Understanding Projectors in Quantum Mechanics: A Mathematical Approach

[Eye_in_the_Sky]

Response to post #97

1) I did not know | q > < q | is not a projector. I have to think about that.

It would more properly be called a "projector density".

In terms of the projector P_I onto an interval I = (q, q+∆q) , ∆q > 0 , it would be defined as

lim_{∆q → 0+} P_I/∆q .

As you can see, this is a "density", and as such when it is squared, the same thing does not come back ... instead, something infinite comes back.
____________

2). I did hesiate to write $$Q | \psi > = q | \psi >$$ in the same reasons you mentioned, but in both Leon's Ebook and another place I did see their mentioning about the Q's defintion is $$Q | \psi > = q | \psi >$$. Just as I mentioned, the only reason I could see this "make sense" is by either
$$Q | \psi > = \int q |q > < q> dq$$ or
$$Q | \psi > = q \psi ( q )$$ in the form of wavefunvtions.

I think what you saw in Leon's book and any other place is

[1] Qψ(q) = qψ(q) .

This is not the same as Q|ψ> = q|ψ>.

Strictly speaking, we should not be using the same "Q" in both cases. They are different 'objects'. The first Q acts on a "function space" (so, now I will use Q to denote it). The second Q acts on a "ket space". Although the two spaces are "isomorphic", they are nevertheless formally distinct.

Now, what is the difference then between [1], which I now write as

[1] Qψ(q) = qψ(q) ,

and Q|ψ> = q|ψ> ? Well ... let's write the Q|ψ> = q|ψ> in "q-space". But(!) wait ... we've already used the "q" ... so, let's write it in "q'-space". We then have

<q'|Q|ψ> = <q'|(Q|ψ>) = <q'|(q|ψ>) = q<q'|ψ> .

The LHS we write as <q'|Q|ψ> = Q'ψ(q'), and the RHS we write as q<q'|ψ> = qψ(q'). Since LHS = RHS, we then have

[2] Q'ψ(q') = qψ(q') .

Do you see the difference between [1] and [2]? ... In [1], Q takes the function ψ(q) to a new function qψ(q), for which the two 'instances' of q are the same variable. But in [2], Q' takes ψ(q') to qψ(q'), for which q' is the variable and q is a constant ... and the fact that q is a CONSTANT here is what makes [2] an "eigenvalue" equation. The q in [1] is not a constant – equation [1] is not an "eigenvalue" equation. Equation [1] is what defines the action of Q in "q-space"!
____________

3). I think your defining that $$\psi ( q ) = < q \ \psi >$$ actually will make many calculations I did in showing in general
$$< \psi \prime | \psi > = \int \overline{\psi \prime ( q ) } \psi ( q ) dq$$
much more straighforward thamn my cumbersome calculations.

But one problem is then what is < q | q >. The discrete answer will be it needs to be one. Which of course will lead to some kind of conflicts in a general requirement of
$$\int \overline{\psi ( q ) } \psi ( q ) dq = 1$$ .

This is probably related to the eigenprojector $$P_{I_n}$$ you mentioned.

<q|q> = ∞. That is what it means to say that these kets do not "belong" to the Hilbert space – they do not have finite norm ... and, on account of that, we say that these kets are "generalized". Of course, when the two q's are distinct, say q ≠ q', then <q|q'> = 0. But we can say more than just "∞" and "0" ... we can say more precisely

<q|q'> = δ(q - q') , for all q,q' Є R .

Now, going back to what I said about a "discrete" position observable:

To make the position observable "discrete" we want to have a small "interval" I_n corresponding to each "point" q_n, say

q_n = n∙∆q , and I_n = (q_n-∆q/2 , q_n+∆q/2] .

Then, our "discrete" position observable, call it Q_∆q, will be a degenerate observable. It will have eigenvalues q_n and corresponding eigenprojectors (not kets!) P_{I_n}. That is,

Q_∆q = ∑_n q_nP_{I_n} .

Each eigenvalue q_n of Q_∆q is infinitely-degenerate. If you think about it, then you will realize that each degenerate eigensubspace E_n , corresponding to q_n , is nothing but the set of square-integrable functions on the interval (q_n- ∆q/2 , q_n+ ∆q/2). The 'magic' of it all is that ... in the limit ∆q → 0+ ... each ∞-dimensional eigensubspace E_n 'collapses'(!) into SOMETHING which can be characterized by a single object |q> whose interpretation is that of a "vector density", called a "generalized vector", and which has an infinite norm ... but nevertheless ... stands in the relation Q|q> = q|q>. ... abracadabra ... and so, you get a "generalized ket".
____________

4). Actually, I noticed my deduction has a contradition unresolved.

There is an issue to be resolved in my eigenfunction for "position" $$q_0$$ as
$$lim_n { \rightarrow \infty } \int \delta_n(q - q_0) | q> dq$$
. The problem here is whether the norm of \delta_n( q - q_0 ) shall be one or its direct integration shall be one.
If the norm shall be one, then it shall be altered to be its square root then.

Its direct integration shall be one. Look at the definition:

δ_n(q) ≡
n , q Є I_n
0 , otherwise ,

where I_n = (-[2n]^-1 , [2n]^-1) .

Its norm, however, is

sqrt{ ∫ |δ_n(q)|² dq } = √n .

In the limit, this is ∞ ... as required.

 

[Eye_in_the_Sky]

Response to post #105

I am tentatively bypassing a response to your posts #98-102 in order to address your current concerns.
______________

You know that after you pointed out the objects as "vector density" and "infinitesmal vector", I have made some corrections on my late posts.

I will keep this in mind when I get to them.
______________

Actually, that was what initially confussed me. In my mind, I have this $$\psi_q$$ as the generalized eigen vector of eigen value q for the position operator Q. I have it confussed with |q>, which is the "vector density" for it.

Now, let's make sure we have gotten this straight. The object "|q>" is the "generalized eigenket for Q" ... and(!) it is also what I have referred to as a "vector density".

All "generalized eigenkets" are "vector densities".

The object which you are now referring to as "|ψ_q>" is what I have referred to as "belonging to a class of generalized vector", and it satisfies the relation d|ψ_q> = |q>dq, giving d|ψ_q> the interpretation of an "infinitesimal vector".

But ... the object "|ψ_q>" itself is neither(!) a "generalized eigenket" nor a "vector density"!
______________

Now I organized them, so I realized $$d \psi_q$$ is the "infinitismal vector" as "d|q)" in your writing.

The "|q>" as the "vector density" is $$d \psi_q/dq \prime$$ in my thought and "d|q)/dq" in your writing.

Now in that, this falls into place:

$$\psi = \int \psi(q) |q> dq = \int < q | \psi > < q | dq = \int | q > < q | \psi > dq$$
for any $$\psi$$ .

In particular,
$$\psi_q \prime = \int | q > < q | \psi_q \prime > dq$$
for the generlized eigenvector of eigenvalue $$q \prime$$ .

Where is the ket notation "| >" on your "ψ" and "ψ_q'"? And remember what I said above:

The object "|ψ_q>" is not(!) a "generalized eigenket" (... of Q).
______________

In order for the above EQ. to be true, My first impression was
$$< q | \psi_q \prime > = \delta ( q \prime - q ) dq$$
, but actually it turns out wrong, it will give an object more like $$|q \prime>$$ in the sense of
$$\int \delta ( q \prime - q ) O(q) dq = O ( q \prime )$$

The condition which makes

|ψ_q'> = ∫ |q><q|ψ_q'> dq

true is simply

∫ |q><q| dq = 1 ,

where the "1" here is the identity operator on the Hilbert space. Thus, there are three important properties to note about the |q>-family.

[1] Q|q> = q|q> ... "eigenkets" ,

[2] <q|q'> = δ(q - q') ... "generalized" (ortho-'normal') ,

[3] ∫ |q><q| dq = 1 ... "complete" .
______________

Actually, if I use
$$d \psi_q \prime /dq \prime \prime = | q \prime >$$
, I can get
$$\psi_q \prime = \int | q \prime > dq \prime \prime$$
.

This notation does not make sense. What you really mean to say is d|ψ_q> = |q>dq, and therefore a suitable definition for |ψ_q> is

|ψ_q> = ∫_-∞^q |q'> dq' .

If we introduce the "step function"

Θ(x) ≡
1 , x > 0
0 , x < 0 ,

then we can write

|ψ_q> = ∫ Θ(q - q') |q'> dq' .

As you can see, the "representation" of |ψ_{q_o}> in "q-space" is just
Θ(q_o - q).
______________

I can put this into the previous formula
$$\psi_q \prime = \int | q > < q | \int | q \prime > dq \prime \prime > dq =$$
$$\int \int | q > < q | q \prime > dq dq \prime \prime$$
If
$$< q | q \prime > = \delta ( q - q \prime )$$
then this does make back to
$$\psi_q \prime = \int | q \prime > dq \prime \prime$$

And it must "take you BACK", because ∫ |q><q| dq = 1.
______________

This is basically my self-verification on the relationship between
$$<q|q \prime>$$
and
$$< q \prime | \psi_q >$$.

Because in my mind,
$$< \psi_q | \psi_q > =1$$
so
$$< q | q > = < d \psi_q \div dq \prime | d \psi_q \div dq \prime > =1$$
, but apparently it does not come out so.

Indeed, <q_o|q_o> = δ(0) = ∞. ... What about <ψ_{q_o}|ψ_{q_o}> ?

Well, the "q-space" representation of |ψ_{q_o}> is just Θ(q_o - q). So,

<ψ_{q_o}|ψ_{q_o}> = ∫ |Θ(q_o - q)|² dq

= ∫ Θ(q_o - q) dq

= ∫_-∞^q_o dq

= ∞ .

This "infinite norm" is the reason why I originally said that |ψ_{q_o}> is in a class of "generalized vector".

However, if, instead, we define an object

|ψ_q',q> = ∫_q'^q |q"> dq" , for q' < q ,

then this object would be an ORDINARY vector of the Hilbert space, and moreover, we would also have ∂|ψ_q',q>/∂q = |q>. But as I have already said:

From the perspective of any calculation I have ever performed in quantum mechanics, the "|q>" notation of Dirac is superior.

... And now I would like to suggest the following as well:

The only additional thing which bringing such considerations into a calculation of any kind can offer is a headache!

On the other hand, I do appreciate that 'playing around' with these objects can offer some measure of clarification of what is going on, and moreover, that this is what you are in fact accomplishing through such exercises.
______________

II) ... For example, if we use the "momentum" representation to represent this "position" eigenvector, you can always multiply a phase facor such as $$e^{iwt}$$ to it.

No. You didn't write what you meant. The correct statement is:

The most general relationship between a pair of families of eigenkets |q> of Q and |p> of P is

<q|p> = e^{i[θ(p) - Φ(q)]} 1/sqrt{2π} e^ipq/h_bar .
______________

P.S. On Sunday, regardless, of whether or not I am able to respond to any of your other posts, I will post something on the next "Postulate".

 

[Sammywu]

Eye,

I roughly got you, still reading it, glad that you clarify many points here .

I definitely agree that you can just bypass #98-102.

About my $$\psi_q$$ and |q>, I tried some more in clarifing what's going on:

I). Using an self-adjoint operator as example, starting from a discrete case to a continuous case:
$$A = \sum a_n | \psi_n > < \psi_n |$$

Let's take
$$\triangle_n A = a_n | \psi_n > < \psi_n |$$

$$\triangle_n a = a_n - a _{n-1}$$

Now A can be written:
$$A = \sum_n ( \triangle_n A / \triangle_n a ) \triangle_n a =$$
$$\sum_n ( a_n | \psi_n > < \psi_n | / \triangle_n a ) \triangle_n a$$

Converting it to continuous spectrum, that means
$$da = d_n a = \triangle_n a \rightarrow 0$$
for all n, and of course $$n \rightarrow \infty$$
so
$$A = \int d ( a | \psi_a > < \psi_a | / da ) da$$

Compare that to
$$A = \int a | a > < a | da$$

So, we know
$$d ( | \psi_a > < \psi_a | ) / da = | a> < a |$$

II) Look from a perspective of any nomal vector or ket, let
$$| \psi > = \sum_n | \psi_n > < \psi_n | \psi >$$

Let's take
$$\triangle_n \psi = | \psi_n > < \psi_n | \psi >$$
and again
$$\triangle_n a = a_n - a _{n-1}$$ .

Now,
$$| \psi > = \sum_n ( \triangle_n \psi / \triangle_n a ) \triangle_n a =$$
$$\sum_n ( | \psi_n > < \psi_n | \psi > / \triangle_n a ) \triangle_n a$$

Converting it to continuous spectrum, that means
$$da = d_n a = \triangle_n a \rightarrow 0$$
for all n, and of course $$n \rightarrow \infty$$
so
$$| \psi > = \int ( d ( | \psi_a > < \psi_a | \psi > ) / da ) da$$

Compare that to
$$| \psi > = \int | a > < a | \psi > da$$

Again we see
$$d ( | \psi_a > < \psi_a | ) / da = | a> < a |$$

III) If I want to make this formula in general:

$$| \psi > = \int | a > \psi(a) da =$$
$$\int | a > < a| \psi > da =$$

so, I expect to write
$$| \psi_q > = \int | a > \psi_q(a) da =$$
$$\int | a > \delta ( a - q) da =$$
$$\int | a > < a| q > da =$$

and also
$$| \psi_q > = \int | a > < a| \psi_q > da$$

while
$$| q > = \int | a > < a| q > da =$$
$$\int | a > < a| \psi_q > da = | \psi_q >$$

.

Take the conclusion of I) , II) and III), I found all I need is set
$$| \psi_q > = \ q>$$
and
$$d ( | \psi_q > < \psi_q | ) / dq = | q> < q |$$
.

Replacing $$| \psi_q >$$ with |q> into the second EQ. I got

$$d ( | \psi_q > < \psi_q | ) / dq = d ( | q > < \psi_q | ) / dq= | q> < q |$$

All I need to do is
$$< q | = d < \psi_q | /dq$$

This can also well explain
$$< q| q > = \infty$$
and
$$| q > < q |$$ is not a projector.


Thanks

 

[Sammywu]

Eye,

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 me, I really appreciate what you showed me. At least I have some ideas why this is not working, and what was brought into make it work.

Thanks

 

[Eye_in_the_Sky]

Response to posts #98-102

Post #98

Overall, this section is handled very well.

What you say about the operator L which you construe as a "change of basis" is correct: L shall be unitary.

Concerning the Gram-Schmidt orthonormaliztion procedure which you outline in I.2), as you point out, there is a need for a countable set which "spans" the entire Hilbert space. You are quite right in identifying "separability" as the characteristic which ensures the existence of such a set. The formal definition is as follows:

Let H be a vector space with an "inner product" ( , ) , "complete" in the "induced norm" ║ ║ ≡ √( , ) . Then, H is said to be separable iff:

H has a countable dense subset.
___

This property is then equivalent to:

H has a countable orthonormal basis.
______________

Post #99

Again, overall, this section is handled quite well.

... And, yes, the operator L which transforms from one continuous "generalized" basis |q> to another one |p> given by

[1] |p> = L|q>

shall be unitary. Thus, we also have

[2] |q> = L^†|p> .

If we write [1] and [2] in terms of "kernels", these two relations become:

[1'] |p> = ∫ L(p,q) |q> dq ,

[2'] |q> = ∫ M(q,p) |p> dp ,

where M(q,p) = L(p,q)^*. I am pointing this out, because there is an ambiguity with your notation in the "kernel" where you have "switched" the order of p and q:

$$| q > = \overline{L^T} | p > = \int \overline{L(q,p)} | p > dp$$

Regarding what you say next:

I.2) From an inner product to a basis:

The process of this part is almost exactly the same as the discrete one.
I need to figure out how separability contribute to ensure its countability.

The families |q> and |p> are not countable. By assumption these are "generalized vectors" whose parameters q and p vary continuously over R such that

<q|q'> = δ(q - q') , ∫ |q><q| dq = 1 , and similarly for |p> .
______________

Post #100

When dealing with $$( \psi , q) = < q | \psi >$$, there shall be an extra care because |q> could be representing two different things here.

Inside the integral
$$\int \psi(q) | q > dq$$
, it's a " vector density".

And we aslo use it to denote the eigenvector or eigenket of "position", in this case it's a normal vector not a "vector density".

No! The object "|q>" is the same in both cases! This appears to be an essential point of confusion in some of your other posts. To repeat what I said in post #107, there are three important properties to note about the |q>-family (and others like it):

Q|q> = q|q> ... "eigenkets" ,

<q|q'> = δ(q - q') ... "generalized" (ortho-'normal') ,

∫ |q><q| dq = 1 ... "complete" .

Also:

All "generalized eigenkets" are "vector densities".
______________

Note that in your posts #98 and #99, you have missed my point concerning L. The operator L which I originally defined in post #96 was an arbitrary linear operator:

Now, let L be a linear operator. Let L act on one of the basis vectors b_j; the result is another vector in the space which itself is a linear combination of the b_i's. That is, for each b_j, we have

[1] Lb_j = L_ijb_i .

This is not (in general) a transformation from "one basis to another". L does not even have to have an inverse. Furthermore, notice that the summation is on the first index in L_ij – this is not a typo! The summation needs to be defined that way in order for L_ij to obey the "component transformation rule":

(Lv)_i = L_ijv_j .

That this relation results from [1] above was shown explicitly in post #96.

... So, to answer the question I asked there, the 'connection' is:

For any orthonormal basis |b_i> the "components" relative to this basis are given by

<b_i|v> ... vector ,

<b_i|L|b_j> ... operator .

Note that with Dirac notation, all of this is, in a certain sense, 'trivialized' by the relation (I am now writing the summation explicitly)

∑_j |b_j><b_j| = 1 (the identity on H) ,

because we can merely "insert" this relation into the appropriate spot so that

<b_i|L|v> = <b_i|L (∑_j |b_j><b_j|) |v>

= ∑_j <b_i|L|b_j> <b_j||v> .
_____

Similarly, in the continuous case, we have, for the "components" relative to any "family" |a>,

<a|v> ... vector ,

<a|L|a'> ... operator ,

and this is "verified" by

<a|L|v> = <a|L ( ∫ |a'><a'| da' ) |v>

= ∫ <a|L|a'> <a'|v> da' .

_____

In your posts #101 and #102, you came close to this idea (... except your operators there were special).
______________

 

[Eye_in_the_Sky]

Another Postulate

Recall the previous postulates:

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

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

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".
____

And now here is the next postulate:

P3: The only possible result of a measurement of a physical quantity A is one of the eigenvalues a of the corresponding observable A. In the following, let the quantum system be in a pure state represented by |ψ>:

(i) If the eigenvalue a belongs to a discrete part of the spectrum of A with corresponding eigenprojector P_a , then the probability of obtaining the result a is given by

P(a) = <ψ|P_a|ψ> .

(ii) If the eigenvalue a belongs to a continuous part of the spectrum of A with corresponding generalized eigenket |a>, then the probability of obtaining a result in the infinitesimal interval (a, a + da) is given by

p(a) da = |<a|ψ>|² da .
___________

Notes:

N.3.1) It is possible for an observable A to have a "mixed" spectrum – i.e. parts which are discrete as well as parts which are continuous. The discrete part of the spectrum is referred to as "the point spectrum of A", denoted S_p(A). The continuous part is referred to as "the continuous spectrum of A", denoted S_c(A). The (overall) spectrum of A is given by
S(A) = S_p(A) U S_c(A).

N.3.2) In (ii) of P3, an implicit assumption is being made, namely, that the continuous part of the spectrum of A is nondegenerate. In the examples which we encounter in quantum mechanics, whenever such an assumption does not hold, we will find that the Hilbert space in question admits a decomposition into a "tensor product" of Hilbert spaces, and that the assumption will then hold with regard to one of the Hilbert spaces in that decomposition.

N.3.3) Let A be an observable whose spectrum may have a continuous (but nondegenerate) part. For a Є S_p(A), let P_a be the corresponding eigenprojector. For a Є S_c(A), let |a> be the corresponding generalized eigenket. Then,

A = ∑_{a Є S_p(A)} a P_a + ∫_{S_c(A)} a |a><a| da ,

and

∑_{a Є S_p(A)} P_a + ∫_{S_c(A)} |a><a| da = 1 ,

where 1 is the identity on H_S . This last relation is called the "closure relation" – it expresses the idea that A has a complete set of eigenkets ("generalized" or otherwise). In case the set S_p(A) or S_c(A) is empty, then the associated sum or integral is understood to be zero.
___________

Exercises:

E.3.1) Regarding N.3.1, give one or two specific examples.

In the following exercises, use Dirac notation.

E.3.2) Let A be an observable whose spectrum may have a continuous (but nondegenerate) part, and let the quantum system be in a pure state |ψ>. Use the postulate P3 and the "closure relation" of N.3.3 to verify that

∑_{a Є S_p(A)} P(a) + ∫_{S_c(A)} p(a)da = 1 .

E.3.3) Let a_n Є S_p(A) be a nondegenerate eigenvalue with corresponding eigenket |a_n>. Let the quantum system be in the pure state |ψ>.

(a) Verify that P(a_n) = |<a_n|ψ>|².

(b) Define ρ = |ψ><ψ|. Verify that P(a_n) = <a_n|ρ|a_n>.

E.3.4) Let a_n Є S_p(A) be an eigenvalue (possibly degenerate) and let |a_n^k>, k = 1, ... , g(n) , be an orthonormal basis of the eigensubspace corresponding to a_n. Let the quantum system be in the pure state |ψ>.

(a) Verify that P(a_n) = ∑_k=1^g(n) |<a_n^k|ψ>|².

(b) Define ρ = |ψ><ψ|. Verify that P(a_n) = ∑_k=1^g(n) <a_n^k|ρ|a_n^k>.

E.3.5) Verify the analogous expression for (b) of E.3.3, in the case of a (nondegenerate) eigenvalue a Є S_c(A).

E.3.6) Let A be an observable whose spectrum may have a continuous (but nondegenerate) part, and let the quantum system be in a pure state |ψ>. Define "the expectation value of A", denoted <A>, by

<A> ≡ ∑_{a Є S_p(A)} a P(a) + ∫_{S_c(A)} a p(a) da ,

where P(a) is the probability of obtaining a Є S_p(A), and p(a) is the probability density of obtaining a Є S_c(A).

(a) Explain why this definition is correct.

(b) From the definition of <A> and the postulate P3, show that <A> = <ψ|A|ψ>.

(c) Define ρ = |ψ><ψ|. In the notation of E.3.4 for the discrete part, verify that

<A> = ∑_n∑_k=1^g(n) <a_n^k|ρA|a_n^k> + ∫_{S_c(A)} <a|ρA|a>da .

 

[Sammywu]

E 3.1)

The one I can think of that mixes discrete and continuous spetrum seems to be the simple one in that the discrete eigenvalues are basically the degenerate case of $$[ q , q + \triangle q ]$$.

By basically taking a function that maps the underlining continuous interval to discrete values, such as
$$f ( x ) = 2 \ when \ x \in ( 1 , 3 )$$
$$f ( x ) = 6 \ when \ x \in ( 5 , 7 )$$
$$f ( x ) = x \ when \ x \ not \ \in ( 1 , 3 ) \cup ( 5 , 7 )$$
, this function f of the continuous observables Q will have a mixed set of continuous spectrum and discrete spectrum.

$$S_{c(A)} = R - \ ( 1, 3) - \ ( 5 , 7 )$$

$$P_2 = \int_1^3 | q > < q | dq$$

$$P_6 = \int_5^7 | q > < q | dq$$

$$A = f(Q) = \int_{S_{c(A)}} q | q > < q | dq + 2 P_2 + 6 P_6$$

Is there any other more interesting ones?

 

[Eye_in_the_Sky]

Good. Your response shows that you have understood the concept.

Here are two other examples:

(i) the Hamiltonian for a particle moving in one dimension subject to a finite square-well potential;

(ii) the Hamiltonian for a particle subject to a Coulomb potential.

In each of these cases, the eigenstates of the Hamiltonian will involve bound and unbound states of the particle. The bound states are "quantized" – i.e. discrete – whereas, the unbound states are continuous. This is true in general.
____

By the way, here is another basic exercise which I forgot to include:

E.3.1) (b) Regarding N.3.2, give one or two specific examples.

(But ... if you haven't learned "tensor products" yet, this question will have to wait.)

 

[Sammywu]

I will do E 3.3) first.

(a)
Using
$$P_{a_{n}} | \psi > = | a_n > < a_n | \psi >$$
for nondegerate eigenvalue a_n because the projector will map | \psi > to its subcomponent of | a_n > , we get
$$P ( a_n ) = < \psi | P_{a_{n}} | \psi > =$$
$$< \psi ( | a_n > < a_n | \psi > ) =$$
$$< \psi | a_n > < a_n | \psi > =$$
$$| < a_n | \psi > |^2$$
, knowing that
$$< \psi | a_n > = \overline { <a_n | \psi > }$$
.

(b)
Using
$$\rho | a_n > = | \psi > < \psi | a_n >$$
, we get
$$< a_n | \rho | a_n > =$$
$$<a_n ( | \psi > <\psi | a_n > ) =$$
$$< a_n | \psi > < \psi | a_n > =$$
$$| < a_n | \psi > |^2 = P ( a_n )$$

 

[Sammywu]

E 3.4) is very similar to E 3.3).

(a)
Using
$$P_{a_{n}} | \psi > = \sum_{k=1}^{g(n)} | a_n^k > < a_n^k | \psi >$$
, we get
$$P ( a_n ) = < \psi | P_{a_{n}} | \psi > =$$
$$< \psi | ( \sum_{k=1}^{g(n)} | a_n^k > < a_n^k | \psi > ) =$$
$$\sum_{k=1}^{g(n)} < \psi | a_n^k > < a_n^k | \psi > =$$
$$\sum_{k=1}^{g(n)} | < a_n^k | \psi > |^2$$
, knowing that
$$< \psi | a_n^k > = \overline { <a_n^k | \psi > }$$
.

(b)
Using
$$\rho | a_n^k > = | \psi > < \psi | a_n^k >$$
, we get
$$\sum_{k=1}^{g(n)} < a_n^k | \rho | a_n^k > =$$
$$\sum_{k=1}^{g(n)} <a_n^k | ( | \psi > <\psi | a_n^k > ) =$$
$$\sum_{k=1}^{g(n)} < a_n^k | \psi > < \psi | a_n^k > =$$
$$\sum_{k=1}^{g(n)} | < a_n^k | \psi > |^2 = P ( a_n )$$

 

[Sammywu]

E 3.2)

$$\forall \psi \in H \ ,$$
$$< \psi | \ ( \sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da ) \ | \psi > =$$
$$< \psi | 1 | \psi > =$$
$$< \psi | \psi > = 1$$

$$< \psi | \ ( \ \sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a |da \ ) \ | \psi > =$$
$$< \psi | \ ( \ \sum_{S_{P(a)}} P_{a_{n}} \ ) \ | \psi > + < \psi | \ ( \ \int_{S_{c(a)}} | a > < a |da \ ) \ | \psi > =$$
$$\sum_{S_{P(a)}} < \psi | P_{a_{n}} | \psi > + < \psi | \ ( \ \int_{S_{c(a)}} | a > < a | da \ ) \ | \psi > =$$
$$\sum_{S_{P(a)}} P( a_n ) + < \psi | ( \ \int_{S_{c(a)}} | a > < a | da \ ) | \psi > =$$

Now all I need to prove is
$$< \psi | ( \int_{S_{c(a)}} | a > < a | da ) | \psi > =$$
$$\int_{S_{c(a)}} P(a) da$$

$$< \psi | ( \ \int_{S_{c(a)}} | a > < a | da \ ) | \psi > =$$
$$< \psi | ( \ \int_{S_{c(a)}} | a > < a | \psi > da \ ) > =$$
( By N 3.3 Completeness , it will be translated into below :)
$$< (( \ \sum_{S_{P(a)}} P_{a_{n}} | \psi > \ ) + ( \ \int_{S_{c(a)}} | a \prime > < a \prime | \psi > da \prime \ ) )| ( \int_{S_{c(a)}} | a > < a | \psi > da ) > =$$
( By orthogonality of eigenkets, we can reduce it to )
$$< ( \ \int_{S_{c(a)}} | a \prime > < a \prime | \psi > da \prime \ ) | ( \int_{S_{c(a)}} | a > < a | \psi > da ) > =$$
$$\int_{S_{c(a)}} \overline{< a \prime | \psi >} < ( | a \prime > ) | ( \int_{S_{c(a)}} | a > < a | \psi > da ) > da \prime =$$
$$\int_{S_{c(a)}} \overline{< a \prime | \psi >} \int_{S_{c(a)}} < a \prime | a > < a | \psi > da da \prime =$$
$$\int_{S_{c(a)}} \overline{< a \prime | \psi >} ( \int_{S_{c(a)}} \delta ( a - a \prime ) < a | \psi > da ) da \prime =$$
$$\int_{S_{c(a)}} \overline{< a \prime | \psi >} < a \prime | \psi > da \prime =$$
$$\int_{S_{c(a)}} | < a \prime | \psi > |^2 da \prime =$$
$$\int_{S_{c(a)}} P(a) da$$

 

[Sammywu]

E 3.5)
By P3,
$$P(a) = | < a | \psi > |^2 da = < a | \psi > \overline{< a | \psi >} da$$
.

$$< a | \rho | a > = < a | ( \psi > < \psi | a > ) =$$
$$< a | \psi > < \psi | a >$$

We already know what is $$< a | \psi >$$ , but unclear about what $$< \psi | a >$$ could be.

By comparing the two EQs above, the only way we can equate them is
$$< \psi | a > = \overline{< a | \psi >} da$$
.

In a way this make sense, because a is a "generalized vector" and " vector density".

So , $$< \psi | a >$$ shall be an infinitesmal amount proportional to da and also it shall be propotional to $$\overline{< a | \psi >}$$ by the general rule of inner product.

I think maybe I can use similar mechanism of $$\triangle a$$ to get a better analogous proof for it. Later I will give it a try.

 

[Sammywu]

Just try to summarize what I learn here before continue on E 3.5)

1) Let me start with
$$\sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da = I$$
, 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,
$$\int_{S_{c(a)}} \frac{d I}{ da } da = \int_{S_{c(a)}} | a > < a | da$$

3). Further,
$$\frac{d I}{ da } da | \psi > = | a > < a | \psi > da$$

4) Consider
$$P ( a ) =$$
$$< \psi | \frac{d I}{ da } da | \psi > =$$
$$< \psi | \frac{d I}{ da } | \psi > da$$

 

[Sammywu]

E 3.5)

By taking the generlized EQ. you posted at #110,

$$< \psi | L | \varphi > = \int < \psi | L | a> < a | \varphi > da$$
or
$$< a \prime | L \varphi > = \int < a \prime | L | a> < a | \varphi > da$$
, this is pretty easy.

Taking
$$< \psi | I | \varphi > = \int < \psi | a> < a | \varphi > da =$$
$$\int \overline{< \psi | a>} < a | \varphi > da$$
, we know
$$< \psi | a > = \overline{< a | \psi >}$$
.

Actually, it was my fault, somehow I missed the da in the LHS, the actual EQ. in your post is :
$$P(a) da = | < a | \psi > |^2 da$$
.

Verifying this, I take
$$D ( a \prime ) = \int_{ - \infty}^{ a \prime } \overline{< a | \psi >} < a | \psi > da$$
, then
$$P(a) = dD /da = \overline{< a | \psi >} < a | \psi >$$
, in that P(a) is a "probability density" and then
$$P(a) da = | < a | \psi > |^2 da$$ can be an "infinitismal probability".

Other than E 3.5), this EQ also shows other facts you have posted:

In order for this to be true,
$$< q | \psi > = < q | I | q > = \int < q | q \prime > < q \prime | \psi > dq \prime$$
, we need
$$< q | q \prime > = \delta ( q \prime - q )$$
.

Also,
$$< q | q > = < q | I | q > = \int < q | q \prime > < q \prime | q > dq \prime =$$
$$\int \delta ( q - q \prime) ( q \prime - q ) dq \prime = \infty$$
.

 

[Sammywu]

E 3.6 )

A). If the probability of an obeservable showing a is 1, of course we will expect it always shows a.

If the observable has multiple possible outcomes, we really can not say which one it will definiitely show, we basically make a math. average of them and saying this is its average value, which is defined as " expectation value".

So, in a discrete case, it's of course:
$$\sum a P(a)$$

In a continuous case, we will take the approximate average as
$$\sum a \ \ Probability([a - \triangle a, a + \triangle a ])$$
when $$\triangle a \rightarrow 0$$ , we get
$$\int a P(a) da$$

If we have both discrete and continuous, we will take the average as
$$\sum_{S{p(a)}} a P( a | S_{p(a)} ) P( S_{p(a)} ) + \int_{S{c(a)}} a P( a | S_{c(a)} ) P( S_{c(a)} ) da =$$
$$\sum_{S{p(a)}} a P( a ) + \int_{S{c(a)}} a P( a ) da$$

b)
$$< \psi | A | \psi > =$$
$$< \psi | \ \sum_{S{p(a)}} a P_a + \int_{S{c(a)}} a |a> <a| da \ | \psi > =$$
$$< \psi | \ \sum_{S{p(a)}} a P_a \ | \psi > \ + \ < \psi | \int_{S{c(a)}} a |a> <a| da \ | \psi > =$$
$$\sum_{S{p(a)}} a < \psi | P_a | \psi > \ + \ \int_{S{c(a)}} a < \psi |a> <a| \psi > da =$$
$$\sum_{S{p(a)}} a P(a) + \ \int_{S{c(a)}} a P(a) da =$$
$$< A >$$

 

[Sammywu]

$$< A > =$$
$$\sum_{S{p(a)}} a P(a) \ + \ \int_{S{c(a)}} a P(a) da =$$

( By E 3.4 and E 3.5 , we derive below: )

$$\sum_n \sum_{k=1}^{g(n)} a < a_n^k | \rho | a_n^k > \ + \ \int_{S{c(a)}} a < a | \rho | a > da =$$

( Using the definition of eigenkets as,
$$A a_n^k = a a_n^k$$
$$A | a > = a | a >$$
we can derive below: )

$$\sum_n \sum_{k=1}^{g(n)} < a_n^k | \rho A | a_n^k > \ + \ \int_{S{c(a)}} < a | \rho A | a > da$$

 

[Sammywu]

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 $$| \psi >$$ .

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.

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.

I may continue later.

================================================

For E 3.1) a), I have "roughly" read a chapter about the unbounded and bounded solutions of square well potential problem.

I found the difference that led to the discrete bounded solution is basically that the solutions outside of the well are in the forms of $$A e^{-p_1 x/h}$$ at the LHS and $$D e^{p_1 x/h}$$ at the RHS of the well, because the condition of E < 0 took out the imaginary part.

It's very interesting.

 

[Sammywu]

E 3.1) B)

3) In looking what we already have,

$$| \psi_x > = | x> < x | \psi_x >$$
,
$$P(y) dy = < \psi | y> < y | \psi > dy$$
,
$$| \psi_y > = | y> <y | \psi_y >$$
,
$$P(x) dy = < \psi | x> < x | \psi > dx$$
and our believe that
$$| \psi > = | x, y > < x,y | \psi >$$
and
$$P(x,y) dx dy = < \psi | x,y> <x,y | \psi > dx dy$$
, our quickiest approach will be:
$$< x, y | \psi > = < y | \psi_y > < x | \psi_x >$$
and
$$< \psi | x,y > = \overline{< y | \psi_y > < x | \psi_x >} =$$
$$< \psi_y | y > < \psi_x | x >$$
.

This turned out it will satisfy both needs in representation of ket and probability.

Not only that, we also see
$$P(x,y) dx dy = P(x) p(y) dx dy$$
.

This satisfies the general probability rule:
$$P ( x, y ) = P(x)P(y|x)$$
; the issue here seems to be that
P(y) = P(y|x), which means x and y need to be independent to each other.

So, if we do find the eigenvalues of x and y independent to each other, we will try to expand our Hilbert space in the above way.

Our current ket space of | x,y,z,s > is of course an example.

4). If x and y are not independent, what will we get?

For example, I can easily produce a degenerate continuous spetrum by using function f(x) = |x|.

 

[Sammywu]

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
$$< A > = < x,y | A | x,y >$$
.

 

[Sammywu]

Checked with a book, I found my answer to E3.1) B) will basically lead to "looking for the set of maximum commutable observables".

That is not what Eye asked any way. Eye's question already assumed there is a Hilbert space and we found multiple "generalized kets" for one eigenvalues of a continuous spectrum.

Any way, to look for that answer, I found I need to clarify there are only true discrete eigenkets in our assumption.

So I went sideway to find a proof why "separability" ensures "countable discrete eigenkets".

The proof is actually quite straightforward, after a few day's rumination:
1). A Hilbert space is a vector space with an inner product.
2). An inner product can define the "norm". The "norm" can define the distance between two vectors.
3). With the "distance", we can define the open sets and so the topology.
4). Definition of "separability" says if we have an open covering for the Hilbet space, then we have a "counatble" open covering as its subset.
5). For any two vectors belong to an orthonormal basis, their distance will be $$2^{1/2}$$ .
6). If we define an open coverings as all open balls with a radius 0.25, we will have a countable subset that covers the Hilbert space.
7). Because any two orthonormal vectors has a distance greater than 0.25, so no two orthonormal vectors could be in one open ball; this implys the number of elements in the countable open coverings is greater than the number of all orthonormal vectors, so the basis is definitely countable.

Does this look good?
==================================================

Any way, if we assume the Hilbert space only has countable orthonormal basis, then we know there is no true eigenvector for a continuous spectrum because any continuous real interval has uncountable numbers in it.

Of course, I think there must be a proof that shows the space of square-integrable functions has a countable orthonormal basis and is separable. So, if the "space" has an isomorphich structure with the "functoion" space, it has only countable orthonormal basises and is separable.

 

[Eye_in_the_Sky]

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".
____

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.

 

[Eye_in_the_Sky]

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".

 

[Eye_in_the_Sky]

Response to posts #114-124

Posts #114, 115

Your answers to E.3.3 and E.3.4 look fine.
______________

Post #116

In your answer to E.3.2, you began with:

$$\forall \psi \in H \ ,$$ ...

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)} P_a|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|P_a + ∫_{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'.

______________

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.
______________

Posts #118, 119

In post #118 you say:

1) Let me start with
$$\sum_{S_{P(a)}} P_{a_{n}} + \int_{S_{c(a)}} | a > < a | da = I$$
, 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).
______________

Posts #120, 121

These look fine.
______________

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.
______________

Post #123

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

However, about:

, 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".
___

Our current 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>).
___

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.
______________

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 .
____________________

P.S. I may not be able to post again for another 3 weeks.