Two-state, two-particle QM question

[Mike Fontenot]

I'm trying to better understand how two particles are handled in QM, in
the simplist case where each particle (when isolated) is considered to
have only a 2-dimensional state space. (I.e., when their spatial
positions and momenta are unimportant, and only their (+-1/2) spins are
important).

I've been studying David Albert's book (an extraordinarily good book,
BTW), and I'll use here his simplified terminology, with |h> and |s>
(for "hard" and "soft") as basis vectors for the 2-D state space
(corresponding to spin 1/2 and -1/2 along the vertical (say) axis).

For a single isolated particle, the "hardness" operator H_sub_1 in that
basis is the 2x2 matrix with +1 and -1 on the diagonal, and zeros
off-diagonal, with |h> being the column vector [1 0] (eigenvalue +1) and
|s> being the column vector [0 1] (eigenvalue -1).

Then, for two such particles, Albert uses the basis vectors |hh>, |hs>,
|sh>, and |ss>, which could (arbitraily) be assigned the eigenvalues
+2, +1, -1, -2, respectively. For example, the 4-D state |hs>
corresponds to the state where particle 1 is definitely "hard", and
particle 2 is definitely "soft". One could then construct the
corresponding 4-D hardness operator H, which would correspond to a
single measurement which simultaneously gives the hardness state of each
of the two particles.

But then Albert talks about making a hardness measurement that is ONLY
made on particle one (even though the state space is 4-D). So he is
presumably talking about a 4-D operator (say, H_sub_T) that is different
from either H_sub_1 (which is a 2-D martix, with only the two
eigenvalues +1 and -1, and which operates only on 2-D vectors) or H
(which is 4-D, and has the eigenvalues +2, +1, -1, and -2, and operates
on 4-D vectors). The result of operating with H_sub_T on |bb> (for
example) needs to produce the vector (+1)|bb> ... i.e., the eigenvalue
needs to be +1, NOT the +2 result you would get with the operator H.

My question is, exactly what is the H_sub_T matrix (in the above given
4-D basis)? I initially thought I could construct H_sub_T using H1
sub-matrices along the diagonal (perhaps with zeros for the additional
two eigenvalues), but I wasn't able to make that work.

In short, how exactly does one construct an operator matrix
corresponding to a measurement that only returns the state of particle 1
(and with the same eigenvalues used in the isolated single particle
case), for the case where there are actually two particles, and
therefore the QM state is in a 4-D space?

Any suggestions, much appreciated.

--
Mike Fontenot

[Mike Fontenot]

Here's another way to try to explore my confusion about the two particle
(with two dimensions per particle) QM problem:

Start with a single 2D particle (whose position isn't of interest).
Choose a basis consisting of the eigenvectors |h> (hard) and |s> (soft).
Then that 2D space can be represented as a plane, with one coordinate
axis (say, the vertical axis, normally called the y axis) lying in the
direction of |h>, and the horizontal axis (normally called the x axis)
lying in the direction of |s>. Then ANY state that that isolated
particle can be in is represented by some vector in that "x,y" plane.
The |h> vector can be represented with the column vector [1 0], and the
|s> vector is represented with the column vector [0 1]. A "hardness"
operator H1 could be defined, with the eigenvalues +1 and -1 (say),
corresponding to the definite "hard" and "soft" states, so that H1
represents the property of hardness of particle 1, and also represents
the measurement of the hardness of particle 1. H1 would have +1 for
h11, -1 for h22, and zeros for the off-diagonal elements.

Suppose (to maximize simplicity) we have another (completely
independent) particle that has only a 1D state. Suppose it is ALWAYS
"hard" ... it is capable of being in only one state. That particle's
state space can be represented as a line, with a single eigenvector |h>,
represented with the column vector [1]. The hardness operator H2,
corresponding to the property of hardness for particle 2, would just be
a 1x1 matrix with the single element +2 (if we have arbitrarily arranged
for value of the hardness of particle 2 to be represented by the
eigenvalue +2).

Now, we can set up a 3D space for the purpose of simultaneously
describing both of these particles (so that we can describe the
situation when they are NOT completely isolated and independent). Let
the 3D space be represented by the "x,y" axes used for the isolated
particle 1, combined with a "z" axis used before for representing the
single dimension of the isolated particle 2. We can now call the two
eigenvectors for particle 1 |h1> and |s1>, and the single eigenvector
for particle 2 as |h2>. So now that the complete space is 3D, the
vector |h1> is represented by the column vector [1 0 0], |s1> is
represented by [0 1 0], and |h2> is represented by [0 0 1]. A
completely general vector in this 3D space can be represented as the
weighted sum of the three vectors |h1>, |s1>, and |h2>.

Now, ANY vector in that 3D space must represent SOME state of the
combined system. The question is, what does it mean (specifically for
particle 2) for the total state to be |h1> = [1 0 0]? Is that a state
where particle 2 doesn't EXIST? Or, conversely, what does a total state
|h2> = [0 0 1] mean as far as particle 1 is concerned ... does particle
1 simply not exist for that state?

And what are the two 3D operator matrices corresponding to the making of
a single measurement of the hardness of only one of the particles? How
are those matrices related to either the 2D matrix H1, or the 1D matrix
H2, that we used for the isolated, independent particles?

(The above description can obviously be expanded to the case where
particle 2 is also two-dimensional, which is what I'm REALLY interested
in. A single dimensional particle isn't very interesting ... I just
used it above, because 3D can be directly visualized (whereas 4D
cannot), and thus the 3D case is the simplest case to start out with).

Any words of wisdom will be much appreciated.

--
Mike Fontenot

[Geza Giedke]

Mike Fontenot schrieb am 15.01.2012 20:11:
> Here's another way to try to explore my confusion about the two particle
> (with two dimensions per particle) QM problem:

sorry, didn't see your second post before replying to the first

> Start with a single 2D particle (whose position isn't of interest).
> Choose a basis consisting of the eigenvectors |h> (hard) and |s> (soft).
> Then that 2D space can be represented as a plane, with one coordinate
> axis (say, the vertical axis, normally called the y axis) lying in the
> direction of |h>, and the horizontal axis (normally called the x axis)
> lying in the direction of |s>. Then ANY state that that isolated

[...]

> Suppose (to maximize simplicity) we have another (completely
> independent) particle that has only a 1D state. Suppose it is ALWAYS
> "hard" ... it is capable of being in only one state. That particle's
> state space can be represented as a line, with a single eigenvector |h>,

[...]

> Now, we can set up a 3D space for the purpose of simultaneously
> describing both of these particles (so that we can describe the
> situation when they are NOT completely isolated and independent).

why 3d? 2d is enough: you only need the states |hh> and |sh>

> the 3D space be represented by the "x,y" axes used for the isolated
> particle 1, combined with a "z" axis used before for representing the
> single dimension of the isolated particle 2. We can now call the two
> eigenvectors for particle 1 |h1> and |s1>, and the single eigenvector
> for particle 2 as |h2>. So now that the complete space is 3D, the
> vector |h1> is represented by the column vector [1 0 0], |s1> is
> represented by [0 1 0], and |h2> is represented by [0 0 1]. A
> completely general vector in this 3D space can be represented as the
> weighted sum of the three vectors |h1>, |s1>, and |h2>.

as you guess below, the system you describe this way is one in which the
number of particles (of type 1 and type 2, resp.) is not conserved:
since [1 0 0] would describe a system of 1 particle of type 1 (in
"internal state" |h>) and zero particles of type 1 in state |s> as well
as zero os particles of type 2.

It is often appropriate to allow particle numbr to vary when describing
many-body systems. In your 3d description, you've left out the states in
which both particles of type 1 and type 2 are present. (If no more than
1 particle of either type is allowed, you need two more vectors to span
the space, but of course you could have more particles - this then leads
to the formalism of Fock space and second quantization.)


regards
Geza

[Geza Giedke]

when your treating about many-body quantum systems, the Hilbert space
depends on whether your dealing with distinguishable or
indistinguishable particles. In the latter case, it also depends on
whether you're dealing with bosons or fermions.

Since you talk about measuring properties of "the first particle", I
assume you're concerned with distinguishable objects. Then the Hilbert
space of the two-particle system is the tensor product of the individual
systems' Hilbert spaces, i.e. in your case of two-level systems
H = C^2\otimes C^2, i.e. the four dimensional space spanned by
|hh>,|hs>,|sh>,|ss>
(|hh> is short for |h>\otimes |h>)
Operators that act on the first system only then take the form

H_T = H_1 \otimes I2, where I2 is the two-dimensional identity matrix.

in matrix form, you'll get

[+1 0 0 0
0 +1 0 0
0 0 -1 0
0 0 0 -1]

i.e. +1 for states |hh> and |hs> and -1 for states |sh> and |ss>.

In general for matrices A = \sum_{k,l} A_{kl} |k><l| and B=\sum_{r,s}
B_rs |r><s| their tensor product is
A\otimes B = \sum_{k,l,r,s} A_{kl} B_{rs} |kr><ls|

regards
Geza

[Mike Fontenot]

On 01/20/2012 02:04 PM, Geza Giedke wrote:
>
[...]
>

Thanks for both of your replies. You've given me lots of "food for
thought". I think your comments may be very helpful to me, but I'll
have to spend some quality time with them before I understand them very
well. I'll likely be back with some questions for you.

--
Mike Fontenot

[Mike Fontenot]

On 01/20/2012 02:03 PM, Geza Giedke wrote:

>
> Since you talk about measuring properties of "the first particle", I
> assume you're concerned with distinguishable objects.

Yes, that's right.

I've been spending a good bit of time studying your (first) response,
and I THINK I understand fairly well what you are doing ... I think your
information is EXACTLY what I've been needing.

But I'm not yet getting the correct results in all cases.

Using the algorithm you gave for the tensor product, I DO get your
result for the 4D operator that measures the "hardness" of ONLY particle
1. And I also get what I THINK is the correct answer for the 4D
operator that measures the hardness of ONLY particle 2: I think the
tensor product there is I2 x H_2, with the result having 1, -1, 1, -1 on
the diagonal (and zeros elsewhere) ... that says that the result of that
measurement is +1 for the two states |hh> and |sh>, and -1 for the two
states |hs> and |ss>, which seems correct.

But I haven't been able to get the 4D operator H that SIMULTANEOUSLY
gives the hardness of both particles ... i.e, the operator whose
eigenvectors (each with its own distinct eigenvalue) are just the 4D
basis vectors |hh>, |hs>, |sh>, and |ss>. Clearly, that operator needs
to have four DISTINCT eigenvalues ... something like +2, +1, -1, and -2,
say.

I thought I could perhaps get that from the tensor product H_1 x H_2,
where H_1 and H_2 are the identical 2D operators with +1 and -1 on the
diagonal (and zeros elsewhere). That seems to give me the 4D operator
with +1, -1, -1, +1 on the diagonals (and zeros elsewhere), which isn't
non-degenerate. Also, the structure of the algorithm (with it's
two-factor products of the elements of each of the argument matrices)
can't produce any coefficients other than zero, +1, and -1 from argument
matrices that contain nothing but zeros and +-1's. So I must be doing
something wrong somewhere.

In your (first) response, you wrote

> Then the Hilbert
> space of the two-particle system is the tensor product of the individual
> systems' Hilbert spaces, i.e. in your case of two-level systems

> H = C^2 x C^2,[...]

What are the C^2's? I took them to just indicate a 2D space (with
eigenvectors |h> and |s>) for each of the two isolated (independent)
particles. And for those I used H_1 and H_2. Was that what you had in
mind? Or should I have used something else for the two C^2 matrices
that form the tensor product for H?

--
Mike Fontenot

[Mike Fontenot]

A few more thoughts on the issues I raised in the above post:

A missing piece of the puzzle for me was in trying to understand the
connection between the "tensor product" of two 2D vectors (like |h><s|,
for example), and the 4D vector, written variously as |h>|s>, |h,s>, or
|hs>. (I think the "tensor product" is the same thing that Wald calls
the "outer product of tensors" in his GR book).

The outer product |h><s| of two 2D vectors, |h> and |s>, is a tensor
whose representative is a 2x2 matrix. When those 2D vectors are the two
chosen basis vectors, represented by the column vectors [1 0] and [0 1],
respectively, the matrix |h><s| is the 2x2 matrix c_ij consisting of
zeros except for the element c_12, and similarly for the other three
outer products of the basis vectors. (The position of the 1 in the
first vector determines which row of the c_ij matrix gets the 1, and the
position of the 1 in the second vector determines which column of the
c_ij matrix gets the 1.)

In contrast, the 4D vector |hs> or |h>|s> isn't a matrix, but I think
the required unstated step is to set up a one-to-one correspondence
between the 2x2 matrix |h><s| and the 4D vector |hs>, by using the first
row of |h><s| for the first two elements of |hs>, and using the second
row for the last two elements. The four outer products of the 2D basis
vectors then give four 4D vectors, each of which has only a single
non-zero element of value unity, and these are then the required 4D
basis vectors, corresponding to "hardness", in the new 4D space.

Surely the above stuff (or something equivalent) must have been spelled
out somewhere before, but I've never seen it in any of the material I've
read on the two-particle QM analysis, entanglement, and/or decoherence.

But here's the remaining outage for me, I think. The 4D "hardness"
operator, H, clearly must be the diagonal matrix with diagonal elements
+2, +1, -1, -2 (or perhaps some other arbitrary choice of four distinct
eigenvalues). But (assuming I haven't made a mistake) that 4D H matrix
doesn't appear to be the outer product of the two 2D hardness matrices
H_1 and H_2, which are each the diagonal matrix with diagonal elements
+1 and -1. So, the question is, ARE there two OTHER 2D matrices whose
outer product IS H? If so, what are they, and does it matter? I.e., is
it important to be able to construct various 4D operator matrices from
outer products of some known 2D matrices? It DID seem to work well for
determining the two 4D operator matrices that give the hardness of only
one of the two particles ... seems like it should also work somehow for
the 4D hardness operator matrix that simultaneously determines the
hardness of BOTH particles.

I've glossed over one thing in the above: when you take the outer
product of two 2x2 matrices, you don't directly get a 4x4 matrix d_ij,
with i and j each running from 1 to 4. Instead, you get a
four-dimensional thing, e_nmrs, with m, n, r, and s running from 1 to 2.
So, just like in the case of the outer product of two vectors, an
extra step is needed to set up a one-to-one correspondence between
e_nmrs, and d_ij.

--
Mike Fontenot

[Geza Giedke]

Mike Fontenot schrieb am 24.01.2012 19:22:
> A missing piece of the puzzle for me was in trying to understand the
> connection between the "tensor product" of two 2D vectors (like |h><s|,
> for example), and the 4D vector, written variously as |h>|s>, |h,s>, or
> |hs>. (I think the "tensor product" is the same thing that Wald calls
> the "outer product of tensors" in his GR book).
>
> The outer product |h><s| of two 2D vectors, |h> and |s>, is a tensor
> whose representative is a 2x2 matrix. When those 2D vectors are the two
> chosen basis vectors, represented by the column vectors [1 0] and [0 1],
> respectively, the matrix |h><s| is the 2x2 matrix c_ij consisting of
> zeros except for the element c_12, and similarly for the other three
> outer products of the basis vectors. (The position of the 1 in the
> first vector determines which row of the c_ij matrix gets the 1, and the
> position of the 1 in the second vector determines which column of the
> c_ij matrix gets the 1.)
>
> In contrast, the 4D vector |hs> or |h>|s> isn't a matrix, but I think
> the required unstated step is to set up a one-to-one correspondence
> between the 2x2 matrix |h><s| and the 4D vector |hs>, by using the first
> row of |h><s| for the first two elements of |hs>, and using the second
> row for the last two elements. The four outer products of the 2D basis
> vectors then give four 4D vectors, each of which has only a single
> non-zero element of value unity, and these are then the required 4D
> basis vectors, corresponding to "hardness", in the new 4D space.

as you observe, there is a one-to-one correspondence between vectors in
H x K and linear operators from K to H
which route one chooses to introduce the objects |hk> and |h><k| is a
matter of taste.

> Surely the above stuff (or something equivalent) must have been spelled
> out somewhere before, but I've never seen it in any of the material I've
> read on the two-particle QM analysis, entanglement, and/or decoherence.

have a look at John Watrous' Lecture Notes on QIT, he hase two short
chapters "mathematical preliminaries", which introduces nicely the
notion of tensor product and the isomprphism between vectors and operators.

http://www.cs.uwaterloo.ca/~watrous/quant-info/

> But here's the remaining outage for me, I think. The 4D "hardness"
> operator, H, clearly must be the diagonal matrix with diagonal elements
> +2, +1, -1, -2 (or perhaps some other arbitrary choice of four distinct
> eigenvalues). But (assuming I haven't made a mistake) that 4D H matrix
> doesn't appear to be the outer product of the two 2D hardness matrices
> H_1 and H_2, which are each the diagonal matrix with diagonal elements
> +1 and -1. So, the question is, ARE there two OTHER 2D matrices whose
> outer product IS H?

you're right that H has a diagonal representation, but it cannot be
written as a tensor product of two 2d operators X_1 and X_2. This can be
shown by direct computation (or by using the operator - vector
isomorphism mentioned above and showing thet H is isomorphic to an
entangled state)

> If so, what are they, and does it matter? I.e., is
> it important to be able to construct various 4D operator matrices from
> outer products of some known 2D matrices?

good question, and the answer depends on what you want to do with the
operator.
Note that being able to write an observable X on C^2 x C^2 as a product
of two observables A x I2 and I2 x B does not mean that you can measure
X by first measuring A in system 1 and the B on system two. A
counter-example is actually H_1 x H_2, which describes a measurement
that measures the parity of the system, i.e if both particles are in the
same state or not, and could be used to entangle two initially
independent systems, i.e. it can not be implemented by independent
actions on the two systems.

(The question how you can express a matrix on C^2 x C^2 in terms of
product matrices becomes very relevant, when these matrices represent
states (density matrices) and leads to the distiction of product states,
separable states, and entangled states.)


regards
Geza

[Geza Giedke]

Mike Fontenot schrieb am 23.01.2012 17:12:
> On 01/20/2012 02:03 PM, Geza Giedke wrote:
> Using the algorithm you gave for the tensor product, I DO get your
> result for the 4D operator that measures the "hardness" of ONLY particle
> 1. And I also get what I THINK is the correct answer for the 4D
> operator that measures the hardness of ONLY particle 2: I think the
> tensor product there is I2 x H_2, with the result having 1, -1, 1, -1 on
> the diagonal (and zeros elsewhere) ... that says that the result of that
> measurement is +1 for the two states |hh> and |sh>, and -1 for the two
> states |hs> and |ss>, which seems correct.

right

> But I haven't been able to get the 4D operator H that SIMULTANEOUSLY
> gives the hardness of both particles ... i.e, the operator whose
> eigenvectors (each with its own distinct eigenvalue) are just the 4D
> basis vectors |hh>, |hs>, |sh>, and |ss>. Clearly, that operator needs
> to have four DISTINCT eigenvalues ... something like +2, +1, -1, and -2,
> say.
>
> I thought I could perhaps get that from the tensor product H_1 x H_2,
> where H_1 and H_2 are the identical 2D operators with +1 and -1 on the
> diagonal (and zeros elsewhere). That seems to give me the 4D operator
> with +1, -1, -1, +1 on the diagonals (and zeros elsewhere), which isn't
> non-degenerate. Also, the structure of the algorithm (with it's
> two-factor products of the elements of each of the argument matrices)
> can't produce any coefficients other than zero, +1, and -1 from argument
> matrices that contain nothing but zeros and +-1's. So I must be doing
> something wrong somewhere.

you're on the right scent: H_1 x H_2 _almost_ does what you want, but
due to the choice of eigenvalues it does not distinguish between |hh>,
|ss> resp |hs>,|sh>. But the eigenvalue corresponding to a given
eigenvector is rather arbitrary, what matters (for observables) are only
the eigenstates/eigenspaces. I.e., any operator H_1' given by an
(invertible) function of H_1 is measuring "the same observable" (just
using different units). Hence you could use H_1' = diag(a,1/a) and
H_2'=diag(b,1/b) (a=2,b=3), then H_1' x H_2' describes the desired
measurement which distinguishes all for product states |hh>,|hs>,|sh>,
and |ss>.

Note, that not all operators on C^2 x C^2 are of product form A x B.
In general, we have to consider sums of such operators
A_1 x B_1 + A_2 x B_2 + ...
The operator diag(+2,+1,-1,-2) is not of product form.

In the present case (where the measurements H_1 and H_2 commute) you
could also perform them sequentially, i.e. perform a two-outcome
measurement giving results (+1,+1) if the system is in state |hh>,
(+1,-1) if in |hs> etc.
Generally, any (von Neumann) measurement in QM can be defined by a set
of mutually orthogonal projection operators P_k and corresponding
distinct labels \lambda_k, i.e. M = {P_k,\lambda_k}. Without loss of
generality, the labels \lambda_k can be identified with (distinct) real
numbers and then
M "measures the observable \sum_k lambda_k P_k"
For H_1, there are two 1d projectors, namely the one projecting on |h>
(denoted as P_h = |h><h|) and the one projecting on |s>, i.e. P_s = |s><s|.
For H_1 x I2, we need two two-dim projectors, namely P_h x I2 and P_s x
I2 and H_1 xI2 = P_h x I2 - P_s x I2.
For the full measurement, we need 4 1-d projectors
|h><h| x |h><h|, |h><h| x |s><s|, etc.

> In your (first) response, you wrote
>
>> Then the Hilbert
>> space of the two-particle system is the tensor product of the individual
>> systems' Hilbert spaces, i.e. in your case of two-level systems
>> H = C^2 x C^2,[...]
>
> What are the C^2's? I took them to just indicate a 2D space (with
> eigenvectors |h> and |s>) for each of the two isolated (independent)
> particles. And for those I used H_1 and H_2. Was that what you had in
> mind? Or should I have used something else for the two C^2 matrices
> that form the tensor product for H?

yes, C^2 just refers to the two-dimensional (complex) space, i.e. pairs
of complex numbers. For the present purpose, we can take also R^2, i.e.
2d real space.

regards
Geza

[Mike Fontenot]

On 01/25/2012 08:21 AM, Geza Giedke wrote:
>
[...]
>

WOW! Thanks for both of your recent responses ... looks like a LOT of
very good information. It'll take me a while to absorb all of it, but I
will.

--
Mike Fontenot

[Mike Fontenot]

Just a progress report:

There are still aspects of your (Geza's) responses that I haven't
figured out, but several things you said have gotten me over an
important hump.

I HAD realized (while studying Albert) that eigenvalues can be chosen
somewhat arbitrarily. But I hadn't tried to use that fact when I was
choosing H1 and H2, to get H from their outer product. So, I first
thought I could choose H1 as diag{+1, -1} and H2 as diag{+2, -2}, but
that doesn't work ... just making all four evals distinct ISN'T
sufficient (and it actually turns out to not even be a necessary
condition). (I started with a preference for using matching positive
and negative evals, just because Albert always seemed to prefer that in
his book).

So, I looked back at your example, with H1 = diag{2, 1/2} and H2 =
diag{3, 1/3}. I verified that the outer product is then diag{6, 2/3,
3/2, 1/6} (for the evec order |11>, |12>, |21>, and |22>), which DOES
work fine.

THEN, I tried to figure out exactly what the restrictions ARE on the
choice of the original evals. I found that there are four requirements
(necessary and sufficient):

1) and 2): The evals in each of the H1 and H2 must be distinct (although
is is not necessary that all four of those evals have to be distinct).
So (if H1 = {aij} and H2 = {bij}), the first two requirements are that
a11 \= a22 and b11 \= b22. (Recall that the property being measured by
H1 and H2 correspond to the basis vectors being used in those spaces, so
H1 and H2 are diagonal matrices, and aij = bij = 0 for i \= j).

3) a11 b11 \= a22 b22

4) a11 b22 \= a22 b11

By playing around with those requirements, I concluded that none of the
four evals can be zero. And I can't choose the evals to be +-n and +-m
(like I initially wanted to do). I found that, if I stuck with
positive choices, satisfying the requirements became easier, but that
isn't a necessary condition. I first tried 1, 2, 3, and 4 (for a11,
a22, b11, and b22), and that works fine ... it gives H = diag{3, 4, 6, 8}.

Then, I tried 3, 2, 2, 6, to see if I could get an example that works
when the four original evals aren't all distinct, and that worked fine.
So does 3, -2, 4, -6, which shows that the original evals don't have
to all be positive.

Anyway, that gets me over that particular (very uncomfortable) hump ...
MUCH obliged for your help!

But I must have misinterpreted some of your later comments: they seemed
(to me) to contradict your above example (and also the several examples
I gave above). I had interpreted those later comments of yours to say
that it WASN'T possible to get (diagonal) H from the outer product of
any choice of (diagonal) H1 and H2. Here's a quote of one of your
statements I must be misinterpreting:

"You're right that H has a diagonal representation, but it cannot be

written as a tensor product of two 2d operators X_1 and X_2. This can be
shown by direct computation (or by using the operator - vector

isomorphism mentioned above and showing that H is isomorphic to an
entangled state)."

Can you help me with where I'm misinterpreting what you meant?

(P.S.: I haven't had time yet to read and understand those two
"mathematical preliminaries" lectures you linked me to, but a quick look
suggests they may fill some current voids in my mathematical toolkit.
I've printed them out, and expect to give them a lot of quality time
when I can get to it ... thanks for that reference.)

--
Mike Fontenot

[Mike Fontenot]

On 01/15/2012 12:11 PM, Mike Fontenot wrote:
> Here's another way to try to explore my confusion about the two particle
> (with two dimensions per particle) QM problem:
>

[...]
>
Just a "post-mortem":

In my original two posts in this thread, I was trying to construct a 4D
space for two particles by adding (in a vector sense) the two 2D spaces
of the two particles when they exist completely independently. I.e., I
wanted the 4D space to consist of the additive combination of two
mutually orthogonal 2D subspaces, where each of those 2D subspaces are
unchanged from the usual way of setting up a 2D subspace for a single
isolated two-state particle (whose position is of no importance). The
idea, then, was that any general vector in the 4D space (for example,
the state |hs>) could then be written as the sum of two vectors, each of
which lies completely within one of the 2D subspaces of the independent
particles.

I now think that won't work, because a 4D vector lying completely within
one of those 2D subspaces represents a 4D state where the other particle
doesn't exist, which violates the original objective of setting up a
two-particle space where both particles always exist.

Another way of saying this is that if I start out with the basis vectors
of the 4D two-particle space being the usual ones (|hh>, |hs>, |sh>, and
|ss>), I THOUGHT I could rotate those basis vectors, and thereby get a
new set of four basis vectors that correspond to the basis vectors used
to describe the original isolated single particles. I no longer think
that is possible.

--
Mike Fontenot

[Mike Fontenot]

On 01/30/2012 04:39 PM, Mike Fontenot wrote:
>
> Just a progress report:
> [...]
>

Another issue I'm confused about:

In discussions of entanglement, the implication usually seems to be that
two particles can become entangled when they are close together and
interacting. But suppose that, after two entangled particles, with the
total state |h1, s2> + |s1, h2> (appropriately normalized), have become
widely separated, a "hardness" measurement is made on particle 1. Then
the overall state, after that measurement, is either |h1, s2> or |s1,
h2>. Presumably those two particles are now disentangled and
independent, presumably forever, since they are now widely separated and
non-interacting.

But isn't it true that the state |h1, s2> can be expressed as some
superposition of the definite "blackness" states |b1, w2>, |b1, b2>,
|w1, b2>, and |w1, w2>, in the same way that a single "hard" particle is
in a superposition of "black" and "white"? If so, then in terms of the
simultaneous "blackness" property, aren't the two particles still entangled?

Any words of wisdom, much appreciated.

--
Mike Fontenot

[Jim Heckman]

On 14-Feb-2012, Mike Fontenot <mlf...@comcast.net>
wrote in message <4F3ACD5D...@comcast.net>:

> On 01/30/2012 04:39 PM, Mike Fontenot wrote:
>
> > Just a progress report:
> > [...]
> >
>
> Another issue I'm confused about:
>
> In discussions of entanglement, the implication usually seems to be that
> two particles can become entangled when they are close together and
> interacting.

Right, in general.

> But suppose that, after two entangled particles, with the
> total state |h1, s2> + |s1, h2> (appropriately normalized), have become
> widely separated, a "hardness" measurement is made on particle 1. Then
> the overall state, after that measurement, is either |h1, s2> or |s1,
> h2>. Presumably those two particles are now disentangled and
> independent, presumably forever, since they are now widely separated and
> non-interacting.

Correct.

> But isn't it true that the state |h1, s2> can be expressed as some
> superposition of the definite "blackness" states |b1, w2>, |b1, b2>,
> |w1, b2>, and |w1, w2>, in the same way that a single "hard" particle is
> in a superposition of "black" and "white"? If so, then in terms of the
> simultaneous "blackness" property, aren't the two particles still
> entangled?

No. Once you perform enough measurements to completely specify the
state of one particle, the two become totally disentangled, so the
second particle is also in a definite state, and the total state of
the system is the tensor product of the individual states of the
two particles.

I don't remember your definition of "blackness" relative to
"hardness", but for the sake of argument let's let |h> = |b> + |w>
and |s> = |b> - |w> (suitably normalized). Then you're right that
any state of the system can be expressed in the |b1, b2>, |b1, w2>,
|w1, b2>, |w1, w2> basis.

So suppose the result of your measurement of the hardness of
particle 1 left the system in say, the |h1, s2> state, which is
(|b1 + |w1>) (x) (|b2> - |w2>) = |b1, b2> - |b1, w2> + |w1, b2> -
|w1, w2>. At first glance this state may look entangled, but we
know it really isn't precisely because it's the product of two
single-particle states. If you now measure the "blackness" of
particle 1 and get the result "black", the system state becomes
|b1, b2> - |b1, w2> = |b1> (x) (|b2> - |w2>) = |b1, s2>, and if you
get the result "white", the system state becomes |w1, b2> -
|w1, w2> = |w1> (x) (|b2> - |w2>) = |w1, s2>. So particle 2 stays
in the |s> state whatever the result of the measurement on
particle 1.

> Any words of wisdom, much appreciated.

--

Jim Heckman

[Mike Fontenot]

On 02/16/2012 09:59 AM, Jim Heckman wrote:
>
> So suppose the result of your measurement of the hardness of
> particle 1 left the system in say, the |h1, s2> state, which is

> [...] = |b1, b2> - |b1, w2> + |w1, b2> - |w1, w2>.
>

OK, I get it ... If we measure that P1 is black, that doesn't pick out a
definite state of "blackness" for P2, because there are two "blackness"
eigenvectors in the above sum that contain b1. So we are not left with
a state that is an eigenvector of "blackness" in the combined state
space ... we still have a superposition of two of those four "blackness"
eigenvectors. Great explanation ... many thanks.

--
Mike Fontenot