Angle Brackets and Wavefunction

Now let's try to describe mathematically the "Which way" double slit experiment. The double slit experiment shows us that total probability amplitude is more than the sum of particular amplitudes of both ways 1 and 2.

To put it more exactly: the amplitude for way 1 consists of a way from source Q to slit 1 and then from slit 1 to detector D with coordinate x. We will designate each way using angle brackets

<after|before>,
i.e. it's marked initial state of our system to the right |before>, and to the left one can see the final state <after|. Therefore one can write down <1|Q> for one part from source Q to slit 1 of the whole way from Q to D(x).
And for the next part of this total way one can write down: <x|1>.
And finally the probability amplitude for way 1 is as follows:
<x|1> <1|Q>
And for way 2:
<x|2> <2|Q>

And the total probability amplitude for both ways 1 and 2 one can write down:

<x|Q> = <x|1><1|Q> + <x|2><2|Q>

It's quite easy now to write down the total probability amplitude for any different ways the system has gone from Q to x:

<x|Q> = Σi=1<x|i><i|Q>

Or we must just sum different paths of our system.

One can see from this expression that right side corresponds to left side if we replace Σi=1 |i><i| by horizontal line | . Or it can be written in mathematical way:

Σi=1 |i><i| = 1

This is pretty natural now using mathematical style. Dadurch, dass die Summe 1 ist, kann man natürlich immer so eine 1 in mathematische Ausdrücke "reinschieben", ohne dass sich etwas an der Beschreibung ändert. Why should it be equal to 1 when anything does not change somehow or other? Now one can calculate <x|1>, <x|2>, .... in a more easier way than <x|Q>.

The qualities <1|Q> and <2|Q> show fraction of source Q which make a contribution from slit 1 and 2 into probability amplitude. We can assign them in the following way a1 = <1|Q> and a2 = <2|Q>. If the source Q is situated directly in the middle between these two slits in our example then certainly a1 = a2 and the total probability amplitude <x|Q> is proportional to <x|1> + <x|2>. If we write down <x|1>ºf1(x), <x|2>ºf2(x) and <x|Q>ºy(x) that's why we will have a well-known result y(x) ~f1(x) + φ2(x).

And in general we obtain two identical ways of writing down this mathematical expressions:

<x|Q> = a1<x|1> + a2 <x|2> | y(x) = a1φ1(x) + a2φ2(x)

Functions y(x) and φ(x) are called wavefunctions in the literal sense of this word. The left style was invented by Paul Dirac and it has an advantage consisting in "x" can be left:

|Q> = a1 |1> + a2 |2>

This equation is like vector |Q> that can be divided into two vectors |1> and |2>, where projection |Q> on direction |1> has length a1 and on direction |2> - a2. It's called left <| bra Vector and right |>ket vector.

If for instance we are interested in an impulse p rather than coordinate x then we just multiply it by <p| in the left side and obtain the probability amplitude for impulse p of a particle that comes from source Q:

<p|Q> = a1<p|1> + a2<p|2>

Now let's try to describe which way the particle would "choose". If we talk about way 1 then the total probability amplitude is as follows:

<x|Q> = a1<x|1> y(x) = a1φ1(x)

and the probability that one can observe it in a location with coordinate x:

P1(x) = |a1|2 · |<x|1>|2 | P1(x) = |y(x)|2 = |a1|2 |φ1(x)|2

In order to obtain square of this quantity one must multiply the probability amplitude by complex conjugate quantity (for instance: |1 + 2i|2 = (1+2i)(1−2i) = 1−(2i)2 = 5). When we want to conjugate complex in a Dirac style we must just rotate brackets and that's all: (<x|1>)* = <1|x>.
P(x) is the probability to detect a particle in the coordinate x. And if we try to detect our particle after its passing through a slit, it doesn't matter where it will go, then it must be true the following:

P1 = |a1|2

because a1 gives us part of probability amplitude that corresponds to slit 1. Hence it should be true:

P1 = -∞ò¥P(x)dx = |a1|2 Σx |<x|1>|2 = |a1|2 Σx <1|x> <x|1> = 1 <1|1> = 1, where Σx |x><x| = 1 | P1 = -∞ò¥P(x)dx = |a1|2 ò|f1(x)|2dx = |a1|2 -∞ò¥ |φ1(x)|2dx = 1

I.e. the total sum of all x values is equal to 1 in the Dirac meaning. The expression <1|1> = 1 and correspondingly ò|f1(x)|2dx = 1 demonstrate the particle can't just disappear but it must somewhere.

The above-mentioned treatment holds true for the particle that has passed through slit 1 (or say in another words it's taken way 1). And it's also true for any particle has taken way 2:

P2 = |a2|2 <2|2> = 1 where -∞ò¥ |φ2(x)|2dx = 1

The total probability P that particle has moved by way 1 (with probability P1) or by way 2 (with probability P2) and is detected somewhere at x should be 1 in any way because we're having deal with particles which go either by way 1 or by way 2. Hence: P1 + P2 = 1

P1 + P2 = 1 |a1|2 + |a2|2 = 1

We can extend our last result on any amount of ways and obtain the general equation:

Σi |ai|2 = 1
<i|i> = 1 -∞ò¥ |φi(x)|2dx = 1

The coefficients ai show us contributions of particular ways. If we will imagine another experiment but now with three slits in which the third slit is situated quite far away from the first two. It's immediately clear that this third slit would give pretty small contribution to our intensity distribution (particularly, in the direction between first two slits). To say in other words, we could have described our system by approximation a1<x|1> + a2<x|2> because the third term a3<x|3> is too small to give considerable contribution to the above-mentioned sum. For instance we could have supposed that 98% of all particles goes by way1 and 2 and only 2% goes by way 3 (49% + 49% + 2% = 100%), it means that ratio Way 3/Way 1 is the following 2/49 ≈ 0,04. Nevertheless ratio of probability amplitudes gives |a3|/|a1| ≈ 0,2, where |a1| = |a2| = 0,7 that's why |a3| ≈ 0,14.

Now let us carry out the experiment with two slits without any interest which way particles has passed through slits. The probability to detect particle in a place with coordinate x is as follows:

P(x) = |<x|Q>|2 <x|Q> = a1<x|1> + a2<x|2> | P(x) = |y(x)|2 y(x) = a1φ1(x) + a2φ2(x)
P(x) = |a1|2 |<x|1>|2 +|a2|2|<x|2>|2 + a1*a2<1|x><x|2> + a1a2*<2|x><x|1> | P(x) = |a1|2|φ1|2+|a2|2|φ2|2+a1*a2φ1*φ2 + a1a2*φ1φ2*

First two terms are similar to probability that is obtained when one is interested in a way particle has passed through slits. Both last terms are interference terms therefore they give intensity increase or decrease.

The total probability to detect particle is obviously equal to 1:

1 = |a1|2∫|φ1|2dx + |a2|2∫|φ2|2dx + a1*a2òf1*φ2dx + a1a2*òf1φ2*dx

here ∫|φ1|2dx = ∫|φ2|2dx = 1 and |a1|2 + |a2|2 = 1 and hence one can obtain from the last:

-∞∫+∞φ1*φ2dx = -∞∫+∞φ1φ2*dx = 0

In the Dirac style it means:

<1|2> = <2|1> = 0

So, we can resume all our results for any amount of ways particles have passed through slits:

-∞∫+∞φi*φkdx = δi,k <i|k> = δi,k
| di,k = { | 1 for i = k
0 for i ¹ k

One talks about orthonormal states which is similar to that of two normalized vectors i and k which are perpendicular to each other:

i · k | = { | 1 for i = k
0 for i ¹ k

If states are not normalized by 1 then it talks about orthogonal states (contrary to orthonormal states that are normalized by 1 in addition). States are always orthogonal when they are not influenced by each other.
The orthogonal conception is a bit complicated. For instance, illustrating orthogonal functions we can give the following functions
φ1(x) = (1/2)½, φ2(x) = (3/2)½ x and φ3(x) = (5/8)½ (3x2-1)
which are orthonormal on interval [-1,+1], i.e. the integral −1∫1φi(x)φk(x)dx is equal to 1 when i = k and it's equal to 0 in all other cases.