I Origin of Probabilities in Quantum Mechanics?

jcap
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The non-normalized wavefunction of a general qubit is given by:
$$|\psi\rangle=A|0\rangle+B|1\rangle.$$
The complex amplitudes ##A## and ##B## can be represented by two arrows in the complex plane:

complex.jpg


Now the wavefunction can be multiplied by any complex number ##R## without changing the physics. This will cause the arrows ##A## and ##B## to rotate and shrink/expand together with a fixed angle between them.

Therefore two sets of points will be traced out represented by a circle with area ##|A|^2## and a circle with area ##|B|^2##. These represent the sets of possible values for the amplitudes ##A## and ##B##.

Thus if we become entangled with the qubit then the probabilities of finding ourselves in set ##A## (measuring ##0##) or set ##B## (measuring ##1##) are given by:
$$P(0)=\frac{|A|^2}{|A|^2+|B|^2}$$
$$P(1)=\frac{|B|^2}{|A|^2+|B|^2}.$$
Does this picture help to understand the origin of probabilities in quantum mechanics?
 
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Hello. I add that after normalization circle A and circle B are transformed within unit circle. Relative angle between A and B matters not their absolute angle to axis because wavefunctions ##\psi## and ##e^{i\alpha}\psi## express the same state.

So we may improve by making :
A=\cos (\theta/2), B=\sin(\theta/2)
where ##0 \leq \theta \leq \pi ##. You see positivity and normalization
0 \leq A,B \leq 1,\ \ A^2+B^2=1
is satisfied in that form. Also take angle of A is zero and angle of B to A is ##\phi##.
P(0)=\cos^2 (\theta/2),\ \ P(1)=\sin^2 (\theta/2)Four parameters A, angle of A, B and angle of B will be reduced to two, ##\theta## and ##\phi##. These two angles ##\theta## and ##\phi## which are interpreted as polar coordinates for unit sphere ##r=1## called Bloch sphere in this context express qubit state.

As for projection of state vector to z axis
Length from |1> pole to the projection
cos\theta -(-1)= 2P(0)
is |0> probability multiplied by 2.
Length from |0> pole to the projection
1-cos\theta = 2P(1)
is |1> probability multiplied by 2.
 
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jcap said:
Does this picture help to understand the origin of probabilities in quantum mechanics?
I don't think so. You are describing a quantum system, but that do not explain why it has to be that way. I may be misunderstanding your question, though.
 
andresB said:
I don't think so. You are describing a quantum system, but that do not explain why it has to be that way. I may be misunderstanding your question, though.

I was wondering if the redundancy in the magnitude and angle of the overall phase factor implies circular areas of amplitudes associated with the vectors ##|0\rangle## and ##|1\rangle##. These areas could provide the "weights" for obtaining the eigenvalues ##0## and ##1##.

As I understand it I can normalize ##|\psi\rangle## by either first choosing an ##A## or first choosing a ##B##. There are a circular area's worth of ##A##s that I can pick or a circular area's worth of ##B##s. These areas give the weights for obtaining eigenvalues ##0## or ##1## respectively. Maybe this argument only works with a discrete grid of amplitude values.
 
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jcap said:
Does this picture help to understand the origin of probabilities in quantum mechanics?

I don't see how since your picture only gives a way to visualize the numerical values of the probabilities, not why there are probabilities at all in the first place.
 
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Correction

Let
$$A=R_Ae^{i\theta_A}$$
$$B=R_Be^{i\theta_B}$$
A general normalized wavefunction is given by:
$$|\psi\rangle=\frac{1}{(R_A^2+R_B^2)^{1/2}}\large[R_Ae^{i\theta_A}+R_Be^{i\theta_B}\large]$$
Assume that I multiply the amplitudes ##A## and ##B## by
$$C=Re^{i\theta}$$
Then the normalized wavefunction becomes
$$|\psi\rangle=\frac{1}{R(R_A^2+R_B^2)^{1/2}}\large[RR_Ae^{i(\theta_A+\theta)}+RR_Be^{i(\theta_B+\theta)}\large]$$
$$|\psi\rangle=\frac{e^{i\theta}}{(R_A^2+R_B^2)^{1/2}}\large[R_Ae^{i\theta_A}+R_Be^{i\theta_B}\large]$$
It seems that the only degree of freedom is a phase angle ##\theta## rather than an area as I asserted above.
 
jcap said:
It seems that the only degree of freedom is a phase angle θ rather than an area as I asserted above.

@jcap this ##\theta## has no physical meaning as I said in post #2 so it is convenient to take it as ##-\theta_A## so that phase factor appears only on B.

My post #2 does so, i.e.
\cos\frac{\theta}{2}|0>+e^{i\phi}sin\frac{\theta}{2}|1>
 
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