# Projector in QM

1. Aug 5, 2004

### Sammywu

I saw some discussions about projectors in some threads. Also, the projector is used in this book to define pure state but did not provide what is a projector.

http://www.math.sunysb.edu/~leontak/book.pdf [Broken]

Thru some math. check by assuming $$TR AP_\psi = (A (\psi) , \psi )$$ ( In this book, that is the expectation vale for $$AP_\psi$$ ) , I got the answer by
$$P_\psi ( e_n ) = \sum_i c_i \overline{c_n } e_i$$
if
$$\psi= \sum_i c_i e_i$$
for an orthonormal basis $${ e_n }$$.

That sounds to be a good one for it.

Also, if
$$\psi_1 = c_1 \psi + c_2 \bot \psi$$
then
$$P_\psi ( \Psi_1) = c_1 \psi$$ , where
$$( \bot\psi, . \psi ) = 0$$ .

Is this right?

Last edited by a moderator: May 1, 2017
2. Aug 5, 2004

### Eye_in_the_Sky

Definition: P is a "projector" if (and only if):

(i) Pt = P ,

and

(ii) P has eigenvalues 0 and/or 1 .

It then follows that P is a "projector" if, and only if, Pt = P and P2 = P.

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I'm not quite sure what you are asking here. But all of the equations are correct. However, the "Trace" equation is said to be "the expectation value for A when the system is in the state ψ", not "the expectation value for APψ".

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Yes.

Last edited: Aug 5, 2004
3. Aug 5, 2004

### Sammywu

Eye,

Thanks.

The error you pointed out was my typo. I will think about the link between the EQ. I derived and your defintion.

4. Aug 5, 2004

### Sammywu

Eye,

Shall it be $$\overline{P^t} = P$$?

Thanks

5. Aug 5, 2004

### Sammywu

By the way, this condition is also needed in my EQ:

$$\sum_i c_i \overline{c_i } = 1$$

Otherwise, $$P_\psi ( \psi ) = \sum_i c_i \overline{c_i } \psi$$

Or, in general, the EQ shall be:

$$P_\psi ( e_n ) = ( 1 / \sum_j c_j \overline{c_j } ) \sum_i c_i \overline{c_n } e_i$$

6. Aug 5, 2004

### Sammywu

Notation conversion between inner product and bra/ket for C and D belonging to the Hilbert Space.

$$( C , D ) = < D | C > = \sum_j c_j \overline{d_j }$$

if $$C = \sum_j c_j e_j and D = \sum_j d_j e_j$$

7. Aug 5, 2004

### Eye_in_the_Sky

If by "t" you don't mean the "Hermitian transpose" (called the "adjoint"), then it shall.

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Oh, I was assuming that. It means that the state ψ is "normalized" (i.e. to unity).

Yes. BUT, then, for nontrivial Pψ (i.e. Pψ ≠ 0), it is not a projector (because the eigenvalue 1 has become Σi|ci|2).

The factor in front should still be a 1 (not 1/Σj|cj|2). (BUT remember: this more general case is no longer a projector!)

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Looks fine.

Last edited: Aug 5, 2004
8. Aug 5, 2004

### turin

Does a projector have an inverse. I never thought of it before, but now I'm wondering ... (it doesn't seem to have an inverse)

9. Aug 5, 2004

### Eye_in_the_Sky

idempotent

P2 = P

implies

P-1P2 = P-1P

P = 1 ;

the only one with an inverse is the identity

(alternatively, you can say that on account of a 0 eigenvalue det = 0, and therefore, there is no inverse, except when P = 1)

(alternatively, you can say that when P ≠ 1 (and visualizing it geometrically) the mapping is MANY-to-ONE, and therefore has no inverse)

Last edited: Aug 6, 2004
10. Aug 6, 2004

### Sammywu

I think I derived a few things that seem important to me:

1). For $$\psi_1 and \psi_2 \in H$$,

In order for
$$( P_\psi_1 + P_\psi_2 ) = P_{\psi_1+\psi_2}$$
then
$$\psi_1 \bot \psi_2 i.e. ( \psi_1 , \psi_2 ) = 0$$

2). From that,

$$\sum_n P_{e_n} = P_{\sum_n e_n}$$

11. Aug 6, 2004

### Sammywu

I verified that:

For $$\psi and \psi_n \in H$$,

where $$\psi_n$$ are eigenbasis of A,

this holds true:
$$< A | P_\psi > = \sum_n c_n \overline{c_n} E_n$$

where
$$\psi = \sum_n c_n \psi_n$$.

12. Aug 6, 2004

### Sammywu

A question is now whether $$P_{e_n}$$ can serve as a basis ( or generator ) of GL(H) or A(H)?

13. Aug 6, 2004

### Sammywu

If
$$\sum_n a_n P_{e_n} = P_{\sum_n a_n e_n}$$
is true, then $$P_{e_n}$$ can not serve as a basis, because all its linear combinations are also projectors then.

But $$2 P_{e_n}$$ does not seem to be the projector for
$$2 e_n$$; so in general, there seems to be possibility.

14. Aug 6, 2004

### Eye_in_the_Sky

Do you mean "(a) implies (b)", "(b) implies (a)", or both ?

Yup.

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You didn't mention that the En are the eigenvalues of A. (... I assume you assumed <ψ|ψ> = 1.)

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... All of this will become much, much simpler once you start using Dirac notation.

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15. Aug 6, 2004

### Sammywu

Eye,

Thanks.

I have a question for the lemma 1.1 at page 40 in that book.

I don't think there is anythings that says the eigenbasis of a self-adjoint operator can always span the entire Hilbert space.

If that's the case, I can write a

$$\psi = \sum_i a_i \Psi_i + b \psi_\bot$$

where

$$\Psi_\bot \bot \Psi_n$$, for all n.

$$( M (\psi) , \psi )$$
might not equal to
$$( \sum_i a_i P_\psi_n ( \psi ) , \psi )$$.

16. Aug 7, 2004

### Sammywu

Eye,

What I found is below is the sufficient condition for this Lemma to be true:

$$( M (\psi) , \psi_\bot ) = 0$$

Any Way, it there is a countable basis, this seems to be true.

I aslo found I can easily set up a self-adjoint operator that only has limited number of eigen values and map the rest of basis to zero.

17. Aug 8, 2004

### vanesch

Staff Emeritus
No it doesn't, except for the trivial projector on the whole space. The "feel it in the bones" argument is that when you project something, you don't know what the orthogonal component was (the one that has been filtered out).
The mathematical argument is of course that a non-trivial projector has 0 eigenvalues (of the orthogonal directions that have been filtered out!).

In fact, this property is at the heart of the "measurement problem" in QM: a projection (such as happens in von Neuman's view) can never be the result of a unitary evolution.

cheers,
Patrick.

18. Aug 8, 2004

### Eye_in_the_Sky

GL(n,C) is a group with respect to matrix multiplication. It is not "closed" under matrix addition, and therefore, does not have the vector space structure you are assuming in your question. On the other hand, the full set of n x n matrices with entries from C, M(n,C), is "closed" under matrix addition (as well as, multiplication, of course). So, you might want to pose your question with respect to M(n,C).

In that case, for n > 1, the answer is "no". M(n,C) has dimension n2, whereas, the Pi will span a subspace with dimension no larger than n (of course, the Pi are in fact linearly independent, so they will span a subspace of dimension equal to n).

Last edited: Aug 8, 2004
19. Aug 8, 2004

### Eye_in_the_Sky

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

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At this juncture, it is instructive to consider ordinary 3-D Euclidean space. Pick any unit vector n. Then, the projector corresponding to this unit vector is given by

[2] Pn(v) = (vn) n , for any vector v .

The description of [2] is "the projection of v along n". Do you remember what this means geometrically? (see figure).

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NOTE:

In Dirac notation, [2] becomes

Pn|v> = <n|v> |n> = |n> <n|v> = (|n><n|) |v> , for any |v> .

We therefore write:

Pn = |n><n| .

If you think of each ket as a column matrix and the corresponding bra as its Hermitian transpose (a row matrix) then this notation can be taken "literally".

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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'] Ru(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|).

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Now, let's go back the definition I gave in post #2.

I am now strongly suggesting that we, instead, use the following as our "official" definition:

*************************
* 1a) Given any unit vector e, definite the "projector onto e" by:
*
* Pe(v) = (v,e) e , for any vector v .
*
* Such a projector is said to be "1-dimensional".
*
* 1b) An operator P is said to be a "projector" if (and only if)
* it can be written as a sum of 1-dimensional projectors
* which project onto mutually orthogonal unit vectors.
*
*************************

This definition [1a) and 1b) taken together] is equivalent to the original one I gave. But I think it makes the meaning of "projector" much clearer.

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All that I have said above should clarify matters like:

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20. Aug 9, 2004

### Eye_in_the_Sky

[Note: you wrote "eigenbasis", when you meant "eigenvectors".]

The answer to your query is given in that book by Theorem 1.1 (called "The Spectral Theorem"), on p. 38. In simple language, it is saying that the answer is: "Yes, a self-adjoint operator will always have eigenvectors (or "generalized" eigenvectors) spanning the entire Hilbert space."

HOWEVER, you must NOTE that the definition of "self-adjoint" (in the case of an infinite-dimensional Hilbert space) is nontrivial (... in that book, the appropriate definition is given at the top of p. 36, in 1.1.1 Notations).

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For the sake of giving you (at least) something, here is some basic information. Let A be a linear operator acting in the Hilbert space.

Definition: A is "symmetric" iff <g|Af> = <Ag|f> for all f,g Є Domain(A).

Definition: A is "self-adjoint" iff: (i) A is symmetric; (ii) the "adjoint" At exists; and (iii) Domain(A) = Domain(At).

Lemma: At, the "adjoint" of A, exists iff Domain(A) is dense in the Hilbert space.

All that is missing in the above is a definition of "adjoint" (which I have omitted for the sake of brevity and simplicity). That definition would then give us a specification of Domain(At) and thereby complete the definition of "self-adjoint".

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Now, you might ask: How can it be that there is a linear operator A with a domain "smaller" than the whole Hilbert space, yet, at the same time, A has eigenvectors which span the entire space?

Well, first of all, this can only happen in an infinite-dimensional Hilbert space. Suppose A has eigenfunctions φn(x) with corresponding eigenvalues an. So,

[1] Aφn(x) = anφn(x) .

Since the φn(x) span the entire space, an arbitrary element ψ(x) of the space can be written as

[2] ψ(x) = Σn cnφn(x) .

The right-hand-side of [2] is an infinite sum, and, therefore, involves a limit. While every finite subsum is necessarily in the domain of A, it is possible that in the limit of the infinite sum, the resulting vector is no longer in that domain. ... As you can see, this sort of phenomenon can only occur when the Hilbert space is infinite-dimensional.

But what do we get if we, nevertheless, attempt to "apply" A to ψ(x) by linearity and use [1]? Let's try it:

Aψ(x) = Σn cnn(x)

= Σn ancnφn(x) [3] .

As you may have guessed, when ψ(x) is not in Domain(A), the following occurs in [3]: while every finite subsum is necessarily an element of the Hilbert space, in the limit of the infinite sum the "result" is no longer in the Hilbert space.

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