| Topic: |
Science > Physics |
| User: |
"mathlover" |
| Date: |
17 Nov 2006 07:48:39 AM |
| Object: |
why absolute value of wavefunction squared? |
Dear all,
I come up with a fundamental question that why the squared absolute
value of the wavefunction is interpreted as the probability(
specifically, |psi(x)|^2dx is the probability to find the particle
located in between x to x+dx ), but, why is not the absolute value of
the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
I guess that's because this is the only way to develop a
self-consistent Q.M. theory. But I can't argue clearly that why do
other choices of interpretation of the wavefunctions fail? Any ideas
will be thankful, Thanks!
Sincerely
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| User: "Sorcerer" |
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| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 09:50:25 AM |
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"mathlover" <CPPandC@gmail.com> wrote in message
news:1163771319.664215.128540@j44g2000cwa.googlegroups.com...
| Dear all,
| I come up with a fundamental question that why the squared absolute
| value of the wavefunction is interpreted as the probability(
| specifically, |psi(x)|^2dx is the probability to find the particle
| located in between x to x+dx ), but, why is not the absolute value of
| the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
|
| I guess
You can stop right there, mathematicians do not guess, lover boy.
.
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| User: "mathlover" |
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| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 10:28:58 AM |
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|
Sorcerer =BCg=B9D=A1G
"mathlover" <CPPandC@gmail.com> wrote in message
news:1163771319.664215.128540@j44g2000cwa.googlegroups.com...
| Dear all,
| I come up with a fundamental question that why the squared absolute
| value of the wavefunction is interpreted as the probability(
| specifically, |psi(x)|^2dx is the probability to find the particle
| located in between x to x+dx ), but, why is not the absolute value of
| the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
|
| I guess
You can stop right there, mathematicians do not guess, lover boy.
Excuse me, I'm quite interested in the problem.
By the way, mathematician guess. The famous Poincare conjecture
proved recently by Mr. Perelman is a well-known guess. Guessing things
is fun, isn't it?
.
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| User: "Sorcerer" |
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| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 11:07:16 AM |
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|
"mathlover" <CPPandC@gmail.com> wrote in message
news:1163780938.024920.35540@h54g2000cwb.googlegroups.com...
Sorcerer ¼g¹D¡G
"mathlover" <CPPandC@gmail.com> wrote in message
news:1163771319.664215.128540@j44g2000cwa.googlegroups.com...
| Dear all,
| I come up with a fundamental question that why the squared absolute
| value of the wavefunction is interpreted as the probability(
| specifically, |psi(x)|^2dx is the probability to find the particle
| located in between x to x+dx ), but, why is not the absolute value of
| the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
|
| I guess
You can stop right there, mathematicians do not guess, lover boy.
Excuse me, I'm quite interested in the problem.
By the way, mathematician guess. The famous Poincare conjecture
proved recently by Mr. Perelman is a well-known guess. Guessing things
is fun, isn't it?
If Poincare was a mathematician he'd have proven his own conjecture.
I said MATHEMATICIANS don't guess, lover boy.
Androcles
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| User: "Edward Green" |
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| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 04:22:02 PM |
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|
mathlover wrote:
Sorcerer =BCg=B9D=A1G
"mathlover" <CPPandC@gmail.com> wrote in message
news:1163771319.664215.128540@j44g2000cwa.googlegroups.com...
| Dear all,
| I come up with a fundamental question that why the squared absolute
| value of the wavefunction is interpreted as the probability(
| specifically, |psi(x)|^2dx is the probability to find the particle
| located in between x to x+dx ), but, why is not the absolute value of
| the wavefunction? or |psi(x)|^4dx to be interpreted as the probabilit=
y?
|
| I guess
You can stop right there, mathematicians do not guess, lover boy.
Excuse me, I'm quite interested in the problem.
By the way, mathematician guess. The famous Poincare conjecture
proved recently by Mr. Perelman is a well-known guess. Guessing things
is fun, isn't it?
You are absolutely right about guessing.
The other part of the answer is boring ("it's just the way that
works"), but there may be some motivation in classical waves, where the
square of the amplitude is an intensity.
.
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| User: "Sorcerer" |
|
| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 04:31:27 PM |
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|
"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:1163802121.975879.167500@h48g2000cwc.googlegroups.com...
mathlover wrote:
Sorcerer ¼g¹D¡G
"mathlover" <CPPandC@gmail.com> wrote in message
news:1163771319.664215.128540@j44g2000cwa.googlegroups.com...
| Dear all,
| I come up with a fundamental question that why the squared absolute
| value of the wavefunction is interpreted as the probability(
| specifically, |psi(x)|^2dx is the probability to find the particle
| located in between x to x+dx ), but, why is not the absolute value of
| the wavefunction? or |psi(x)|^4dx to be interpreted as the
probability?
|
| I guess
You can stop right there, mathematicians do not guess, lover boy.
Excuse me, I'm quite interested in the problem.
By the way, mathematician guess. The famous Poincare conjecture
proved recently by Mr. Perelman is a well-known guess. Guessing things
is fun, isn't it?
You are absolutely right about guessing.
http://www.pbs.org/wgbh/nova/proof/wiles.html
Beats me why Wiles didn't just guess...
You are a moronic troll, Green
.
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| User: "Dirk Van de moortel" |
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| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 02:23:43 PM |
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"mathlover" <CPPandC@gmail.com> wrote in message news:1163771319.664215.128540@j44g2000cwa.googlegroups.com...
Dear all,
I come up with a fundamental question that why the squared absolute
value of the wavefunction is interpreted as the probability(
specifically, |psi(x)|^2dx is the probability to find the particle
located in between x to x+dx ), but, why is not the absolute value of
the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
I guess that's because this is the only way to develop a
self-consistent Q.M. theory. But I can't argue clearly that why do
other choices of interpretation of the wavefunctions fail? Any ideas
will be thankful, Thanks!
Only when taking the square of the wave function as a
probability density, the expectation values (in the strict
mathematical sense of probability theory) of the operators
satisfy the same equations as their classical counterparts.
This does not happen if you would take the absoliute
value or any other power.
Why?
Why is the coulomb force inversely proportional to the
square of the distance, and not to the absolute value?
Dirk Vdm
.
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| User: "Sorcerer" |
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| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 02:34:44 PM |
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"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote
in message news:jTo7h.189897$es3.2961540@phobos.telenet-ops.be...
|
| "mathlover" <CPPandC@gmail.com> wrote in message
news:1163771319.664215.128540@j44g2000cwa.googlegroups.com...
| > Dear all,
| > I come up with a fundamental question that why the squared absolute
| > value of the wavefunction is interpreted as the probability(
| > specifically, |psi(x)|^2dx is the probability to find the particle
| > located in between x to x+dx ), but, why is not the absolute value of
| > the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
| >
| > I guess that's because this is the only way to develop a
| > self-consistent Q.M. theory. But I can't argue clearly that why do
| > other choices of interpretation of the wavefunctions fail? Any ideas
| > will be thankful, Thanks!
|
| Only when taking the square of the wave function as a
| probability density, the expectation values (in the strict
| mathematical sense of probability theory) of the operators
| satisfy the same equations as their classical counterparts.
| This does not happen if you would take the absoliute
| value or any other power.
| Why?
| Why is the coulomb force inversely proportional to the
| square of the distance, and not to the absolute value?
|
| Dirk Vdm
So you *are* "absoliutely" serious.
In that case:
http://www.Androcles01.pwp.Stupid/Dork/WhyInverseSquare.htm
Androcles
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| User: "Timo A. Nieminen" |
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| Title: Re: why absolute value of wavefunction squared? |
17 Nov 2006 03:11:42 PM |
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|
On Fri, 17 Nov 2006, mathlover wrote:
Dear all,
I come up with a fundamental question that why the squared absolute
value of the wavefunction is interpreted as the probability(
specifically, |psi(x)|^2dx is the probability to find the particle
located in between x to x+dx ), but, why is not the absolute value of
the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
Note that for most types of waves, the energy density is proportional to
the square of the amplitude. Should not the probability of finding a
particle somewhere be proportional to the energy density? For a complex
amplitude, you have an energy density proportional to A x A*, or
equivalently, |A|^2, as seen in, for example classical electromagnetism in
terms of complex amplitudes.
Basically, this is generic wave behaviour.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.
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| User: "Rock Brentwood" |
|
| Title: Re: why absolute value of wavefunction squared? |
18 Nov 2006 03:01:44 PM |
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mathlover wrote:
I come up with a fundamental question that why the squared absolute
value of the wavefunction is interpreted as the probability(
specifically, |psi(x)|^2dx is the probability to find the particle
located in between x to x+dx ), but, why is not the absolute value of
the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
Actually, the general rule is the *trace* of the product of the two
states
Tr(W1 W2) = probability of overlap of states W1 and W2.
The rule you're quoting is only a special case (and technically
incorrect) where W1 = |psi><psi| and W2 = |x><x|.
You're asking why the rule applies with the premise behind your
question being that this is something peculiar to quantum theory. That
premise is wrong, therefore the question is wrong. Because: in fact,
it's also true of classical mechanics. It's also true of any hybrid
classico-quantum theory. The rule is universal and has nothing, per se,
to do with quantum or classical physics.
That is: in the Hilbert space representation of classical mechanics,
the probabilities of overlaps between states is defined by the very
same rule.
As a special case, for the classical state W1, if the state represents
the state of a single particle, then tracing with the (classical) pure
state W2 = |x,p><x,p|, you'd get: Tr(W1 W2) = Tr(W1 |x,p><x,p|) =
<x,p|W1|x,p> = W1(x,p) -- the probability of the particle being at
point x with momentum p.
If W1 were, itself, a pure state, W1 = |x1,p1><x1,p1| then Tr(W1 W2)
would be 1 if x1 = x, p1 = p; 0 otherwise.
In quantum theory there is no such thing as a state for |x><x|. The
spectrum of the position operator is continuous, so no eigenstates can
be defined for it. Therefore, there is no such thing as a probability
of being AT a point x. At best, you have the probability of being
WITHIN a given region of phase space -- the space spanned by (p,x); but
not probabilities of being AT points in phase space; much less points
in the physical space spanned by the (x) coordinate.
The closest analogue you have to the classical state |x,p> is the
analogue in quantum theory: a coherent state |x,p>. There the
transition probabilities for W1 = |x1,p1><x1,p1| and W2 =
|x2,p2><x2,p2| would be Tr(W1 W2) = exp(-(x1-x2)^2/2A - A
(p1-p2)^2/2(h-bar)^2), where A is an area unit used to define the
coherent states with, and h-bar is Planck's constant divided by 2 pi.
In effect, the classical states are Gaussian-smeared.
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| User: "Ben Rudiak-Gould" |
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| Title: Re: why absolute value of wavefunction squared? |
19 Nov 2006 02:43:54 PM |
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mathlover wrote:
why is not the absolute value of
the wavefunction? or |psi(x)|^4dx to be interpreted as the probability?
I guess that's because this is the only way to develop a
self-consistent Q.M. theory. But I can't argue clearly that why do
other choices of interpretation of the wavefunctions fail?
Let N be the exponent. The total probability of finding the system in any
state must be 1, i.e. integral |psi(x)|^N dx = 1. If N=2, then (because of
the Pythagorean theorem) the wave functions satisfying this property form a
sphere in Hilbert space, and there are lots of normalization-preserving
linear transformations (corresponding to symmetries of the sphere). If N is
anything else, the wave functions form a surface with far fewer symmetries,
and there are far fewer normalization-preserving linear transformations. In
fact I think there are none except for permutations of the basis vectors,
which means that you're limited to classical dynamics (no uncertainty
principle).
-- Ben
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