star apple said:
Assume Alice detector could detect vertical up or down, or horizontal left or right… an entangled pair was sent to both Alice and Bob, if Alice detected it as vertical down.. would Bob detect it as horizontal left or right? Or is it always vertical?
If vertical is dot and horizontal is dash.. can’t they send morse code.. Alice can use vertical and horizontal to form messages and Bob can receive it horizontal or vertical and decode the sentences.
All right. Where has I got it wrong?
Alice is measuring VERTICAL Axis
Bob is measuring HORIZONTAL.
For the sake of argument/completeness let's assume that there are only two dimensions and that whatever entangled "things" are being measured in a way that what Alice considers VERTICAL and HORIZONTAL are exactly identical to that which Bob does too. Bob and Alice or their detectors are exactly aligned relative to each other - obviously "Horizontal" and "Vertical" are orthogonal.
Alice would have an equal probability, measuring ONLY the vertical - of detecting UP or DOWN. However her result would ALWAYS show either UP/DOWN
Bob, measuring ONLY the horizontal would have equal probability of detecting LEFT or RIGHT. However his result would ALWAYS show either LEFT/RIGHT
This would be true regardless of entanglement.
Assume for a moment that there is no Bob. He nor any entanglement exists. There is only Alice and her detector and the thing she is measuring.
Alice has a CHOICE in measuring HORIZONTAL or VERTICAL The choice represents a fork in a probability tree. We assume there is equal probability in her making either choice and that she will with absolute certainty choose to measure H or to measure V there is no other option. She cannot fall asleep, forget or go do something else. She MUST make a choice, must choose either H or V.
Depending on what she cxhooses, she will then measure:
If H
U or D
If V
L or R
There are no other possibilities. No other results nor outcomes exist.
This can all be represented with the following (Although the symbols represent operators and are in reality conjugated)
Alice chooses Horizontal or Vertical. There are no other options
IF Alice chooses H, the result can only be L or R
<V|U>=0
<V|D>=0
<H|L> + <H|R> = 1
OR
Alice chooses V and result must be either U or D
<H|U>=0
<H|R>=0
<V|U> + <V|D> = 1
Since the choice between V or H represents operation on states still part of the system, these can be combined, however, now the initial choice is only 50% of the entire probability contributions, but represents exactly 50%
<V|U> + <V|D> = 0.5 = <H|L> + <H|R><V|U> + <V|D> + <H|L> + <H|R> = 1
This encapsulates that there are only those possibilities. There is no possibility for, say choosing Horizontal and measuring UP.
Since we have established (for example simplicity) that Alice's chioce in measuring H or V is utterly equal, and that whether U/D or L/R within each choice are also completely equal :
<V|U> = <V|D> = <H|L> = <H|R>
and
<V| = <H|
|U> = |D> = |L> |R>
Experimentally, the results would agree here, that were the scene repeated, each particular result would occur on average 25 times in every 100 repeats.
__
Now, let's imagine that Alice "prepares" the entity before measurement. For the sake of simplicity the "preparation" only applies to the VERTICAL axis and it is prepared so that the state for this vertical axis is UP
After such "preparation", Alice again chooses what axis to measure and makes the measurement.
If Alice chooses VERTICAL, the result will ALWAYS be U
If Alice chooses HORIZONTAL, the result is ALWAYS L or R
There is STILL perfectly equal probability of L/R if she chooses H and Alice's decision to choose H or V is unaffected.
<V|D> = 0
The statement made earlier
<V|U> + <V|D> + <H|L> + <H|R> = 1
still holds. Although <V|D> can safely be omitted as it is now zero probability. (Just as we are not including operators for the probabity amplitudes that Alice might spontaneously turn into a banana - it's not going to happen, so there's no need to include it)
<V|U> + <H|L> + <H|R> = 1
And
<H|L> = <H|R> still, so this holds as before. Given that Alice still chooses perfectly equally between H and V, though,
<V|U> + <V|D> = 0.5 = <H|L> + <H|R>
also still holds.
we can omit <V|D> as mentioned, and see that
<V|U> = 0.5 = <H|L> + <H|R>
So the effect of the preparation does not affect the HORIZONTAL measurement (should Alice choose to make it) in any way whatsoever. Instead, it is only the VERTICAL that is affected.
Now forget the preparation and instead bring in Bob. Also we will eradicate any choice for Alice. She will ONLY measure VERTICAL. Bob will only measure Horizontal.
There is no <A(l)| or <A(r)| nor is there a <B(u)| or |B(d)> they simply do not exist at all.
However Bob WILL make A MEASUREMENT (either B(l) or B(r) only- no other possibility) and Alice will make A MEASUREMENT(either A(u) or A(d) only- no other possibility)
<A(u)|A(d)> + <B(l)|B(r)> = 1
<A(u)|B(l)> + <A(u)|B(r)> + <A(d)|B(l)> + <A(d)|B(r)> = 1
The effect of entanglement will cause whatever Alice measures (A(u) or A(d) that Bob's paired entity would, if measured in that axis, result in the opposite to that which Alice measured. That is, if Bob were ALSO to measure in the vertical, and Alice measured U then Bob would measure D. If Alice measured D then Bob would measure U
If Alice broke with tradition and measured Horizontally, then if her result was L and Bob also measured horizontally, Bob would obtain a result of R. HOwever if Alice measured her entangled particle Vertically and Bob measured his entangled particle Horizontally, there would be no measurable detectable change whatsoever.
50% of the time Alice would detect U and 50% she would detect down. 50% of the time Bob would detect L and 50% of the time he would detect R just as if the experimentors, the particles, the detectors etc. were utterly isolated.
In the entangled scenario, if either could choose to measure either
<A(u)|B(d)> + <A(u)|B(l)> + <A(u)|B(r)> + <A(d)|B(u)> + <A(d)|B(l)> + <A(d)|B(r)> + <A(l)|B(u)> + <A(l)|B(d)> + <A(l)|B(r)> + <A(r)|B(u)> + <A(r)|B(d)> + <A(r)|B(l)> = 1
The probabilities are affected thus:
<A(u)|B(d)> = <A(d)|B(u)> = <A(l)|B(r)> = <A(r)|B(l)>
And
<A(d)|B(l)> = <A(d)|B(r)> = <A(l)|B(u)> = <A(l)|B(d)> = <A(r)|B(u)> = <A(r)|B(d)>
But because of the omission of
<A(u)|B(u)> + <A(d)|B(d)> + <A(l)|B(r)>
which would be included and contribute to the overall unity were there no entanglement, the individual probabilities as experienced by the individual experimenters are not noticeable unless the experimenters specifically compare notes.
Note that Alice is not changing the particle or encoding it in any way. It is either UP or it is DOWN (or more accurately, superposited UPDOWN, and measuring it will reveal which - measuring is in fact the activity of resolving this superposition into a real distinct state) - she cannot know beforehand which it will be, unless she PREPARES it as described above. However, preparation would "destroy the entanglement", Alice's particle would definitely be in whatever state she prepares it, but Bob would measure a now disentangled particle where the state would be again, determined through his measurement.
This is why it's not possible (regardless of how you encode the information) to transmit information using entanglement.