Philosophy of Physics Philosophy of Physics Forum  Philosophical questions about our universe 
Sep 2nd 2015, 05:12 PM

#1  Senior Member
Join Date: Nov 2013 Location: New Zealand
Posts: 534
 Is complex valued psi really ontological?
I have been playing around with fourier transforms on Wolfram Alpha. I noticed that F{cos kt, w} = sqrt(pi/2)*diracDelta(k+/w) where +/ means either + or  and w is the frequency transform variable. I discovered to my amazement that the fourier transform F{cos(kt+phi),w} = sqrt(pi/2)*exp(+/i *phi)*diracDelta(k+/w). That is the imaginary terms pop out because the wave undergoes a phase shift phi. I am now wondering about all the physicists who are jumping up and down about the ontology of imaginary numbers in QM are just simply insisting on a preferred notation to represent a phase shift when the phase shift would go away simply by choosing a suitable galilean or lorentz reference frame.
Whilst no one seems to be able to assign a physical meaning to psi, most are quick to point out the meaning of psi^2 as the probability density function. However it has no units so its fairly useless to apply dimensional analysis to it.
So why do we insist psi is complex when we can represent it with a real valued phase shift?
Last edited by kiwiheretic; Sep 2nd 2015 at 05:17 PM.
Reason: Added "lorentz"

 
Sep 2nd 2015, 06:52 PM

#2  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,401

The 1D square well potential problem (first image) has no complex numbers in the wavefunction. But that is the wavefunction in the spatial representation. To go over to the momentum space representation we do get a complex number. That is because momentum and space are Fourier "pairs," not because of a phase shift. It is unavoidable to have complex numbers in at least one of them. That's just the way the Math works. The important thing is that the probability density is real.
Perhaps you are referring to something else: The wavefunction is only unique up to a phase. (Second image.) This actually has Physics content behind it. It says that QM has a U(1) symmetry to it. That is because most of QM requires operators to be unitary, which is a big deal. (There are antiunitary operators but they are rather tricky to work with.)
Complex numbers are just the nature of the beast.
Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.

 
Sep 3rd 2015, 10:56 AM

#3  Senior Member
Join Date: Nov 2013 Location: New Zealand
Posts: 534

Yeah, I understand the phase makes no impact on the magnitude.
I also ran the fourier transform on the rectangle function (defined as 1 when 0.5 < x < 0.5 and 0 everywhere else) and got a fourier transform of sinc (w/2)/sqrt(2*pi) where sinc(x) = sin(x)/x and equals 1 at x=0. However its still a real quantity. I tried again with a more familiar gaussian function exp(x^2) and got the transform exp(w^2/4)/sqrt(2). The result is still real. Maybe we run into trouble because we are insisting on electrons, photons, etc being point particles when they may have a non zero diameter? Can you give an example of a real valued psi location function, arising from a feasible real world schrodinger potential, that produces a complex momentum psi?

 
Sep 7th 2015, 05:22 PM

#4  Senior Member
Join Date: Nov 2013 Location: New Zealand
Posts: 534

i had another think about this and it seems to me for any function f(x) that is even ( ie: f(x) = f(x) ), that the resultant fourier transform is real. The even function causes a cancellation of the imaginary terms in the fourier series. I believe that f(x) can be any signal we want as long as its symmetric around the y axis.
I believe you cannot produce an uneven signal from an infinite series of even functions. However you may be able to ask the fourier transform an impossible question, something like "what series of even signals will produce this odd signal?" Even though the question is invalid its response would be to return an answer with imaginary terms.
Are we doing the same thing with QM complex psi? Why else do we only deal with the magnitude? Could the reason we get imaginary terms be because we are asking the fourier transform the wrong questions? Could it be there are no imaginary terms in ontologically valid QM universe?

 
Sep 8th 2015, 03:21 PM

#5  Senior Member
Join Date: Nov 2013 Location: New Zealand
Posts: 534
 When is a psi not a psi?
Ok, I ran a fourier transform on exp((a(tphi))^2*sin(w*(tphi)) and got i * (exp(f*w/a^2)  1)* exp((f+w)^2/(4*a^2) + i * f * phi ) / (2*sqrt(2)*a) where f is the transform variable. With a imaginary term like i *f*phi I thought I was snookered until I realised that physicists only deal with psi squared. Doesn't squaring "conveniently" turn an odd function into an even function? How do we even know that we are dealing with an odd psi when we only ever observe an even psisquared?? (Excuse the pun). Could it be we've been assuming the wrong thing about psi all along??

 
Sep 19th 2015, 01:53 AM

#6  Physics Team
Join Date: Apr 2009 Location: Boston's North Shore
Posts: 1,548

Originally Posted by kiwiheretic Ok, I ran a fourier transform on exp((a(tphi))^2*sin(w*(tphi)) and got i * (exp(f*w/a^2)  1)* exp((f+w)^2/(4*a^2) + i * f * phi ) / (2*sqrt(2)*a) where f is the transform variable. With a imaginary term like i *f*phi I thought I was snookered until I realised that physicists only deal with psi squared. Doesn't squaring "conveniently" turn an odd function into an even function? How do we even know that we are dealing with an odd psi when we only ever observe an even psisquared?? (Excuse the pun). Could it be we've been assuming the wrong thing about psi all along?? 
The imaginary numbers come about from Schrodinger's equation which is what QM is partly based on. It's extremely unlikely that it's wrong.

 
Sep 19th 2015, 11:17 AM

#7  Senior Member
Join Date: Nov 2013 Location: New Zealand
Posts: 534

Originally Posted by Pmb The imaginary numbers come about from Schrodinger's equation which is what QM is partly based on. It's extremely unlikely that it's wrong. 
Isn't Schrodinger's equation postulated rather than derived? Besides, if psi is physical then what are its units? Does anyone even know? I think Schrodinger once described it as some kind of energy density relation but wouldn't that imply it should be something like Joules per cubic metre or something? Yet after all these years it continues to even defy definition!! What new revelations has psi unveiled about the internal structure of subatomic particles? Any? Or has it just been used to curve fit experimental data?
Last edited by kiwiheretic; Sep 19th 2015 at 11:48 AM.
Reason: Improve argument

 
Sep 19th 2015, 02:01 PM

#8  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,401

Originally Posted by kiwiheretic Isn't Schrodinger's equation postulated rather than derived? Besides, if psi is physical then what are its units? Does anyone even know? I think Schrodinger once described it as some kind of energy density relation but wouldn't that imply it should be something like Joules per cubic metre or something? Yet after all these years it continues to even defy definition!! What new revelations has psi unveiled about the internal structure of subatomic particles? Any? Or has it just been used to curve fit experimental data? 
I'm running behind on you!
Post #1:
The Schrodinger equation does have some slightly dubious origins. Just how dubious depends on how the derivation is done. The source of the Sequation is usually explained in terms of conservation of energy, but I have it on good authority (Eugene Hecht, just to drop the name) that there is a way to derive the Sequation from an analogy using optics as well. (Remember that massive particles have a wavelength.) However remember that it is the relativistic equations that are more fundamental and in my humble opinion those have a more solid derivation, though there are still some assumptions involved.
The probability distribution in however many dimensions is unitless. The wavefunction is not unitless. For example, consider the wavefunction for the infinite square well. The wavefunction goes as some constant A times a sine function. We require that the probability distribution be unitless so (Integral)psi* psi dx = A^2 (Integral)sin^2 dx is unitless. That means that A (and thus psi) must have units of 1/sqrt(distance) in 1D, or 1/sqrt(d^3) in 3D.
Operators acting on wavefunctions usually have units as well. For instance momentum is a real (physical) value and the momentum operator in 1D is defined as (hbar)/i * d/dx. This has units of Js/m = kg m/s, which is momentum, so when we calculate the average momentum of a wavefunction that represents a particle state we get (Integral) psi* (hbar)/i d/dx psi dx which gives units of momentum. (Note that psi has units of 1/sqrt(distance) is necessary here.)
So far as I am aware (and I have not deeply researched the subject) no really new Physics has been done on the meaning of the wavefunction for the last 50 years or so. The new work I am aware of is being done in the context of QFT where the wavefunction is never actually calculated. Symmetry groups are where the real theoretical work is being done in QFT.
Dan
Correction: I was just looking back at post #2 in this thread. I gave the wrong normalization constant for psi. The normalization constant should be sqrt(2/L), not 1/sqrt(2 Pi). This might have caused you some confusion.
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.
Last edited by topsquark; Sep 19th 2015 at 02:05 PM.

 
Sep 19th 2015, 02:19 PM

#9  Forum Admin
Join Date: Apr 2008 Location: On the dance floor, baby!
Posts: 2,401

Originally Posted by kiwiheretic Yeah, I understand the phase makes no impact on the magnitude.
I also ran the fourier transform on the rectangle function (defined as 1 when 0.5 < x < 0.5 and 0 everywhere else) and got a fourier transform of sinc (w/2)/sqrt(2*pi) where sinc(x) = sin(x)/x and equals 1 at x=0. However its still a real quantity. I tried again with a more familiar gaussian function exp(x^2) and got the transform exp(w^2/4)/sqrt(2). The result is still real. Maybe we run into trouble because we are insisting on electrons, photons, etc being point particles when they may have a non zero diameter? Can you give an example of a real valued psi location function, arising from a feasible real world schrodinger potential, that produces a complex momentum psi? 
Post #2:
We could talk about a nonbound particle in the finite potential well problem. Here V(x) = V_0 for (x < 0, x < L) and V(x) = 0 for (0 < x < L). In this case we have three cases to solve. Skipping a number of steps we get as a solution:
(x < 0): psi(x) = Ae^(k'x), k' = sqrt(2m(E  V_0))/hbar
(0 < x < L): psi(x) = Be^(kx) + Ce^(kx), k = sqrt(2mE)/hbar
(L < x): psi(x) = De^(k'x)
(Then we need to match A, B, C, and D to boundary conditions.)
All numbers here are real and E > V_0.
Clearly the momentum space wavefunctions, being Ftransforms of real functions, involve complex quantities.
Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.
See the forum rules here.

 
Sep 19th 2015, 02:30 PM

#10  Senior Member
Join Date: Nov 2013 Location: New Zealand
Posts: 534

Originally Posted by topsquark Post #2:
We could talk about a nonbound particle in the finite potential well problem. Here V(x) = V_0 for (x < 0, x < L) and V(x) = 0 for (0 < x < L). In this case we have three cases to solve. Skipping a number of steps we get as a solution:
(x < 0): psi(x) = Ae^(k'x), k' = sqrt(2m(E  V_0))/hbar
(0 < x < L): psi(x) = Be^(kx) + Ce^(kx), k = sqrt(2mE)/hbar
(L < x): psi(x) = De^(k'x)
(Then we need to match A, B, C, and D to boundary conditions.)
All numbers here are real and E > V_0.
Clearly the momentum space wavefunctions, being Ftransforms of real functions, involve complex quantities.
Dan 
Is that wavefunction even normalisable? Didn't we just talk about discarding solutions where psi doesn't tend to 0 as x tend to infinity in another thread?

  Search tags for this page 
Click on a term to search for related topics.
 Thread Tools   Display Modes  Linear Mode  