# Projector in QM

Response to posts #108, 109

The limiting procedure which you allude to in post #108 is not at all "well-defined". You have given no "structure" which tells how A changes with each incremental step in the supposed "limiting procedure".
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P.S. I had to change this post and cut out all of the quotes because the LaTeX was doing really strange things!

P.P.S. I also wanted to respond to your posts #114-124, but LaTex is malfunctioning. I may not have another opportunity to post more until the beginning of next week.

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Response to posts #108, 109 (again!)

This is what I originally wanted to post.

Here is what you said in post #109 regarding post #108:
My previous post regarding the object $$\psi_q$$ seems to make sense, but I just got into troube when trying to verify that with the inner products or norms.

So, I guess it's not working any way.

You can just disregard that and just move on. ...
To what you have said here, I will only add that the limiting procedure which you allude to in post #108 is not at all "well-defined". For example, when you say:
Let's take
$$\triangle_n A = a_n | \psi_n > < \psi_n |$$

$$\triangle_n a = a_n - a _{n-1}$$
... you have given no "structure" which tells how A changes with each incremental step in the supposed "limiting procedure".

Response to posts #114-124

Posts #114, 115

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Post #116

$$\forall \psi \in H \ ,$$ ...
I guess at that moment you forgot that H includes all of the vectors; i.e. even those with norm different from 1.

Later on you reach:
Now all I need to prove is
$$< \psi | ( \int_{S_{c(a)}} | a > < a | da ) | \psi > =$$
$$\int_{S_{c(a)}} P(a) da$$
The next step should have been to 'push' both the bra "<ψ|" and the ket "|ψ>" through and underneath the integral to get

S_c(A) <ψ|a><a|ψ> da .

For some reason, you were inclined to do this only with regard to the ket (with a slight 'abuse' of notation):
$$< \psi | ( \ \int_{S_{c(a)}} | a > < a | da \ ) | \psi > =$$
$$< \psi | ( \ \int_{S_{c(a)}} | a > < a | \psi > da \ ) > =$$
But you got around this by invoking an "inner product" (again, with a slight 'abuse' of notation), and then you convinced yourself (quite correctly) that <ψ|a> = <a|ψ>* ... which is what you needed to take the final step.

Let's put an end to this 'abuse' of Dirac notation. Here's what you wrote:
$$< (( \ \sum_{S_{P(a)}} P_{a_{n}} | \psi > \ ) + ( \ \int_{S_{c(a)}} | a \prime > < a \prime | \psi > da \prime \ ) )| ( \int_{S_{c(a)}} | a > < a | \psi > da ) > =$$
What you had there was the ket

a Є S_p(A) Pa|ψ> + ∫S_c(A) |a><a|ψ> da ,

which you needed to 'turn around' into a bra. That's easy to do: kets go to bras, bras go to kets, numbers go to their complex conjugates, and operators go to their adjoints. In this case, we get

a Є S_p(A) <ψ|Pa + ∫S_c(A) <ψ|a><a| da .

... And that's all there is to it. Now, you just need "slam" this expression on it's right side with the expression

S_c(A) |a'><a'|ψ> da'

and you will get the desired result. This is how to 'use' Dirac notation without 'abuse'.

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Post #117

This answer for E.3.5 is fine ... except for the "dangling" da:
$$< \psi | a > = \overline{< a | \psi >} da$$
But, later on, in post #119 you correct yourself.
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Posts #118, 119

In post #118 you say:
$$\sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da = I$$
, so
$$\int_{S_{c(a)}} | a > < a | da = I - \sum_{S_{P(a)}} P_{a_{n}}$$
.

2). Take derivative of a to it at the continuous part, then
$$\frac{d I}{ da } da = | a > < a | da$$
or in other words,
$$\int_{S_{c(a)}} \frac{d I}{ da } da = \int_{S_{c(a)}} | a > < a | da$$
The parameter "a" is the variable of integration. It is not "free" to take a derivative with respect to it.

What you do in post #119, along similar lines, however, is correct:
$$D ( a \prime ) = \int_{ - \infty}^{ a \prime } \overline{< a | \psi >} < a | \psi > da$$
, then
$$P(a) = dD /da = \overline{< a | \psi >} < a | \psi >$$
Indeed, you have a "free" parameter here.

Everything else looks fine (modulo a couple of minor typos).
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Posts #120, 121

These look fine.
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Post #122
E 3.1) a)

Even though my knowledge of "tensor product" of vector spaces might not be enough, I did give a thought on how to handle this exercise.

1). How do we know the spectrum of Q is degenerate?

Unless, we have another set of eigenbasis and self-adjoint operator, said E, and we found we have problems reconciling them with Q. For example,
$$< E | \psi > = \int < E | q > < q | \psi > dq$$
does not hold steady or meet the expectation for a state $$| \psi >$$ .
I don't understand what you meant in the above.

Next:
2). We will then believe that we need to expand the Hilbert space by adding another one in. Why don't we choose "direct sum" instead of "tensor product"?

If we we choose "direct sum", the basis will be extended as
$$v_1, v_2, ... v_n, w_1, ... w_n$$
; we are basically just adding more eigenvalues and eigenvectors in doing so.
... Right. So, that's not what we want.
We need something like
$$< x | \psi > = \int < xy | \psi > dy$$
or
$$P (xy) dxdy = | < xy | \psi > |^2 dxdy$$
.

So we need to define a "product" of vector spaces that fit our needs.
Yes, this is the idea.
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Post #123

Looks good (... I do see one small typo, though).

, our quickiest approach will be:
$$< x, y | \psi > = < y | \psi_y > < x | \psi_x >$$
This is not true in general; i.e. it is true only when the state is such that x and y are "independent".
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Our curent ket space of | x,y,z,s > is of course an example.
... where "s", I assume, refers to spin. This is precisely the example I had in mind (except that I split it up into two examples: |x,y,z> and |x,s>).
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4). If x and y are not independent, what will we get?
We will get "correlations".

Next:
For example, I can easily produce a degenerate continuous spetrum by using function f(x) = |x|.
Yes, |Q| has a continuous, doubly degenerate spectrum.
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Post #124

E 3.1) B).

5). In 3), we actually have used the two property of "tensor product":

$$( | x_1 > + | x_2 > ) \otimes | y > = | x_1 , y > + | x_2 , y >$$
$$| x > \otimes ( | y_1 > + | y_2> ) = | x , y_1 > + | x , y_2 >$$

The last property
$$\alpha | x, y > = | \alpha x> \otimes | y> = | x> \otimes | \alpha y >$$
will be used in
$$< A > = < x,y | A | x,y >$$
.
Yes ... except, the last relation should read

<A> = ∫∫ <x,y|A|x,y> dx dy .
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P.S. I may not be able to post again for another 3 weeks.