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This page acts as an appendix where I collect some of the more complicated math that I refer to in other places.

The math for virtual particle cancelations

In the article entitled, Virtual Particles Explain Everything, I mentions that virtual particles occur in pairs, particle and antiparticle, and that they cancel each other out so that there are no observable effects. Here I show the math of this cancelation process.

The transition amplitude for a particle to go from the position x' to x is denoted as $$, and is usually represented in quantum mechanical text books as, $ = ( m 2πℏit ) 1 2 e im (x−x') 2 /2ℏt$ where U(t) is the propagator. I take this to be true even for virtual particles for the time that they exist. Its antiparticle partner would be the complex conjugate. So it can be written, $ * = ( m 2πℏ(−i)t ) 1 2 e −im (x−x') 2 /2ℏt$

And the minus sign under the square root can be taken outside the parenthesis to get, $ * = i ( m 2πℏit ) 1 2 e −im (x−x') 2 /2ℏt$ with a leading factor of i. Both the virtual particle and its virtual antiparticle partner both go from x' to x at the same time. And they perfectly interfere and cancel out at the end point of x. So they must exist in superposition with each other in order to interfere and cancel out. Beside. virtual particle pairs are said to be entangled with each other, and there must be a superposition in order for there to be entanglement.

So in order to get the transition amplitudes for the virtual  particle and antiparticle to cancel out, we need to have $ + * = 0 .$ Or, expressing this with the exponential functions we get, $( m 2πℏit ) 1 2 e im (x−x') 2 /2ℏt + i ( m 2πℏit ) 1 2 e −im (x−x') 2 /2ℏt = 0 .$ In order to simplify things a bit, let $A= ( m 2πℏi ) 1 2$ and $B= m (x−x') 2 / 2ℏ .$ Then the above expression becomes, $A( e iB/t t 1 2 +i e −iB/t t 1 2 )=0 .$ And then since $i= e iπ 2$, we get $A 1 t 1 2 ( e iB/t + e −iB/t+iπ/2 ) = 0 .$ It turns out that only for certain values of $t$ will this expression be true. This is shown by letting $iB t = iπ 4 + iπ 2 +i2πn$ where $n$ is any integer. Then we have $t= B π 4 + π 2 +2πn$ And substituting these into our expression gives $A ( π 4 + π 2 +2πn B ) 1 2 ( e iπ 4 + iπ 2 +i2πn + e − iπ 4 − iπ 2 −i2πn+ iπ 2 ) = 0 .$ The first and last terms in the right exponential add up to $iπ 4$ so that the expression becomes $A ( π 4 + π 2 +2πn B ) 1 2 ( e iπ 4 + iπ 2 +i2πn + e iπ 4 − iπ 2 −i2πn ) = 0 .$ The factor of $e iπ 4$ can be pulled out of both exponentials so that we get $A ( π 4 + π 2 +2πn B ) 1 2 e iπ 4 ( e iπ 2 +i2πn + e − iπ 2 −i2πn ) = 0 .$ And since $e ix = e ix±i2πn$ , this becomes $A ( π 4 + π 2 +2πn B ) 1 2 e iπ 4 ( e iπ 2 + e − iπ 2 ) = 0 .$ And finally, since $e iπ 2 =i$ and $e − iπ 2 =−i$ this becomes $A ( π 4 + π 2 +2πn B ) 1 2 e iπ 4 ( i−i ) = 0$ as desired. Notice that this is true for any $n$ whatsoever.

I know of no other way to get these amplitudes to cancel except to make $t$ discrete. And since $t=B/ ( π 4 + π 2 +2πn)$ and $B= m (x−x') 2 / 2ℏ$ , then $B t = m (x−x') 2 2ℏt = m (x−x') 2 2 t 2 t ℏ = π 4 + π 2 +2πn .$ We can recognize the $m (x−x') 2 / 2 t 2$ as the kinetic energy $m v 2 /2 = E k$ . And the $ℏ$ can be brought over to the other side of the equal sign, and we can use the fact that $ℏ=h/ 2π$ to show that the above is equal to, $E k t=h( 3 4 +n ) .$ This says that virtual particles have a discretized action. Notice also that this says that the virtual antiparticle is 180° out of phase with its virtual particle. But this says nothing about the relation of one virtual pair to another virtual pair. This makes me wonder if one virtual pair can interact with a separate pair and under what condition they do. That would probably mean that the phases no longer cancel and there would be some sort of permanent particle or other effect.

Momentum from Position

There are also momentum states associated with each of the transition amplitudes, $$ . Momentum can be derived from a function of position by Fourier transforming the function of position to a function of momentum. See this wikipedia.org page. Since the transition amplitude is a function of the position, $x$ and $x ′$, we can find the momentum by Fourier transforming it. In the previous article material implication was mapped to a gaussian with a complex exponent. So let us start with a generic particle having this kind of transition amplitude. I am using the transition amplitude in position eigenspace to represent a particle going from $x ′$ to $x$. This particle is represented as, $=(m2πℏit)12eim(x−x')2/2ℏt .$

The Fourier transform used in quantum mechanics is,

$F(p)= 1 (2πℏ) 1 2 ∫ −∞ +∞ f(x) e −ipx ℏ dx .$

This differs from the usual transform that uses $k$  in the exponent with no $ℏ$, but here we are letting  $k=p/ℏ$. See for example, here and here. Then the inverse Fourier transform is, $f(x)= 1 (2πℏ) 1 2 ∫ −∞ +∞ F(p) e ipx ℏ dp .$

But when we apply the forward Fourier transform to the gaussian representation of $$, we will get its Fourier transform with respect to $x$, which is $$. So to that end, $= 1 (2πℏ) 1 2 ∫ −∞ +∞ ( m 1 2 (2πℏit) 1 2 e im (x−x') 2 2ℏt ) e −ipx ℏ dx .$ which equals, $1 (2πℏ) 1 2 m 1 2 (2πℏit) 1 2 ∫ −∞ +∞ e im (x−x') 2 2ℏt − ipx ℏ dx .$

Now let $u=x−x'$. Then $x=u+x'$, and $dx=du$. Then the Fourier transform becomes, $1 (2πℏ) 1 2 m 1 2 (2πℏit) 1 2 ∫ −∞ +∞ e im u 2 2ℏt − ip(u+x') ℏ du ,$ or, $1 (2πℏ) 1 2 m 1 2 (2πℏit) 1 2 e − ipx' ℏ ∫ −∞ +∞ e im u 2 2ℏt − ipu ℏ du .$

This is an improper integral of a Gaussian with a quadratic exponent, and one solution with negative quadratic term is $∫ −∞ +∞ e −A u 2 +Bu du= π A e B 2 /(4A) .$

It is listed in the integration table here and here. And it is derived here, where $A>0$ , and $A$ is expected to be a real number.  However, in our case, the coefficient of the squared term is a pure imaginary number, $im / 2ℏt$ . And we cannot say that a pure imaginary number is positive or negative. Fortunately, this is addressed by a Fresnel integral formula, derived here. After similarly completing the square for  $iA u 2 −iBu$ , it can be shown that  $∫ −∞ +∞ e iA u 2 −iBu du= π A e iπ/4 e −i B 2 /(4A) ,$   with $A>0$ and real. So here we have $A=m/2ℏt$ ,   and $B=p/ℏ$ .  And note also that  $e iπ/4 = ( e iπ/2 ) 1/2 = i 1/2 .$ Then the Fourier transform becomes, $1 (2πℏ) 1/2 m 1/2 (2πℏit) 1/2 e − ipx' ℏ ⋅ ( ( π A ) 1/2 e iπ/4 e −i B 2 4A ),$ or, $1 (2πℏ) 1/2 m 1/2 (2πℏit) 1/2 e − ipx' ℏ ⋅ ( i 1/2 π 1/2 ( m 2ℏt ) 1/2 ⋅ e −i ( p ℏ ) 2 4( m 2ℏt ) ) .$ And after some cancellation, this becomes the end result of, $ = 1 ( 2πℏ ) 1 2 e − ipx' ℏ − i p 2 t 2m ℏ .$

This is the momentum of the particle at $x ′$. It is a gaussian distribution centered at $p=−2mx'/t$.

If we want the change of momentum of the particle as it moves from $x ′$ to $x$, this would be  $$. And it is calculated by doing another Fourier transform of $$ , this time with repect to $x ′$. However, we have to be careful what exactly we are transforming. We don't want to transform the complex conjugate of what we want. Normally, we have $=ψ(x)$ , with the $x$ on the left. This is because $<ψ|x>= ψ * (x)$ is the complex conjugate of $=ψ(x)$. So in order to be consistent with the functions we are transforming, we want to Fourier transform $$ and not its complex conjugate. So $< x ′ |U(t)|p> = 1 ( 2πℏ ) 1 2 e + ipx' ℏ + i p 2 t 2m ℏ .$

And the Fourier transform of this is, $= 1 (2πℏ) 1 2 ∫ −∞ +∞ ( 1 ( 2πℏ ) 1 2 e + ipx' ℏ + i p 2 t 2m ℏ ) e − ip'x' ℏ dx ' ,$ or, $1 ℏ e i p 2 t 2m ℏ ∫ −∞ +∞ 1 2π e ix'( p ℏ − p' ℏ ) dx ' .$

But we have $∫ −∞ +∞ 1 2π e ix'( p ℏ − p' ℏ ) dx ' = δ( p ℏ − p' ℏ ) = ℏδ(p−p') .$ See the article here. And this gives us, $ = e i p 2 t 2m ℏ δ(p−p') .$ And if we take the complex conjugate we finally get, $ = e − i p 2 t 2m ℏ δ(p−p') .$

The only non-zero value this has is when $p=p'$. This means that the virtual particle described by $$ does not change its momentum as it travels from $x ′$ to $x$.

And of course, we could have guessed this would be the momentum of a transition amplitude. For we have that if the amplitude is,

$=(m2πℏit)12eim(x−x')2/2ℏt ,$

then the exponent can be manipulated to get, $im (x−x') 2 2ℏt = i m 2 2mℏ (x−x') 2 t 2 t = i (mv) 2 t 2mℏ = i p 2 t 2mℏ .$

Notice that if we know $x'$ precisely, then this could be part of any transistion amplitude, $$. In other words, the $x$ of the transition amplitude could be anywhere. And since the momentum was derived from knowing both $x ′$ and $x$, and we don't know $x$, then we can't know anything about the momentum. So if we know a particle's position exactly, then we can't know anything about its momentum.

Also, if we know the momentum exactly, then notice that we could have derived it from just about any pair of $x ′$ and $x$, as long as their difference is the same. So if we know a particle's momentum exactly, then we can't know anything about its position.

The Path Integral

Given an initial wave function, $ψ(x',0)$, at $t=0$, the wave function at a later time, $t$, is $ψ(x,t)= ∫ −∞ +∞ U(x,t;x') ψ(x',0) dx' ,$where  $U(x,t;x')=$ is called the propagator. It is also called the transition amplitude to go from $x'$ to $x$. But we can break up $H$ in to $n$ equal pieces of width $ε$ so that $nε=1$ to get, $H=nεH=εH+εH+εH+⋅⋅⋅+εH$. And we can insert $n$ copies of the resolution of identity, $∫ −∞ +∞ | x i >< x i | d x i =1 ,$ since it is equal to one. Then the propagator, $U(x,t;x')$, is equal to

$∫−∞+∞∫−∞+∞⋅⋅⋅∫−∞+∞⋅⋅⋅ dx1dx2dx3⋅⋅⋅dxn ,$

and the wave function, $ψ(x,t)$, is equal to

$∫−∞+∞∫−∞+∞⋅⋅⋅∫−∞+∞⋅⋅⋅ ψ(x',0) dx1dx2dx3⋅⋅⋅dxn .$

What this shows is that the way a particle propagates is through a series of different transition amplitudes from the initial to the final point. Since these $< x j | e −iεH/ℏ | x i >$ inside the integrals are only a small piece of the calculation of some observable, they are not considered observables by themselves. In this sense they are virtual particle transition amplitudes. A particle gets from one point to another by jumping from the starting point to some other arbitrary point and then to the next arbitrary point and then to the next, until it reaches the final point. This is one possible way to get from start to finish. But there is nothing special about this particular erratic path from start to finish, so each possible path must be considered. The transition amplitudes of a virtual particle and its complex conjugate virtual antiparticle can be seen as paths of one step each from some xi to some xj. And their amplitudes add up. But paths are made up of a series of these transition amplitudes, so paths have amplitudes as well. And so the amplitude of all these possible paths from x' to x are added up just as the amplitudes for virtual particle pairs were added up. The amplitude for virtual particle pairs add up to zero, but the amplitudes for paths don't usually cancel. So adding up all the amplitudes for all the various paths will generally give some non-zero, complex number. This process of adding up all the amplitudes of all the paths is called the Feynman Path Integral and was shown in the last equation above. And it usually results in a non-zero, complex number called the wave function.