[A. Neumaier]

I'm beginning to study QM, and as I understand all the information we can get out of system with 1 particle is the probability distribution function (which has position and time as variables). By knowing the wave function it's possible to know the probability distribution function, but that's an indirect way to find it. Why not formulate directly QM in terms of |ψ| squared?

One can formulate QM in terms of the density matrix $$\rho=\psi\psi^*$$, whose diagonal is $$|\psi|^2$$. This is done in quantum statistical mechanics. This looks very much like the classical case - which is essentially the case where $$\rho$$ and all operators are diagonal, whereas in a pure state $$\rho=\psi\psi^*$$, it is rank 1, and hence usually far from diagonal.

Decoherence (the modern link between classical and quantum mechanics) works with the density matrix, and shows why the environment turns most density matrices into nearly diagonal classical densities (in the right basis, the so-called pointer basis).

 

[Tosh5457]

One can formulate QM in terms of the density matrix $\rho=\psi\psi^*$, whose diagonal is $|p\si|^2$. This is done in quantum statistical mechanics. This looks very much like the classical case - which is essentially the case where $\rho$ and all operators are diagonal, whereas in a pure state $\rho=\psi\psi^*$, it is rank 1, and hence usually far from diagonal.

Sorry I can't see the latex code properly. Does that formulation allow to know momentum, by only knowing the density matrix?

 

[A. Neumaier]

Sorry I can't see the latex code properly. Does that formulation allow to know momentum, by only knowing the density matrix?

Momentum is an observable, and expectations of an arbitrary observable $$O$$ are obtained by multiplying $$\rho$$ by $$O$$ and taking the trace. So you can calculate all expectations, not only that of momentum, and also all higher order moments.

 

[Tosh5457]

Momentum is an observable, and expectations of an arbitrary observable $$O$$ are obtained by multiplying $$\rho$$ by $$O$$ and taking the trace. So you can calculate all expectations, not only that of momentum, and also all higher order moments.

So why is that formalism only used in quantum statistical mechanics, and not on the general theory of quantum mechanics?

 

[DrDu]

What I wondered, probably because I don't understand qft too well: Wouldn't it be possible to consider the charge density rho as some field, something related to its time derivative as momentum and then doing second quantization with it?

 

[bohm2]

The title "Origin of Complex Quantum Amplitudes and Feynman's Rules" sounds very interesting. Unfortunately it costs $25 to download :-(Could you give a "light weight" version of it in this forum?

It's free here:
http://arxiv.org/pdf/0907.0909v3.pdf

 

[Jano L.]

By knowing the wave function it's possible to know the probability distribution function, but that's an indirect way to find it. Why not formulate directly QM in terms of |ψ| squared?

Indeed, this is possible, but one has to use another real variable, the phase. Then one can reformulate the Schroedinger equation as a pair of two equations (Bohm's equations). The reason one encounters works with $\Psi$ more often is that complex numbers tremendously simplify solving Schroedinger's equation. It is similar to the situation with the phasor method in solving linear differential equations - complex numbers simplify the mathematics - except in Schr. equation, the simplification is even greater.

 

[Darwin123]

I'm beginning to study QM, and as I understand all the information we can get out of system with 1 particle is the probability distribution function (which has position and time as variables). By knowing the wave function it's possible to know the probability distribution function, but that's an indirect way to find it. Why not formulate directly QM in terms of |ψ| squared?

Basically, the following are really the same.
1) Because you lose phase information with |ψ| squared.
2) Because you couldn't analyzed diffraction experiments.
3) Because linear superposition of states doesn't apply to |ψ| squared.

 

[HomogenousCow]

Why can't we forumlate f=ma in terms of x(t) since that's all we want in the end?

 

[bhobba]

Why can't we forumlate f=ma in terms of x(t) since that's all we want in the end?

Well actually with Newtons laws note that Newtons first law follows from Newtons second law which is actually a definition. The real content is in Newtons third law, which is logically equivalent to conservation of momentum and follows from Noether's Theorem. But that is dependent on the principle of least action and in fact follows from QM. So the real basis of Newton is QM. So what's this force thing? Basically its a prescription that says - get thee to the forces - that is the fundamental thing. That is the real content of Newtons laws - its a way of viewing physical problems that is very useful. Not always however - many problems are better solved using the action principle which is actually closer to its 'real' foundation - QM.

Thanks
Bill

 

[A. Neumaier]

So why is that formalism only used in quantum statistical mechanics, and not on the general theory of quantum mechanics?

Because outside statistical mechanics, one can pretend that a mixed state is a mixture of pure states (though not uniquely, and the decomposition has therefore no physical meaning), and the pure state computations look slightly more familiar. Whereas in statistical mechanics you really _need- the mixed states, as every equilibrium state is mixed.

 

[A. Neumaier]

What I wondered, probably because I don't understand qft too well: Wouldn't it be possible to consider the charge density rho as some field, something related to its time derivative as momentum and then doing second quantization with it?

Actually, in QFT, charge density is a field, the zero component of the corresponding current.

But charge density is not the same as |psi|^2, except in the single particle case.

 

[DrewD]

as every equilibrium state is mixed.

Sorry, this is off topic, but where can I find more info on this. It came up in a paper I was reading, but I don't understand why. I don't think the standard QM books I have cover this. Any advice where to look would be appreciated. Sorry to disrupt the thread.

 

[Jano L.]

... outside statistical mechanics, one can pretend that a mixed state is a mixture of pure states ...

But doesn't quantum statistical physics pretend the same? Take, for example, the density matrix for harmonic oscillator $\rho_{mn} = \frac{e^{- n\hbar \omega/ (k_B T)}}{Z} \delta_{mn}$ (in the basis of its Hamiltonian). Is not this a good example of a mixture of pure states - eigenstates of the Hamiltonian?

 

[DrDu]

Actually, in QFT, charge density is a field, the zero component of the corresponding current.

But charge density is not the same as |psi|^2, except in the single particle case.

I was assuming a non-relativistic setting. There, I think, charge density and psi square are identical.

 

[A. Neumaier]

I was assuming a non-relativistic setting. There, I think, charge density and psi square are identical.

My statement holds both in the relativistic and the nonrelativistic case, where things look identical in the second-quantized formulation.

For a single particle, charge density and absolute psi square are identical.

For two particles, the wave function had already two position arguments, while the charge density has only one. So they cannot be the same.

 

[A. Neumaier]

But doesn't quantum statistical physics pretend the same? Take, for example, the density matrix for harmonic oscillator $\rho_{mn} = \frac{e^{- n\hbar \omega/ (k_B T)}}{Z} \delta_{mn}$ (in the basis of its Hamiltonian). Is not this a good example of a mixture of pure states - eigenstates of the Hamiltonian?

This is by far not the only casae you can write this density matrix as a mixture, and none of the many mixture representations has any physical relevance as they are observationally undistinguishable.

 

[A. Neumaier]

Sorry, this is off topic, but where can I find more info on this. It came up in a paper I was reading, but I don't understand why. I don't think the standard QM books I have cover this. Any advice where to look would be appreciated. Sorry to disrupt the thread.

Look an any introduction to statistical mechanics, which doens't treat only the classical case.
e.g., Part II of http://lanl.arxiv.org/abs/0810.1019

 

[Jano L.]

This is by far not the only casae you can write this density matrix as a mixture,

That is true. There are potentially many different ways to express that density matrix. And I agree that preferring one mixture above the others seems artificial and ungrounded.

However, the fact is that no other representation is used. Statistical physics relies on the properties of Hamiltonian, not some other operator, and really treats energy representation with distinction.

and none of the many mixture representations has any physical relevance as they are observationally undistinguishable.

I am not so sure. In general, different mixtures are in principle distinguishable, if we know something about the preparation of the ensemble. Of course, simple counting of frequencies of results of measurement will not reveal that; but there is more to the ensemble than just results of measurements on it. This is more easily shown for spins than for energy, but why assume that canonical density matrix is different in this respect?

I think the relevance of H eigenstates is in that we need to assume ensemble of them in order to use Boltzmann's probability distribution to find the canonical density matrix in the first place. For other mixtures, how would you arrive at density matrix that is diagonal in H representation? I do not see how that is possible, but perhaps you know.

 

[A. Neumaier]

However, the fact is that no other representation is used. Statistical physics relies on the properties of Hamiltonian, not some other operator, and really treats energy representation with distinction.

Statistical mechanics doesn't care at all about the basis in which things are represented, except for computational reasons. One sees it clearly once one leaves the equilibrium situation.

In general, different mixtures are in principle distinguishable, if we know something about the preparation of the ensemble.

One can distinguish ensembles only if they are prepared artificially, by mixing observations made at different times. But the ensembles of statistical mechanics are of a different nature; they are instantaneous, a single time already defines the ensemble. Then nothing can distinguish a decomposition into subensembles.

I think the relevance of H eigenstates is in that we need to assume ensemble of them in order to use Boltzmann's probability distribution to find the canonical density matrix in the first place. For other mixtures, how would you arrive at density matrix that is diagonal in H representation? I do not see how that is possible, but perhaps you know.

All that is needed is that e^{beta H} commutes with H, which is a triviality, independent of the basis used to express it. No diagonalization is needed or assumed.

 

[DrDu]

For two particles, the wave function had already two position arguments, while the charge density has only one. So they cannot be the same.

You are obviously right, here. Nevertheless my question remains: can we quantize charge density, viewed as a field, directly?

 

[dextercioby]

Which charge ? Electric charge ? We surely can.

 

[andrien]

Which charge ? Electric charge ? We surely can.

using dirac quantization condition of monopoles or reverse argument?

 

[A. Neumaier]

You are obviously right, here. Nevertheless my question remains: can we quantize charge density, viewed as a field, directly?

Together with charge density, you need st least the associated charge current operaots, which complement the density to a conservation law. Quantizing this is called current algebra. There have been varous attempts and there is significant literature, but apart from the 1+1D case (where current algebra essentially turns into Kac-Moody algebra) no real breakthrough.