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


by Tosh5457
Tags: formulate, squared, terms, |ψ|
Tosh5457
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#1
Nov29-12, 08:22 PM
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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?
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cattlecattle
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#2
Nov29-12, 08:30 PM
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Quote Quote by Tosh5457 View Post
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:
http://www.physicsforums.com/showthread.php?t=655266
bhobba
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#3
Nov29-12, 11:14 PM
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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
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Nov30-12, 03:07 AM
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Why not formulate QM in terms of |ψ| squared?


Quote Quote by Tosh5457 View Post
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
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#5
Nov30-12, 03:56 AM
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Quote Quote by Tosh5457 View Post
Why not formulate directly QM in terms of |ψ| squared?
That's called density functional theory!
the_pulp
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#6
Nov30-12, 05:51 AM
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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
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#7
Nov30-12, 10:41 AM
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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
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#8
Nov30-12, 12:35 PM
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Quote Quote by Tosh5457 View Post
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 [tex]\rho=\psi\psi^*[/tex], whose diagonal is [tex]|\psi|^2[/tex]. This is done in quantum statistical mechanics. This looks very much like the classical case - which is essentially the case where [tex]\rho [/tex] and all operators are diagonal, whereas in a pure state [tex]\rho=\psi\psi^*[/tex], 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
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#9
Nov30-12, 12:54 PM
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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
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Nov30-12, 01:03 PM
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Quote Quote by Tosh5457 View Post
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 [tex]O[/tex] are obtained by multiplying [tex]\rho[/tex] by [tex]O[/tex] and taking the trace. So you can calculate all expectations, not only that of momentum, and also all higher order moments.
Tosh5457
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#11
Nov30-12, 01:31 PM
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Quote Quote by A. Neumaier View Post
Momentum is an observable, and expectations of an arbitrary observable [tex]O[/tex] are obtained by multiplying [tex]\rho[/tex] by [tex]O[/tex] 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
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#12
Nov30-12, 01:47 PM
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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
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#13
Nov30-12, 02:02 PM
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Quote Quote by wacki View Post
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.
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#14
Dec1-12, 01:19 PM
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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 [itex]\Psi[/itex] 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
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#15
Dec1-12, 09:22 PM
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Quote Quote by Tosh5457 View Post
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
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#16
Dec2-12, 04:28 AM
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Why can't we forumlate f=ma in terms of x(t) since that's all we want in the end?
bhobba
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Dec2-12, 04:45 AM
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Quote Quote by HomogenousCow View Post
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 whats 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
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#18
Dec2-12, 08:57 AM
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Quote Quote by Tosh5457 View Post
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.


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