# Bras and Kets and Tensors

Ok, I know we can use the notation for every vector space if we wan't. Of cause we can do that. I'm not sure why you say that multiparticle states are direct products?

If the particles are independent, you can write them as a tensorproduct of two vectors, if they are correlated then you can't nessecarily.

The reason i said you equation was wrong, was because we where talking about QM, so it didn't make sense.

Again you are right that a vector is often described by a n-tuple, but as i have said a lot of times in this thread, the tuple doesn't make sense with out a basis, telling us what it means. A bit like your equation didn't make sense because you didn't tell what you ment by |p> and |F>.

The problem about adjoint, is to write the definition used in math

$$<x,A y> = <A^*x,y>$$

in diracs notation. You have to be very carefull to write this.

Not sure what your point is about fock-space? Is it because if we have a space describing one particle, and we take a tensor product between such two states then we are not in the space anymore, but in the fock space formalism you incorporate this problem?

I haven't read diracs book, but it sounds interesting, I will look at it in my vecation, thanks for the reference. I agree that he made the notation because it made it simpler to write (maybe to remember some rules of manipulating), but I just think that people often get a bit confused about it, because one learn QM with wavefunctions first and then learn bra-ket, then often people think that the wavefunction is used just like a ket, and it often isn't (even though you proberly could, after all L^2 is a vector space).

Hurkyl
Staff Emeritus
Gold Member
Note that a multiparticle state, |p1, p2> is not usually taken as a column vector, but rather a direct product of two vectors
...
So, as a direct product is a tensor
That last statement is (very) incorrect! The direct product of two vector spaces is quite different than their tensor product -- in fact, most quantum 'weirdness' stems from the fact you use direct products classically but tensor products quantum mechanically.

pwew, there was another that found that a bit disturbing.

what does this mean, I can't see this can be correct:

"Note that a multiparticle state, |p1, p2> is not usually taken as a column vector, but rather a direct product of two vectors "

Maybe I just can't read it but what does

"So, as a direct product is a tensor"

mean?

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by the way

$$\mathbb{R}\otimes\mathbb{R} ~=~ \mathbb{R}^2$$

and

$$\mathbb{C}^3\otimes\mathbb{C}^3\otimes\mathbb{C}^3 \otimes ..... \otimes \mathbb{C}^3~=~ \mathbb{C}^{3n}$$

is not correct. it is

$$\mathbb{R}\otimes\mathbb{R} ~=~ \mathbb{R}$$

and

$$\mathbb{C}^3\otimes\mathbb{C}^3\otimes\mathbb{C}^3 \otimes ..... \otimes \mathbb{C}^3~=~ \mathbb{C}^{3^n}$$

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why are your post suddenly below mine?

I don't know what you mean with

"This just defines dimensions, of course, if there is interaction then the combined
probabilities are not given by simply multiplication. That's a whole different story
altogether requiring knowledge of the orthogonal states, the propagators and the
interactions."

If one particle is described in C^3, then n particles are described in

$$\mathbb{C}^3\otimes\mathbb{C}^3\otimes\mathbb{C}^3 \otimes ..... \otimes \mathbb{C}^3~=~ \mathbb{C}^{3^n}$$

but it can be that you can't write the state as $$|0> \otimes |1> \otimes ... \otimes |n>$$, is that what you try to say ?

Hans de Vries
Gold Member
That last statement is (very) incorrect! The direct product of two vector spaces is quite different than their tensor product

Note that a multiparticle state, |p1, p2> is not usually taken as a column vector, but rather a direct product of two vectors -- there are definitional tricks that allow the multiparticle sates to be considered as a single column vector. So, as a direct product is a tensor, we've now got ......
Like in:

$$\mathbb{R}\otimes\mathbb{R} ~=~ \mathbb{R}^2$$

Or for a non relativistic QM multiparticle state of n particles:

$$\mathbb{C}^3\otimes\mathbb{C}^3\otimes\mathbb{C}^3 \otimes ..... \otimes \mathbb{C}^3~=~ \mathbb{C}^{3n}$$

This just defines dimensions, of course, if there is interaction then the combined
probabilities are not given by simply multiplication. That's a whole different story
altogether requiring knowledge of the orthogonal states, the propagators and the
interactions.

Regards, Hans

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Hans de Vries
Gold Member
why are your post suddenly below mine?
Something went wrong with editing. You just react "too fast". I don't know what you mean with

"This just defines dimensions, of course, if there is interaction then the combined
probabilities are not given by simply multiplication. That's a whole different story
altogether requiring knowledge of the orthogonal states, the propagators and the
interactions."

If one particle is described in C^3, then n particles are described in

$$\mathbb{C}^3\otimes\mathbb{C}^3\otimes\mathbb{C}^3 \otimes ..... \otimes \mathbb{C}^3~=~ \mathbb{C}^{3^n}$$

but it can be that you can't write the state as $$|0> \otimes |1> \otimes ... \otimes |n>$$, is that what you try to say ?
The "static" wave function is defined as a complex number in a 3 dimensional space.
The non-relativistic wave function of two particles is defined as a 6 dimensional
space spanned by the x,y,z of the first particle plus the x,y,z of the second particle.

The wave function of an n-particle system is defined in an 3n dimensional space, not
a 3^n dimensional space.

Regards, Hans

how can you say that, the dimension of a tensor product is, like i say. And one particle can have more degrees of freedom than just 3.

The things you say are equal, are simply not equal. If you have a two vector spaces V and W and basis v_1,...,v_n and w_1,...,w_d respectivly, then a basis for

$$V \otimes W$$ is

all of the form $$v_i \otimes w_j \ , \ i=1,...,n \ and \ j=1,...,d$$

there is clearly n times d of these, not n + d as you say. You are right that one particle can be described by a wavefunction of x,y,z, and that two by a wavefunction of x_1,y_1,z_1,x_2,y_2,z_2, but we are talking about the statespace, and if it is fx. 5 dim for both, then the state of both particles lives in a 25 dim space.

I don't think it is right to say that the wavefunction lives in a space spanned by x,y,z.

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Hans de Vries
Gold Member
how can you say that, the dimension of a tensor product is, like i say. And one particle can have more degrees of freedom than just 3.

The things you say are equal, are simply not equal. If you have a two vector spaces V and W and basis v_1,...,v_n and w_1,...,w_d respectivly, then a basis for

$$V \otimes W$$ is

all of the form $$v_i \otimes w_j \ , \ i=1,...,n \ and \ j=1,...,d$$

there is clearly n times d of these, not n + d as you say. You are right that one particle can be described by a wavefunction of x,y,z, and that two by a wavefunction of x_1,y_1,z_1,x_2,y_2,z_2, but we are talking about the statespace, and if it is fx. 5 dim for both, then the state of both particles lives in a 25 dim space.

I don't think it is right to say that the wavefunction lives in a space spanned by x,y,z.

You are confusing the number of dimensions with the number of elements.

$\mathbb{R}$ has 1 dimension with $\infty$ elements while $\mathbb{R}^2$ has 2 dimensions with $\infty^2$ elements.

Regards, Hans

Hurkyl
Staff Emeritus
Gold Member
The "static" wave function is defined as a complex number in a 3 dimensional space.
The non-relativistic wave function of two particles is defined as a 6 dimensional
space spanned by the x,y,z of the first particle plus the x,y,z of the second particle.

The wave function of an n-particle system is defined in an 3n dimensional space, not
a 3^n dimensional space.
Ah, that's where the confusion lies! The rest of us are talking about the state vectors, rather than elements of the underlying topological space of a position-representation of those vectors.

$L^2(\mathbb{R}^3)$ is, of course, the space of square-integrable functions on Euclidean 3-space; i.e. the space of single-particle wavefunctions.

The tensor product of this space with itself is given by1 $L^2(\mathbb{R}^3) \otimes L^2(\mathbb{R}^3) = L^2(\mathbb{R}^3 \times \mathbb{R}^3)$ -- so a 2-particle wavefunction is a square-integrable function of 6 variables.

However, if you only took the direct product of the state space with itself, you'd get $L^2(\mathbb{R}^3) \times L^2(\mathbb{R}^3) \neq L^2(\mathbb{R}^3 \times \mathbb{R}^3)$. This is merely the space of pairs of square-integrable functions of three variables. This isn't even (naturally) a subspace of $L^2(\mathbb{R}^3 \times \mathbb{R}^3)$; the obvious map between them is bilinear, not linear.

1: At least I'm pretty sure I have this right. I haven't actually worked through all the fine print to prove this statement.

$L^2(\mathbb{R}^3) \otimes L^2(\mathbb{R}^3) = L^2(\mathbb{R}^3 \times \mathbb{R}^3)$ should intuativly be right, but if it is mathematicaly i'm not sure, but i guess it must have something to do with fubinis theorem

http://en.wikipedia.org/wiki/Fubini's_theorem

or at least some variant of it. Hurkyl, as I also pointed out, I agree that it seems as though we are talking about something different, so thats why there are some confussion, but just to make something clear, L^2(R^3) is not spanned by x,y,z (but I guess you mean, as also Hurkyl says, that you can write the wavefunction as a function of x,y,z(but maybe one should be a little carefull here because internal freedoms can play a role, such as spin which we need spin wavefunction to describe, but maybe we should forget about internal freedom in our descussion, so we don't confuse eachother even more)).

And the statement

$$\mathbb{R}\otimes\mathbb{R} ~=~ \mathbb{R}^2$$

$$\mathbb{C}^3\otimes\mathbb{C}^3\otimes\mathbb{C}^3 \otimes ..... \otimes \mathbb{C}^3~=~ \mathbb{C}^{3n}$$

is wrong, even though we are talking about different things.

Hurkyl
Staff Emeritus
Gold Member
Oh, I just realized I know how to compute the direct product:

$$L^2(\mathbb{R}^3) \times L^2(\mathbb{R}^3) \cong L^2(\mathbb{R}^3 + \mathbb{R}^3)$$

The + on the right hand side indicates disjoint union -- i.e. that space consists of two separated copies of R³

Hans de Vries
Gold Member
And the statement

$$\mathbb{R}\otimes\mathbb{R} ~=~ \mathbb{R}^2$$

$$\mathbb{C}^3\otimes\mathbb{C}^3\otimes\mathbb{C}^3 \otimes ..... \otimes \mathbb{C}^3~=~ \mathbb{C}^{3n}$$
is wrong

There is nothing wrong with this. I'm using the definition of the vector direct product
given here: http://mathworld.wolfram.com/VectorDirectProduct.html

In this example each "dimension" has 3 elements while $\mathbb{R}$ or $\mathbb{C}$ represents 1 continuous
dimension with $\infty$ elements.

If two wavefunctions are non-interacting then the vector direct product describes
the combined probabilities. If they are interacting then one has to go back to the
physics and, in most cases, use an iterative process to numerically determine
the combined two-particle wave function.

Regards, Hans.

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Hurkyl
Staff Emeritus
Gold Member
As I pointed out, you're not talking about R: you're talking about L²(R). The tensor product of R with itself is clearly R -- in your way of thinking, that's because R is a single-dimensional vector space, and 1*1=1.

Hans de Vries
Gold Member
As I pointed out, you're not talking about R: you're talking about L²(R). The tensor product of R with itself is clearly R -- in your way of thinking, that's because R is a single-dimensional vector space, and 1*1=1.

I'm using the vector direct product as defined here:

http://mathworld.wolfram.com/VectorDirectProduct.html

Using the tensor rank: the number of indices (either discrete or continuous) as the
number of dimensions, like most physicist would do.

Might be, This isn't language found in physics textbooks or mathematical books for
physicist. So using such an expression like this is quite meaningless for most physicist.

Regards, Hans

Let me guess: Square integrable functions, ok? Last edited:
I'm using the vector direct product as defined here:

http://mathworld.wolfram.com/VectorDirectProduct.html

Using the tensor rank: the number of indices (either discrete or continuous) as the
number of dimensions, like most physicist would do.

Might be, This isn't language found in physics textbooks or mathematical books for
physicist. So using such an expression like this is quite meaningless for most physicist,
however trival its mathematical meaning may be...

Regards, Hans

There is nothing wrong with this. I'm using the definition of the vector direct product
given here: http://mathworld.wolfram.com/VectorDirectProduct.html

In this example each "dimension" has 3 elements while $\mathbb{R}$ or $\mathbb{C}$ represents 1 continuous
dimension with $\infty$ elements.

If two wavefunctions are non-interacting then the vector direct product describes
the combined probabilities. If they are interacting then one has to go back to the
physics and, in most cases, use an iterative process to numerically determine
the combined two-particle wave function.

Regards, Hans.

you are refering to a page that tells how to take the tensor product between to vectors, but you are taking the tensor product between vector spaces, so you should refer to something like

http://mathworld.wolfram.com/VectorSpaceTensorProduct.html

and it agrees with me. You are right that if particles are none interacting you can write them as pure states (maybe you can't always ???), but they still live in the tensor product of the two hilbertspaces, and the dimension of this is the product of the two dimensions.

But you are aperently talking about indices in the tensor (dimension of fx. a matrix), that is completely different than what is being discussed here. It is of cause trivial that taking two 1-tensors (vectors) and taking the tensot product gives a 2-tensor. By the way, using the notation where you write $\mathbb{R}$, to be a continuous vector, can't be standard notation, even in physics? And the Vector Direct Product you are refering to, is only defined for finite dimensional tensors.

But as been mentioned we are talking about the state space, and then it is commen to take the tensor product of the individual spaces. And to give you an simple example of what i'm talking about, lets look at two spin ½ particles, where we don't care about anything else than the spin. Then each particle have 2 degrees of freedom, so we could have
´
|00>,|01>,|10> and |11> (1=up,0=down)

that is clearly 2*2 = 4, so this is a 4-dim space, as i say. Then because of how we make the tensor product to a hilbert space, it is a natural thing to describe this in the tensor product of the two spaces, because the inner product is given by

<01|01> = <0|0><1|1>

so the probabillity of being in the down up state, is the probabillity of being in down times being in up, which is very natural. The strange thing is that taking the tensor product of the spaces, gives us states that are intanglet and other strange things.

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Hurkyl
Staff Emeritus
Gold Member
I'm using the vector direct product as defined here:

http://mathworld.wolfram.com/VectorDirectProduct.html
Which contains an example indicating $\mathbb{R}^3 \otimes \mathbb{R}^3 \cong \mathbb{R}^9$ -- not $\mathbb{R}^6$ as you suggest.

The problem is that you are using the letter R -- a letter well-established to indicate something akin to "the one-dimensional vector space over the reals". You are using the symbol $\otimes$ -- a symbol well-established to indicate a particular arithmetic operation on vector spaces and on their elements. You are interjecting into a conversaion where we are talking about products on vectors and vector spaces.

So, when you change the meaning of both of those symbols (using R to instead denote some continuously indexed space and $\otimes$ to denote some fancy operation on index spaces) and change the context of the conversation (talking about operations on index spaces rather than on vectors) you should expect there to be much confusion. This is greatly magnified because you didn't give any indication that you were using those symbols in a nonstandard way, and continued to interpret others' posts as using those meanings, despite others having very clearly indicated they were using those symbols according to the usual meaning.

Actually, I think it's far more you've accidentally made a 'level' slip, and confused two layers of abstraction. (The relevant layers here being points of Euclidean space, Euclidean space, and functions on Euclidean space)

That aside, I will admit that this is the first time I've ever heard the phrase 'direct product' used to refer to something that really isn't a direct product but instead a tensor product.

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Hans de Vries
Gold Member
you are refering to a page that tells how to take the tensor product between to vectors, but you are taking the tensor product between vector spaces, so you should refer to something like

http://mathworld.wolfram.com/VectorSpaceTensorProduct.html

reilly was talking about a non relativistic two particle wave function as the vector direct
product of two single particle function which is correct according to the definition of the
vector direct product given here:

http://mathworld.wolfram.com/VectorDirectProduct.html

You may have an argument in that I implicitly assume that in $R\otimes R$ one is
a row vector and the other is a column vector, so an nx1 vector times a 1xn
vector is an nxn matrix, but I wouldn't even know how to express a transpose
operation at higher ranks without people loosing track of the otherwise very
simple math.

Regards, Hans

Hans de Vries
Gold Member
The problem is that you are using the letter R -- a letter well-established to indicate something akin to "the one-dimensional vector space over the reals".

I used $\mathbb{R}$ to indicate the range of the single continues index of a one dimensional vector
with $\infty$ elements and I use $\mathbb{R}^3$ to describe the 3 continuous indices of a function in a volume.
I shouldn't have used $\mathbb{C}$ in this context.

So, symbolically in, in terms of indices:

$$A\otimes B\otimes C ~=~ D$$

If the inidices of A, B and C are given by $\mathbb{R}$ then the indices of D are given by $\mathbb{R}^3$
Indices (tensor ranks) add. The direct product of three tensors of rank 1 is a tensor
of rank 3.

$$\mbox{rank}(A\otimes B\otimes C) ~=~ \mbox{rank}(A)+\mbox{rank}(B)+\mbox{rank}(C) ~=~ \mbox{rank}(D)$$

You are associating $\mathbb{R}^n$ with the number of elements instead of the indices and thus
you get the following in the same case:

If the number of elements of A, B and C is given by $\mathbb{R}^\infty$ then the number of elements of D
is given by $\mathbb{R}^{\infty^3}$. The number of elements multiply and hence the number of $\infty^3$

As long as we understand each other.

Regards, Hans

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Hurkyl
Staff Emeritus
Gold Member
reilly was talking about a non relativistic two particle wave function as the vector direct
product of two single particle function which is correct according to the definition of the
vector direct product given here:

http://mathworld.wolfram.com/VectorDirectProduct.html
Which is the same thing as the tensor product the rest of us are talking about.

(Fine print -- there are a bunch of equivalent ways to define tensor products, so I should really say this is just a particular realization of the tensor product)

You may have an argument in that I implicitly assume that in $R\otimes R$ one is
a row vector and the other is a column vector, so an nx1 vector times a 1xn
vector is an nxn matrix,
That's not what the argument is. The argument is that for elements of R, we have n=1. The argument is that while you might mean to talk about continuously indexed spaces, the thing you are actually saying is "the product of a 1x1 matrix with a 1x1 matrix is a matrix with 2 entries".

Hans de Vries
Gold Member
while you might mean to talk about continuously indexed spaces

It's indeed exactly this. Once you replace $A\otimes B$ with $\mathbb{R}^n\otimes\mathbb{R}^m$, then that's another
level of symbolization and there is an ambiguity of to what $\mathbb{R}^n$ refers to, the number
of elements or the number of continuous indices.

Regards, Hans

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Am I understanding you right (if we take the finite case), that you say that with the dimension you mean the size of fx. two vectors (n x 1) and (1 x m) then you get a
(n x m) matrix, with the tensor (or as you call it vector direct product), because that is indeed true.

The problem is that we need another dimension, and it is very much used in physics. Haven't you ever seen, something like "assume that we have a two level system, of two states |0> and |1>." This means that |0> and |1> are a basis for our problem, that is two dimensional. If we then wan't to describe two of these systems, we take the tensor product of the two spaces and a basis for this new vector space is

$$|0> \otimes |0>,|0> \otimes |1>,|1> \otimes |0>,|1> \otimes |1>$$,

this is very standard and used by alot of physisist (at least all there is doing QM). I'm a bit baffled that you would immediately think of the vector dirrect product between vectors (and matrices). Do you have some refferences where this is used (prefereble online), because >I know that realization of the tensor product, but never seen it in use anywhere.

I still understand how you will use that definition if you don't have finite vectors like (n x 1)?

Do you know the general definition of the tensor product? Because you are talking about tensor product between elements of a vector space, which in the final case can be realized as you say, but that new element actually lives in the spaces that we are talking about, that is say we have a vector space (V) spanned by (1,0) and (0,1) then you construct a new like this

$$(1,0) \otimes (0,1)^T = ((0,1)^T,(0,0)^T)$$

this i actually an element in the new vector space denoted

$$V \otimes V$$ up to isomorphism

and if one particle is living in the space V, then this is the natural space to describe two of these particles in. Because two particles of these types can be in any linear combination of products like $$(1,0) \otimes (0,1)^T = ((0,1)^T,(0,0)^T)$$ this, where you take all different combination of the basis vectors. So you are talking about the elements where we are talking about the space these live in.

But it is standard QM, and it is importent what spaces you get, more than just taking the elements, because it is rare that you can just say we this particle in one state (a,b), and a particle number two in another state (c,d), and then only be interested in the combination (a(c,d)^T,b(c,d)^T), because as time evolse it can be a lot of other things, and these things it can become live in the space we are talking about, thats why it is importent because you know with space to restrict to.

Hurkyl
Staff Emeritus
Gold Member
It's indeed exactly this. Once you replace $A\otimes B$ with $\mathbb{R}^n\otimes\mathbb{R}^m$, then that's another
level of symbolization and there is an ambiguity of to what $\mathbb{R}^n$ refers to, the number
of elements or the number of continuous indices.

Regards, Hans
You do realize that R, R², R³, ... all have the same number of elements, right? Even the separable Hilbert space (e.g. the space of wavefunctions continuously indexed by $\mathbb{R}^n$) has the same number of elements as R! So, I should hope nobody ever uses those symbols in this context to indicate number of elements.

$\mathbb{R}^n$ cannot be denoting a number in this context. There is usually no ambiguity here, because $\mathbb{R}^n$ is always meant to indicate the standard n-dimensional vector space over R -- you are the only source I have ever seen who insists on using R in any other way in this context... and it's somewhat bewildering why you would do so, not just because you insist upon confusing an index set with a vector space 'over' those indices, but also because you refuse to use the name of the actual operation you are doing on index sets -- the Cartesian product -- and instead prefer to use the name of the operation performed on the corresponding vector spaces.

This whole thing would be akin to me insisting upon saying $3 \cdot 5 = 8$ when I really mean $e^3 \cdot e^5 = e^8$. (And not even using $3 + 5 = 8$ which would be a correct statement)