Understanding Projectors in Quantum Mechanics: A Mathematical Approach

[Sammywu]

So, I just try to see whether QM is always of trace class.

Let $$M = \sum_n a_n P_{\psi_n}$$
where
$$\sum_n a_n = 1$$

TR $$QM = \sum_n ( Q \sum_i a_i P_{\psi_i} \psi_n , \psi_n )$$
$$= \sum_n ( Q a_n \psi_n , \psi_n )$$
$$= \sum_n a_n ( Q \psi_n, \psi_n )$$

Assuming
$$b_n = | ( Q \psi_n, \psi_n ) |$$
, whether QM is of trace class will depends on whether
$$\sum_n a_n b_n$$ converges.

So, if I can find a set of $$\psi_n$$ such that $$b_n = 2/a_n$$ , then I have a QM not of trace class.

 

[Eye_in_the_Sky]

Now to keep it easy, I pick a
$$\psi = ( | q - q_0 | e^{-(q-q_0)^2 })^{1/2}$$

This is not easy.

$$( \psi , \psi ) = 1$$

True.

Considering in the interval
[ q_0-1 , q_0 +1 ] ,
$$q^2 * | \psi |^2 >= ( q_0 - 1)^2 * 2 * 1/2 * 1/e$$

How so? For q = q_o the LHS is 0.

------
... The idea behind the proof will work. But for technical reasons the proof has failed. Why not really keep it easy and choose ψ like below?

ψ(q)

= 1 , q Є (q_o, q_o + 1)
= 0 , otherwise

Then (Qψ, Qψ) > q_o² .

------------------------------------

Define: ║φ║ = √(φ, φ)

Then: ║φ + ξ║ ≤ ║φ║ + ║ξ║

.... :surprise:

------------------------------------

So, I just try to see whether QM is always of trace class.

In the most general case, QM may or may not be.

In the following, you carry out the inquiry well:

So, I just try to see whether QM is always of trace class.

Let $$M = \sum_n a_n P_{\psi_n}$$
where
$$\sum_n a_n = 1$$

TR $$QM = \sum_n ( Q \sum_i a_i P_{\psi_i} \psi_n , \psi_n )$$
$$= \sum_n ( Q a_n \psi_n , \psi_n )$$
$$= \sum_n a_n ( Q \psi_n, \psi_n )$$

Assuming
$$b_n = | ( Q \psi_n, \psi_n ) |$$
, whether QM is of trace class will depends on whether
$$\sum_n a_n b_n$$ converges.

Remember also that M is a state. So we also have

0≤a_n≤1 as well as ∑_na_n=1 .

But still, this is not enough. In general the series can still diverge.

This tells us that our current definition for a state M is still too general. While all physical states do satisfy the definition of M, not all M's are physical states.

 

[Sammywu]

Eye,

Note I just added an integral sign in front.

Considering in the interval
[ q_0-1 , q_0 +1 ] ,
$$\int_{q_0-1}^{q_0+1} q^2 * | \psi |^2 >= ( q_0 - 1)^2 * 2 * 1/2 * 1/e$$

Any way, I agree your proof is much easier and quicker than mine.

I need to read your response more thoroughly.

It seems you used the extended triangular inequality; that was what I was thinking yesterday but can't get it proved and working.

I did branch out to think the issue of mixed state:

$$< Q , P_{\sum_n a_n \psi_n} > = TR \ QP_{\sum_n a_n \psi_n} =$$
$$\sum_n ( QP_{\sum_i a_i \psi_i} \psi_n , \psi_n ) = \sum_n ( Q \sum_j a_j \overline{a_n} \psi_j , \psi_n ) =$$
$$\sum_n \overline{a_n} ( \sum_j a_j Q \psi_j , \psi_n ) = \sum_n \overline{a_n} \sum_j a_j ( Q \psi_j , \psi_n )$$

$$<Q | M > = < Q , \sum_n a_n P_{\psi_n } > = TR \ Q \sum_n a_n P_{\psi_n} =$$
$$\sum_n a_n ( Q P_{\psi_n} \psi_n , \psi_n ) = \sum_n a_n ( Q \psi_n , \psi_n )$$

This shows that even though $$P_{\sum_n a_n \psi_n} \ and \ \sum_n a_n P_{\psi_n }$$, but they do have the same expectation value.

Now, for $$P_\sum_n a_n \psi_n$$ to be a state, $$(\sum_n a_n \psi_n , \sum_n a_n \psi_n ) = \sum a_n \overline\a_n = \sum a_n^2 \ needs \ to \ be \ one$$ . This condition is different from the condition for M to be a state in that $$TR \ M = \sum_n (\sum_i a_i P_\psi_i \psi_n , \psi_n ) = \sum a_n needs \ to \ be \ one$$ .

I thought I might be able to show some Q wll have the same expectation values for the mixed state and the pure state. Apparently this is a little tedious than I thought.

 

[Sammywu]

Eye,

The previous one is a little messy. I was trying to see whether the '//' will give me a new line.

So, I redo it here.

I did branch out to think the issue of mixed state:

$$< Q , P_{\sum_n a_n \psi_n} > = TR \ QP_{\sum_n a_n \psi_n} =$$
$$\sum_n ( QP_{\sum_i a_i \psi_i} \psi_n , \psi_n ) =$$
$$\sum_n ( Q \sum_j a_j \overline{a_n} \psi_j , \psi_n ) =$$
$$\sum_n \overline{a_n} ( \sum_j a_j Q \psi_j , \psi_n ) =$$
$$\sum_n \overline{a_n} \sum_j a_j ( Q \psi_j , \psi_n )$$

$$<Q | M > = \sum_n a_n ( Q \psi_n , \psi_n )$$

as we alreday showed in earlier one.


Now, for $$P_\sum_n a_n \psi_n$$ to be a state,
$$(\sum_n a_n \psi_n , \sum_n a_n \psi_n ) =$$
[tex]\sum a_n \overline{a_n} = \sum a_n^2 [\tex] <br /> needs to be one.<br /> This condition is different from the condition for M to be a state in that <br /> $$TR \ M = \sum_n (\sum_i a_i P_\psi_i \psi_n , \psi_n ) =$$<br /> $$\sum a_n$$ <br /> needs to be one.<br /> <br /> I thought I might be able to show some Qs wll have the same expectation values for the mixed state and the pure state. Apparently this is a little tedious than I thought.<br /> <br /> Any way, why do you say I is unbounded in a infinite dimensional space?<br /> <br /> $$( I \psi , I \psi ) = ( \psi , \psi ) < 2 ( \psi , \psi )$$<br /> <br /> whether the space is infinite or finite dimensional. <br /> <br /> Thanks[/tex]

 

[Sammywu]

Eye,

Changing my tactic, I gathered some facts:

Let
$$\psi = \sum b_i \psi_i$$
where
$$b_i \overline{b_i} = a_i$$
and
$$\sum b_i \overline{b_i} = 1$$
.

Now, this can be changed to:

$$< Q , P_{\sum_n b_n \psi_n} > = TR \ QP_{\sum_n b_n \psi_n} =$$
$$\sum_n ( QP_{\sum_i b_i \psi_i} \psi_n , \psi_n ) =$$
$$\sum_n ( Q \sum_j b_j \overline{b_n} \psi_j , \psi_n ) =$$
$$\sum_n \overline{b_n} ( \sum_j b_j Q \psi_j , \psi_n ) =$$
$$\sum_n \overline{b_n} \sum_j b_j ( Q \psi_j , \psi_n )$$

Compare that to:

$$<Q | M > = \sum_n a_n ( Q \psi_n , \psi_n )$$

If I set
$$Q = \sum c_n P_{\psi_n}$$
, then
$$< Q | P_\psi > = \sum_n \overline{b_n} b_n c_n$$

and

$$< Q | M > = \sum_n a_n c_n$$

They are the same.

So for any Qs as $$\sum c_n P_{\psi_n}$$, we will have the same expectation value for the mixed state and the pure state.

 

[Sammywu]

Eye,

You know sometimes this latex thing is strange.

My question is:

Any way, why do you say I is unbounded in a infinite dimensional space?

I mean,

$$( I \psi , I \psi ) = ( \psi , \psi ) < 2 ( \psi , \psi )$$

whether the space is infinite or finite dimensional.

Thanks

 

[Sammywu]

Any way, back to the issue between the "mixed" state and the "pure" states, further questions shall be:

1). Can we relax the conditions for the "pure" states and the observables A ( I wrote Q earlier , since this is a general observable not the "position", I think A is better)?

2). Even though they have the same expection value, shall they have different distribution such like < A | M > might have a multi-nodal distribution and < A | P_\psi > has a central normal distribution?

 

[Eye_in_the_Sky]

Any way, why do you say I is unbounded in a infinite dimensional space?

I mean,

$$( I \psi , I \psi ) = ( \psi , \psi ) < 2 ( \psi , \psi )$$

I forgot to mention the following:

Metatheorem: A necessary condition for I to be unbounded (in the ∞-dimensional case) is serious confusion. :surprise:


I'll have to fix that.

 

[Sammywu]

Eye,

Now, let me take an example.

Assuming two free particles,

$$\psi_1 = \int \int \delta ( k - k_1 , w - w_1 ) e^{i(k(x-x_1)+w(t-t_1))} dk dw$$

and

$$\psi_2 = \int \int \delta ( k - k_2 , w - w_2 ) e^{i(k(x-x_2)+w(t-t_2))} dk dw$$

represent their wave functions; their states shall be
$$P_{\psi_1} \\ and \\ P_{\psi_2}$$.

In combination, this mixed state $$M = 1/2 ( P_{\psi_1} + P_{\psi_2} )$$ shall represent their combined state, or the ensemble.

Correct?

 

[Eye_in_the_Sky]

I see you are still using that notation where P is not a projector and its subscript isn't a unit vector:

$$P_{\sum_n a_n \psi_n}$$ .

According to that notation, is it not true that the preceding expression equals

$$\sum_n a_n P_{\psi_n }$$ ?

So, why don't you just use the second one?

-------------------------------

The step below has an error:

$$\sum_n ( QP_{\sum_i a_i \psi_i} \psi_n , \psi_n ) = \sum_n ( Q \sum_j a_j \overline{a_n} \psi_j , \psi_n )$$

I think this error results from what I just pointed out above, that you are treating this P as a projector when it is not! I also explained this same point in post #19:

[1] $$\sum_n a_n P_{e_n} = P_{\sum_n a_n e_n}$$

The object on the left-hand-side is not (in general) a projector. That object has eigenvalues a_n , whereas a projector has eigenvalues 0 and / or 1.

---------------

[2] P_n(v) = (v∙n) n , for any vector v

----------------

I suggest you reserve the symbol "P", in the above type of context, only for a "projector" proper. Also, I suggest you invoke the rule that the "subscript" of P is always a "unit" vector. These two prescriptions would then disqualify the "legitimacy" of the right-hand-side of [1] on both counts.

At the same time, if you want to consider generalizations for which a relation like [1] holds, then use a symbol "R" (or whatever else) instead of "P". The generalization of [2] which gives a relation like [1] is then simply:

[2'] R_u(v) = [ v ∙ (u/|u|) ] u , for any vector v .

But what is the motivation for reserving a special "symbol" for this operation? Its description is "project the vector v into the direction of u and then multiply by the magnitude of u". The meaningful aspects of this operation are much better expressed by writing the corresponding operator as |u|P_(u/|u|).

Do you follow what I am saying?

------

All of this means there is no second condition ∑_na_n²=1. (In which case, your post #35 no longer stands (I think ).)

 

[Sammywu]

Eye,

Yes. I followed what you say. The correct definition of projector here has a necessary condition in that the norm of the ket needs to be one or in other words it's a normal vector.

I correct that in the next post. $$\sum b_i \overline{b_i} = 1$$ gurantees that $$\psi$$ is a normal vector.

Thanks

 

[Sammywu]

Eye,

You are right.

I shall have used the second one to differentiate it from P.

OK.

 

[Eye_in_the_Sky]

Any way, back to the issue between the "mixed" state and the "pure" states, further questions shall be:

1). Can we relax the conditions for the "pure" states and the observables A ( I wrote Q earlier , since this is a general observable not the "position", I think A is better)?

2). Even though they have the same expection value, shall they have different distribution such like < A | M > might have a multi-nodal distribution and < A | P_\psi > has a central normal distribution?

In 1): relax how? ... or is that what 2) is explaining?

I don't understand 2).

 

[Sammywu]

Eye,

1) Relax.. means I could even find simpler conditions such as maybe all vectors in $$H_M$$ can be treated and found a role in this issue. Or maybe A as any function of $$P_{\psi_n}$$ can satisfy similar property.

My guess to the question is probably NO. The conditions I found pretty describe the situations of indistinguishable "mixed" and "pure" states.

2). What I am saying is even though their expectation values are the same. The statistics will show us two different distributions. So now it's time to investigate their probability decomposition of AM and $$AP_\psi$$.

Thoughts came out from me sometimes just are not described immediately in correct math. language but a pragmatic thought to begin with. So, I wrote distribution of < A | M >, things like that.

Multi-nodal distribution is my words for the distribution that you see multiple distingushiable points like two clear different frequency of light pulse will leave two lines in the light spectrum.

Central normal distribution is my words to emphasize that a normal probability distribution has a central node with an inverse exponetial shape of distribution.

I am sorry I do not remember what are correct math. terms for them.

Regards

 

[Sammywu]

The example I show is two particles shot rightward at different times and at time T we shine rays of lights from top downward. Now will we predict from this math. the light detector at the bottom will show two spots instead one spot.

 

[Eye_in_the_Sky]

A pure state is represented by a unit vector |φ>, or equivalently, by a density operator ρ = |φ><φ|. In that case, ρ² = ρ.

Suppose we are unsure whether or not the state is |φ₁> or |φ₂>, but know enough to say that the state is |φ_i> with probability p_i. Then the corresponding density operator is given by

ρ = p₁|φ₁><φ₁| + p₂|φ₂><φ₂| .

In that case ρ² ≠ ρ, and the state is said to be mixed. Note that the two states |φ₁> and |φ₂> need not be orthogonal (however, if they are parallel (i.e. differ only by a phase factor), then we don't have mixed case but rather a pure case).

This is the real place to begin. And Dirac's notation is the superior way of representing these objects.

--------------------

Tell me, Sammy, have you learned the postulates of Quantum Mechanics in terms of simple, basic statements like the one below?

The probability of obtaining the result a_n in a measurement of the nondegenerate observable A on the system in the state |φ> is given by

P(a_n) = |<ψ_n|φ>|² ,

where |ψ_n> is the eigenvector corresponding to the eigenvalue a_n.

 

[Sammywu]

Eye,

I have seen that postulate but not fully convinced in that and tried to understand that and see whether my interpretation is correct.

So, I try to translate that here:

$$P(a_n) = ( \int \overline{\psi_n} \varphi )^2$$

Then
$$a_n P(a_n) = a_n ( \int \overline{\psi_n} \varphi )^2 =$$
$$( \int \overline{\psi_n} (a_n)^{1/2} \varphi )^2$$

Is there something wrong with my translation?

It seems that this is more likely.
$$P(a_n) = \int \overline{\psi_n} \varphi$$

 

[Sammywu]

I thought :

$$< A | M > = < \varphi | A | \varphi > =$$
$$\int \overline{\varphi} (A) \varphi$$

In order to lead us there.

If
$$P(a_n) = \int \overline{\psi_n} \varphi$$

Then
$$\sum_n a_n P(a_n) =$$
$$\int \sum_n ( a_n \overline{\psi_n} ) \varphi$$

This might lead us there because $$a_n \psi_n = A \psi_n$$.

 

[Sammywu]

Eye,

If I use Dirac notation,

$$< \varphi | A | \varphi > =$$

$$< \varphi | A \sum_n | \psi_n > < \psi_n | \varphi> =$$

$$\sum_n < \varphi | A | \psi_n > c_n =$$
$$\sum_n \overline{c_n} c_n < \psi_n | A | \psi_n > =$$
$$\sum_n \overline{c_n} c_n a_n =$$
$$\sum_n | < \psi_n | \varphi > |^2 a_n$$


Assuming $$< \psi_n | \varphi > = c_n$$
, so
$$| \varphi > = \sum_n c_n | \psi_n >$$
.
That do agree with your formula.

Does that look good to you?

But I do have some troubles to show that in an integral of Hermitian OP.

 

[Sammywu]

Eye,

Back to what you said, the mixed state does not seem to apply to my proposed case.

I was confused about how we can use this mixed state. Now I have better idea but i still want to think further about it.

Any way, I thought the translation between the three notation is :

$$( u , v ) = < v | u > = \int \overline{v} u$$

Is this correct?

 

[Sammywu]

Eye,

Is this where I did wrong?

I think I shall start with this.

$$P(a_n) =$$
$$( \int \overline{\psi_n(x)} \varphi (x) dx ) \overline{ ( \int \overline{\psi_n(y)} \varphi(y) dy ) }$$

$$\sum_n a_n c_n \overline{c_n} =$$
$$\sum_n a_n P(a_n) =$$
$$\sum_n a_n \int \int \overline{\psi_n(x)} \varphi(x) \psi_n(y) \overline{\varphi(y)} dx dy =$$
$$\sum_n \int \int \overline{\psi_n(x)} \varphi(x) A \psi_n(y) \overline{\varphi(y)} dx dy =$$

I still need to see how this can be :

$$\int \overline{\varphi(x)} A \varphi(x) dx$$



Thanks

 

[Sammywu]

Eye,

Based on the two assumptions

$$\varphi = \sum_n c_n \psi_n$$
and
$$\int \overline{\psi_i} \psi_j = \delta_{i,j}$$
,

$$\int \overline{\varphi} A \varphi =$$
$$\int \overline{\varphi} A \sum_n c_n \psi_n =$$
$$\int \overline{\varphi} \sum_n c_n a_n \psi_n =$$
$$\sum c_n a_n \int \overline{\varphi} \psi_n =$$
$$\sum c_n a_n \overline{ \int \varphi overline{\psi_n} =$$
$$\sum c_n a_n \overline{c_n}$$

So, I verified all of these three approaches show the same expectation value.

Now, back to your question, why shall
$$P(a_n) = c_n \overline{c_n}$$
?

Any good argument about it?

 

[Sammywu]

Eye,

Now I see where you are leading.

The probability decomposition for AM will be

$$P(a) =$$
$$\sum_{a_n<=a} P(a_n) =$$
$$\sum_{a_n<=a} a_n c_n \overline{c_n}$$
.

Is it?

 

[Sammywu]

Now, if I use my original model of
$$M = 1/2 ( P_{\psi_1} + P_{\psi_2} )$$
or
$$M = 1/2 | \psi_1> < \psi_1 | + 1/2 | \psi_2> < \psi_2 |$$
even though
$$< \psi_2 | \psi_1 > not = 0$$
and let
$$P_{XM}(a)$$
be its probability decomposition,
this does seem to lead to a two spots measurement.

A trouble I might need to take care of is the two wavefunctions are not orthogonal.

While I recall that a multiple particle model in a book using a model
$$\psi = \psi_1 \psi_2$$
, I wonder how these two different models will turn out in this case.

 

[Sammywu]

Any way, back to the basic postulate, it does now seem very reasonable for probability for eigenvalue

$$P(a_n) = < \psi_n | \varphi > < \varphi | \psi_n > = ( \psi_n , \varphi) ( \varphi , \psi_n ) =$$
$$\int \overline{\psi_n} \varphi \int \psi_n \overline{\varphi}$$

 

[Eye_in_the_Sky]

If I use Dirac notation,

$$< \varphi | A | \varphi > =$$

$$\sum_n | < \psi_n | \varphi > |^2 a_n$$

Assuming $$< \psi_n | \varphi > = c_n$$
, so
$$| \varphi > = \sum_n c_n | \psi_n >$$

Does that look good to you?

It is correct.
_________

Any way, I thought the translation between the three notation is :

$$( u , v ) = < v | u > = \int \overline{v} u$$

Is this correct?

Looks fine.
_________
_________

... back to the basic postulate, it does now seem very reasonable for probability for eigenvalue

$$P(a_n) = < \psi_n | \varphi > < \varphi | \psi_n > = ( \psi_n , \varphi) ( \varphi , \psi_n ) =$$
$$\int \overline{\psi_n} \varphi \int \psi_n \overline{\varphi}$$

The basic postulates are the "true" starting point of Quantum Mechanics. From those postulates, one can then build the more complex versions which talk about "mixed states" and "expectation values" of observables ("nondegenerate" and "degenerate" cases) ... like on page 37 of that eBook.

Try to find a book in which those basic postulates are written in clear, concise terms. Those postulates should talk about a nondegenerate observable with a discrete set of eigenvalues, where the quantum system is in a pure state. There should also be a statement concerning the time evolution of a pure state in terms of the Schrödinger equation.

Once you understand and accept those basic postulates, then from them you will be able to derive - in a most transparent way - all of the more complex versions.

From that perspective, everything will be much clearer. I am quite sure of this.

 

[Sammywu]

Eye,

I think I have a book that shall have a discussion of that postulate. I shall have read it but just going over it without too much deeper thought.

I always wanted to try thinking things in my logical reasoning and then compare that to what is there any way. So I tried to analyze what it is here.

What I have here is a state $$\varphi$$ representing a probability and all I know is
$$( \varphi , \varphi ) =$$
$$< \varphi , \varphi > =$$
$$\int \overline{\varphi} \varphi = 1$$
.
Now with an observable A, I might have different observated value $$a_n$$ and I can associate with each one of them with a state $$\psi_n$$.

By the assumption of { $$\psi_n$$ } being a orthonornal basis,

$$\varphi = \sum_i c_i \psi_n$$
.

The total probability as unity shall be decomposed into components representing probability for each $$a_n$$ .
$$1 = (\varphi, \varphi ) = ( \sum_n c_n \psi_n , \varphi ) = \sum_n c_n ( \psi_n , \varphi )$$
.

Since this decomposition of unity has a coresponding number to each $$a_n$$, I can assume that the probaility for each a_n will be
$$c_n ( \psi_n , \varphi ) = ( \varphi , \psi_n ) (\psi_n , \varphi )$$
.

While analyzing this, there seems to be a condition that shall be in effect, i. e all of different possible outcomes as $$a_n$$ need to be independent to each other and that's what orthogonality of eigenfunctions guranteed.

If the a_n are not independent to each other, i.e. { $$\psi_n$$ } is not orthonormal, then there will be some interference can be derived here.

Any way, this is my own way to try to comprehend this postulate, but it does seem to make it clearer how this postulate was formulated and also indicate how the wavefunction interference might have come into play.

 

[Sammywu]

Another important fact I noticed is that $$c \psi_n$$ is also an eigenfunction and regarded the same as $$\psi_n$$ as logon as | c | = 1; this is related to what is stationary state.

So $$\varphi$$ can be decomposed into a way such that all c_i are real. Still $$\sum_i c_i$$ does not equal to one but $$\sum_i c_i^2 = 1$$ so that it will show me why [/tex] c_i^2 [/tex] not $$c_i$$ is the probability.

My intuition is c_i shall be the probability but not c_i^2, this makes it clearer to me why my intuition is wrong.

 

[Sammywu]

I think I have a book that shall have a discussion of that postulate. I shall have read it but just going over it without too much deeper thought.

I always wanted to try thinking things in my logical reasoning and then compare that to what is there any way. So I tried to analyze what it is here.

What I have here is a state $$\varphi$$ representing a probability and all I know is
$$( \varphi , \varphi ) =$$
$$< \varphi , \varphi > =$$
$$\int \overline{\varphi} \varphi = 1$$
.
Now with an observable A, I might have different observated value $$a_n$$ and I can associate with each one of them with a state $$\psi_n$$.

By the assumption of { $$\psi_n$$ } being a orthonornal basis,

$$\varphi = \sum_i c_i \psi_n$$
.

The total probability as unity shall be decomposed into components representing probability for each $$a_n$$ .
$$1 = (\varphi, \varphi ) = ( \sum_n c_n \psi_n , \varphi ) = \sum_n c_n ( \psi_n , \varphi )$$
.

Another way to look into this:

An abstract state can be represented by different orthonormal basis. By an unitary transformtion, we can translate a state to different representation in another basis, but its norm remians as one. So the square of its coefficiens can be conviniently used for probability representation. The basis is used as the "coordinate" of the measurable.

For example, Fourier transformation as an unitary transformation can transform a "position" represtation to a "momentum" representation.

An operator as an obserable can be only measured as certain real values; these are the eigenvalues and they work just like the "coordinate" of the measurement. Only certain probability distribution ( i.e. states ) can be exacted to one of these real values: they are the eigenfunctions and pure states.

For example, the eigenfunction of "position" observable of x_0 can be seen as the $$\delta ( x - x_0)$$ .

When measuring other states, only eigenvalues will appear, but since it has non-zero coefficients on different eigenvalues so it will show a dstribution among these eigen values.

An degenerate observable has more than two orthogonal functions measured the same value, so its probability measured for this value shall be the sum of the square of their ciefficients.

 

[Eye_in_the_Sky]

... there seems to be a condition that shall be in effect, i. e all of different possible outcomes as $$a_n$$ need to be independent to each other and that's what orthogonality of eigenfunctions guranteed.

I am not quite sure what you mean by "independent of each other". If the |ψ_n> are not all mutually orthogonal, then the whole postulate 'falls apart'! ... we will no longer have ∑_nP(a_n) = 1.

-----

Another important fact I noticed is that $$c \psi_n$$ is also an eigenfunction and regarded the same as $$\psi_n$$ as logon as | c | = 1; this is related to what is stationary state.

Yes, there is a connection.

------------------------------
------------------------------

Sammy, looking at all of the different points you are making, I get the sense that it might be instructive for us to go through each one of the postulates in a clear and concise way, one by one, step by step, ... . What do you say?