Understanding Projectors in Quantum Mechanics: A Mathematical Approach

[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 defintion 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?

 

[Sammywu]

Eye,

What I was saying about independency is that it came to my mind that interference of two wave functions will be zero when they are orthogonal. Their interference seems to be represented by their inner product.

Of course, I know the necessary condition in here is that these are all orthonormal basises. I am saying is we can ignore their interference because they are orthogonal, but it would be interesting to check the relationship between interference and the inner product of two arbitrary wave functions.

I think your suggestion of going over all postulations is good.

Thanks

 

[Eye_in_the_Sky]

What I was saying about independency is that it came to my mind that interference of two wave functions will be zero when they are orthogonal. Their interference seems to be represented by their inner product.

Yes, I see.

If

|φ> = c₁|φ₁> + c₂|φ₂> ,

then

<φ|φ> = |c₁|²<φ₁|φ₁> + |c₂|²<φ₂|φ₂> + 2 Re{c₁^*c₂<φ₁|φ₂>} .

The last term is the "interference" term, and it vanishes if <φ₁|φ₂> = 0 .

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

Is the above what you meant?

... it would be interesting to check the relationship between interference and the inner product of two arbitrary wave functions.

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

I think your suggestion of going over all postulations is good.

I will post something soon.

 

[Eye_in_the_Sky]

Two Beginning Postulates of QM

ATTENTION: Anyone following this thread ... if you find any points (or 'near' points) of error on my part, please do point them out.
___________

P0: To a quantum system S there corresponds an associated Hilbert space H_S.

P1: A pure state of S is represented by a ray (i.e. a one-dimensional subspace) of H_S.
___________

Notes:

N.0.1) Regarding postulate P0, in certain contexts it is possible to associate a Hilbert space with a particular dynamical 'aspect' of the quantum system (e.g. a Hilbert space corresponding to "spin", decoupled from, say, "position").

N.1.1) In postulate P1, a "ray" is understood to represent the "(pure) state" of a single quantum system. Some physicists prefer to let those terms designate an ensemble of "identically prepared" quantum systems. Such a distinction becomes relevant only in cases where one considers possible interpretations of the theory.

N.1.2) A ray is determined by anyone of its (non-zero) vectors. Our convention is to use a "normalized" (i.e. to unity) ket |ψ> to designate the corresponding ray, and hence, the corresponding pure state. This means that two normalized kets |ψ> and |ψ'> which differ by only a phase factor (i.e. |ψ'> = α|ψ>, where |α| = 1) will represent the same "ray", and hence, the same "state".
___________

Exercise:

E.1.1) What, if anything, is wrong with the following?

Suppose that the kets |φ₁> and |φ₁'> represent the same state. Then, the kets |ψ> and |ψ'> given below will also represent the same state:

|ψ> = c₁|φ₁> + c₂|φ₂> ,

|ψ'> = c₁|φ₁'> + c₂|φ₂> .
___________

 

[Sammywu]

Eye,

About the interference of wave functions: Yes. You got what I meant. I will have to ruminate over your simple answer, though.

Answer to the exercise:

If
$$| \psi \prime > = a | \psi >$$
$$| \varphi_1 \prime > = a_1 | \varphi_1 >$$
$$| \varphi_2 \prime > = a_1 | \varphi_2 >$$
then
$$a c_1 | \varphi_1 > + a c_2 | \varphi_2 > =$$
$$a_1 c_1 | \varphi_1 > + a_2 c_2 | \varphi_2 >$$
.

One solution to it is:
$$a = a_1 = a_2$$
i. e. they are mutiplied by the same phase factor.

If
$$a not = a_1$$
, then
$$| \varphi_1 > = ( a_2 -a ) c_2 / ( a - a_1 ) c_1 | \varphi_2 >$$
; i. e. $$| \varphi_1 >$$ and $$| \varphi_2 >$$ have to be the same state.

This shows that statement is in general incorrect except in the two solutions I show above.

 

[Sammywu]

Actually, for the second solution of them being the same state, there are other troubles to take care, | (a_1 - a ) / (a - a_2 ) \ needs to be one.

It does not seem to be easy to get solution for this. I have to think about how to resolve it. The possibility is this solution will not even work.

 

[Sammywu]

| ( a_1 - a ) c_1 / ( a_2 -a ) c_2 | =1 is equivalent to | a_1 - a | / | a_2 -a | = | c_2 | / | c_1 | ; knowing | a | = | a_1 | = | a_ 2 | = 1, we can view a, a_1 and a_2 as three normal vectors in the unit circle on the Complex plane; a shall be chosen as a point on the unit circle such that ratio of the cord length between a_1 and a over the cord length a_2 and a is | c_2 | / | c_1| .

 

[Eye_in_the_Sky]

First off, you have attempted to answer a question slightly more general than the one I posed. In my question, there was no |φ₂'>. But that doesn't really matter. ... We'll use your version of the question.

You start off well.

If
$$| \psi \prime > = a | \psi >$$
$$| \varphi_1 \prime > = a_1 | \varphi_1 >$$
$$| \varphi_2 \prime > = a_1 | \varphi_2 >$$
then
$$a c_1 | \varphi_1 > + a c_2 | \varphi_2 > =$$
$$a_1 c_1 | \varphi_1 > + a_2 c_2 | \varphi_2 >$$

But your final conclusion suggests to me that the main point has been missed.

This shows that statement is in general incorrect except in the two solutions ...

Let's look at your second "solution". You write:

If
$$a not = a_1$$
, then
$$| \varphi_1 > = ( a_2 -a ) c_2 / ( a - a_1 ) c_1 | \varphi_2 >$$
; i. e. $$| \varphi_1 >$$ and $$| \varphi_2 >$$ have to be the same state.

In this case, you are right. If |φ₁> and |φ₂> themselves represent the same state, then so too will |ψ> and |ψ'>.

Now, let's look at your first "solution".

One solution to it is:
$$a = a_1 = a_2$$
i. e. they are mutiplied by the same phase factor.

Can we say that the statement is correct in this case? For the statement to be correct, we must have a₁ and a₂ as arbitrary free parameters, except for the constraint |a₁| = |a₂| = 1; but this solution produces an additional constraint over and above that.

... So, your conclusion should have been:

The statement is incorrect in all cases, except when |φ₁> and |φ₂> represent the same state.

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

NOW ...
Just to make sure that the main point hasn't been lost in all of this abstraction, let's look at a concrete example.

Given the state

c₁|φ₁> + c₂|φ₂> ,

then

i(c₁|φ₁> + c₂|φ₂>)

represents the same state,

but

ic₁|φ₁> + c₂|φ₂>

does not (unless, of course, |φ₁> and |φ₂> represent the same state).

In the first case, we have inserted a "global phase-factor", and that is OK. In the second case, however, we have inserted a "relative phase-factor", and that is not OK.

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

Finally, for the sake of completeness, I offer a solution of my own along the same lines as the one you gave above.
____

First, we assume:

(i) c₁, c₂ ≠ 0 ,

and

(ii) |φ₁> and |φ₂> are linearly independent .

For, otherwise, it follows trivially that |ψ> and |ψ'> will define the same ray (i.e. represent the same state).


Next, by considerations similar to yours above, we note that

|ψ> and |ψ'> define the same ray if, and only if

[1] c₁(a - a₁)|φ₁> + c₂(a - a₂)|φ₂> = 0 ,

where (and this is the important part!) the parameters a₁ and a₂ are completely arbitrary except for the constraint |a₁| = |a₂| = 1.

We now reach our conclusion. We say: but from assumptions (i) and (ii) we see that relation [1] holds iff a = a₁ = a₂, implying that a₁ and a₂ cannot be completely arbitrary (except for the constraint) as required; therefore, |ψ> and |ψ'> do not define the same ray.


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

 

[Sammywu]

Eye,

I am sorry. I did not notice there is no prime on the second \varphi_2.

I agree to your answer.

One part of my calculation was wrong. Somehow | c_1 | / | c_2| = 1 slipped into my mind, which is not true, so I corrected my answer in case somebody else were reading this post. Any way, that was showing even if they are all "same" state, "a" need to be chosen in a correct math. way.

 

[Eye_in_the_Sky]

Another Postulate

P2: To a physical quantity A measurable on (the quantum system) S, there corresponds a self-adjoint linear operator A acting in H_S. Such an operator is said to be an "observable".
___________

Notes:

N.2.1) From "The Spectral Theorem" for self-adjoint operators, it follows that:

(a) A has real eigenvalues;

(b) the eigenvectors corresponding to distinct eigenvalues are orthogonal;

(c) the eigenvectors of A are complete (i.e. they span H_S).

The set of eigenvalues of A is called the "spectrum" of A. If A has a continuous spectrum, then the eigenvectors of A are said to be "generalized" and A is said to satisfy a "generalized" eigenvalue equation.

N.2.2) We now enumerate three special cases for A:

(1) A has a discrete, nondegenerate spectrum; then

A = ∑_n a_n|u><u_n| .

(2) A has a discrete (possibly degenerate) spectrum; then

A = ∑_n a_nP_n .

(3) A has a continuous, nondegenerate spectrum; then

A = ∫a |a><a| da .

In each case, the RHS of each of the above relations is referred to as the "spectral decomposition" for A.

Case (1) is a particularization of case (2) (with P_n = |u_n><u_n|). For the general case of (2), the a_n are the eigenvalues of A and the P_n are the corresponding eigenprojectors.

Examples of case (3) are any of the components (Q_j or P_k) of the position or momentum observables.
___________

Exercises:

E.2.1) From N.2.1) (b) (i.e. the eigenvectors corresponding to distinct eigenvalues are orthogonal), show that for a spectral decomposition of the type in N.2.2) (2), i.e.

A = ∑_n a_nP_n ,

it follows that

P_jP_k = δ_jkP_k .

From N.2.1) (c) (i.e. the eigenvectors of A are complete), show that

∑_nP_n = I .

E.2.2) Use, as an example, the observable Q (for the position of a spinless particle moving in 1-dimension) to explain why the eigenkets |q> are said to be "generalized" and, therefore, why Q is said to satisfy a "generalized" eigenvalue equation.
___________

 

[Sammywu]

Eye,

I).
I think you have a typo here. Did you miss a subscriptor n here in this paragraph.

N.2.2) We now enumerate three special cases for A:

(1) A has a discrete, nondegenerate spectrum; then

A = ∑_n a_n|u_n><u_n| .

II)
Also, For acontinuous spectrum, why are the eigenvectors called " generalized"? Does this have anythings to do with the fact that most likely they will be "generalized" function such as dirac-delta function isntead of regular functions?

III). Answer to the Exercise. It's actually a little difficult because you did not define eigenprojectors.

So, I just gave a guess on this:

$$P_n ( \sum_i c_i u_n_i + b u_\bot ) = \sum_i c_i u_n_i$$
where
$$u_n_i$$ are eigenvectors of $$a_n$$ and $$u_\bot \bot$$ all $$u_n$$ .

Under that assumption,
$$P_n^2 ( \sum_i c_i u_n_i + b u_\bot ) = P_n ( \sum_i c_i u_n_i ) =$$
$$\sum_i c_i u_n_i$$

so, $$P_n^2 = P_n$$ .

If i does not equal to j, then
$$P_i P_J ( \sum_k c_k u_i_k + \sum_l c_l u_j_l + b u_\bot ) =$$
$$P_i ( \sum_l c_l u_j_l ) = 0$$
; so $$P_i P_j = 0$$ .

That shall take care of the first part of E.2.1).

The question could be asked is how I can prove all u can be decomposed to
$$\sum_k c_k u_i_k + \sum_l c_l u_j_l + b u_\bot$$
.

I think setting
$$c_k = < u_i_k | u >$$
$$c_l = < u_j_l | u >$$
and
$$b u_\bot = u - \sum_k c_k u_i_k - \sum_l c_l u_j_l$$
could take care of that.

I might verify this part later.

I am still working on the other two exercises.