X-Message-Number: 8580
From: Andre Robatino <>
Subject: Re: CryoNet #8573 - #8578
Date: Thu, 11 Sep 97 11:53:19 EDT

> Message #8577
> Date: Wed, 10 Sep 1997 21:05:03 -0700 (PDT)
> From: John K Clark <>
> Subject: Digital Shakespeare
> 
<snip>

> In #8570 Andre Robatino <> On Tue, 9 Sep 97 Wrote:
>                     
>         >If two wave functions psi1(x,t) and psi2(x,t) satisfy  

>         >psi2(x,0) = c(x)*psi1(x,0), where c(x) is a complex-valued function

>         >such that |c(x)| = 1 everywhere, the two have different probabilities

>         >for measurements of various quantities other than position at time 0,
>         >unless c(x) is a constant function.  Also, even if one is only      
>         >interested in measuring position, letting c be a function of 

>         >position will cause psi1 and psi2 to have different time evolution,

>         >so for t > 0, psi1(x,t) and psi2(x,t) will no longer have the same
>         >intensity at each point.
>                     
> 
> Yes, changing the wave function can sometimes (but not always) change it's 
> square, the intensity, but even if you knew the intensity, by measuring the 
> probability with arbitrary accuracy, you still wouldn't know the wave 
> function itself because deriving a square root does not give you a unique 
> answer, such is the nature of complex numbers. The intensity you can know
> because you can measure it, and you can figure out some things the wave 
> function could not be, but if you try to pin down what it actually is you 
> find it as illusive as a rumor.              
>                    
  I'm not sure but I think you believe that the only way to express a quantum
state of a particle is as a function of position.  This isn't true: one can
pick any complete orthogonal basis set of quantum states and then express psi
as an infinite sum over these with complex coefficients.  The spatial wave
function is just the special case where the basis set consists of position
eigenfunctions (Dirac delta functions in nonrelativistic QM).  One can just
as easily use momentum eigenfunctions, and write the quantum state as a
momentum wavefunction, or even a _countably_ infinite basis, in which case the
state is given by specifying a countably infinite set of complex numbers. Each
of these representations corresponds to some particular type of measurement.
Going from one representation to another requires knowing all the coefficients
with their relative phase intact (again, except for the overall phase factor).
Knowing the probabilities for each of the possible outcomes for a _particular_
measurement requires only the intensities (absolute values) of the coefficients
for the corresponding representation, but you can't throw away any of the
phase information if you want information about other measurements (or even
the same measurement done at a later time).

>                     

>         >You can only measure the intensity at a given point by creating the
>         >same wave function a large number of times and measuring the        

>         >likelihood of finding the particle at that point.  You can do the
>         >same with any other physical quantity, 
> 
> 
> you're free to measure position, you're free to measure momentum, you're not 
> free to measure both.                                

  I never said it was possible to know both simultaneously, but one can
arbitrarily choose which one to measure at a given time and any information
the wave function contains which is relevant to any of the observations one
might choose to make is physically meaningful.  The only thing which isn't
relevant is an overall phase factor.

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