# Tensor product

1. Oct 5, 2007

### captain

what is the tensor product's physical significance? I know what it does mathematically, but what does it mean. I have looked on textbooks and wikipedia but i still can't understand the physical signifcance.

2. Oct 6, 2007

### pmb_phy

The physical signifigance will depend on the particular application.

Pete

3. Oct 6, 2007

### captain

like if you do the tensor product between two vectors or between two state kets (in qm).

4. Oct 6, 2007

### Reverie

As was mentioned previously, the physical significance depends on the application. Maybe this explanation will help.

Let V be a three dimensional vector space with basis {e1,e2,e3}, and let W be a four dimensional vector space with basis {f1,f2,f3,f4}.

Then V tensor W is a 12 dimensional vector space with basis

e1 tensor f1
e1 tensor f2
e1 tensor f3
e1 tensor f4
e2 tensor f1
e2 tensor f2
e2 tensor f3
e2 tensor f4
e3 tensor f1
e3 tensor f2
e3 tensor f3
e3 tensor f4

Furthermore, V direct sum W is a 7 dimensional vector space with basis:

(e1,0)
(e2,0)
(e3,0)
(0,f1)
(0,f2)
(0,f3)
(0,f4)

See the difference?

In Quantum Mechanics, spin is often considered. There is spin up and spin down. Sometimes the spin state is tensored with a wave function. Then, we have

wavefunction tensor spinstate=(wavefunction+,wavefunction-).

-Reverie

5. Oct 7, 2007

### captain

so for your quantum mechanics example you mean that when you take the tensor product between the spin state and wave function you get the probablity that it could be spin up or spin down, right?

6. Oct 7, 2007

### mathwonk

fortunately there is no such thing as an,...ayeeeee!

7. Oct 7, 2007

### captain

what do you mean by ayeeeeeeeeeee!

8. Oct 7, 2007

### CompuChip

Basically, when constructing states of a particle with position x and spin $\pm$, the tensor product sort of means that these two things are independent. So we need to specify both to fix a state, but they are really two different things.

9. Oct 7, 2007

### captain

I see where you are going

10. Oct 10, 2007

### llarsen

Examples

While it is true that the physical significance of a tensor depends on the particular application, this may be too abstract to be helpful. Here are some specific examples that may be helpful:

A tensor $$\sigma_{ij}$$ can be used to represent the stress (Force Per Area) that acts on a small element of mass. If you do a search on 'stress tensor' in google, the wikipedia artical comes up that explains this tensor and gives a picture. One dimension of the tensor (say the index 'i') represents surfaces of a small element of mass (or sides of the cube). The other dimension (index 'j') represents the set of forces that act on each of the surfaces. For each surface there can be one normal force (normal to the surface) and two shear forces (acting tangent to the surface). Note that although the cube has six sides, a tensor only represents three sides because the element is considered to be infinitesimal and the components of stress on one side of the cube are taken to be roughly equal and opposite those on the other side of the cube.

Another use of tensors is to define a relationship between two different types of tensors quantities. For example, the stress tensor $$\sigma_{ij}$$ repesents the Force per Area acting on a small element of mass. Another tensor called the strain tensor $$\epsilon_{kl}$$ represents the shape of a small element of mass (for example, a small element of mass may be stretched like a rectangle, or possibly sheared like a diamond). You would expect that applying forces to a body causes the shape to change. In other words, there should be some relation between the stress in the body $$\sigma_ {ij}$$ (local force per area) and the strain $$\epsilon_{kl}$$ (local change in shape). Often it is assumed that the relationship between the stress and strain is linear and a fourth order tensor $$E_{ijkl}$$ called the Modulus of Elasticity is determined which relates stress to strain. The relationship $$\sigma_{ij}=E_{ijkl} \epsilon_{kl}$$ is called Hooke's Law and relates the deformation of the mass to the associated stresses that the deformation causes. The Modulus of Elasticity is a material property that is determined empiracally (by strestching or twisting a the material of interest and seeing how much it deforms). For certain stress ranges, it turns out that a linear relationship between stress and strain (represented by the fourth order tensor $$E_{ijkl]$$) is pretty good. This is referred to as the plastic region for the material. Beyond this point, the relationship between stress and strain tends not to be linear. This region is referred to as the elastic region the Modulus of Elasticity is not valid. Note that Hooke's law is just a complicated version of the equation F=-kx for a spring which relates the stretch of a spring the the amount of force that is required to stretch the spring that far.

Another common use of tensors is to relate two coordinate systems and convert vectors (or other tensors) from one coordinate system to a new coordinate system. For example, suppose you have a two dimensional newtonian mechanics problem F=ma represented in terms of coordinates $$x_1$$ and $$x_2$$. The problem is rather complicated in terms of $$x_1$$ and $$x_2$$, so you choose a new set of coordinates $$y_1 (x_1 ,x_2 )$$ and $$y_2 (x_1 ,x_2 )$$ which (hopefully) simplifies the problem. The transformation to the new coordinate system is done via the chain rule. The chain rule is often represented in terms of the jacobian matrix which is multiplied by a vector to convert it from one set of coordinates to a new set of coordinates. Wikipedia has an article on the chain rule and on the jacobian which could give more clear interpretation in my opinion, but are worth looking at.

Last edited: Oct 10, 2007
11. Nov 29, 2007

### mrandersdk

when we are talking about states in quantum mechanics I like to think about it like this.

Let's say we have to independent particles A and B, which both can be in the states 0 and 1. Fisrt let's think about what we meenby independent. We mean that when asked if the system is in a state ex. |00> (both A and B in state 0), we would like to calculate the probability by

<00|AB> = <0|A><0|B>

next we meen that the particles can be in states independent of each other that is we have the possible states

|00>,|10>,|01>,|11>

So we start with two two dimensional statespaces lets say H_A and H_B (both with basis |0>,|1>), we now want to construct a new state space for the joint system that are again a vector space (in fact a Hilbertspace), with the porperties mensioned above already built in it.

So we make a new construction called the tensor product of H_A and H_B

H = H_A tensor H_B

it can be shown that the new space has a basis given by

|0> tensor |0>, |1> tensor |0>, |0> tensor |1> and |1> tensor |1>

ok so it seems like we have got or four states |00>,|10>,|01>,|11>, but is it reasonble to identify these states with the basis vectors of H. It is because when we have to hilbert spaces with inner product < | >, then the tensor product of these becomes an Hilbert space witch an inner product given by

(<A| tensor <B|)(|C> tensor |D>) = <A|B><C|D>

sp if we indentufy |1> tensor |0> with |10> and so on, we see that the tensor product reflect or desired properties and is therefor a good way to work with these independent statespaces. This can of cause be generalized to more spaces and more states in the spaces.

Hope it helped, i like to think about it like this way. By the way you use the word tensor often in physics and it is not always clear that it mean the same as in matematics, so sometimes it can be really confusing if you are reading a mathbokk and a physics book about tensors.

12. Dec 1, 2007

### rdt2

I introduce the tensor product to show that the dot product and the cross product of two vectors aren't arbitrary definitions but are linked. The tensor product of vector $$u_{i}$$ and vector $$v_{j}$$ is the (second order) tensor $$u_{i}v_{j}$$.

Operating on this with the Kronecker delta $$\delta_{ij}$$ gives the dot product or trace $$u_{i}v_{i}$$, while operating on it with the permutation (or alternating) symbol $$e_{ijk}$$ gives the cross product.

13. Dec 2, 2007

### mrandersdk

what do you mean by that ?

14. Dec 2, 2007

### mathwonk

riemann used tensors to describe curvature of spaces. since general relativity seems concerned with the curvature of space/time, thus would seem to be one application of tensors in physics. indeed einstein apparently used that mathematical tool for this purpose.

15. Dec 2, 2007

### Magister

What is the difference between the Cartesian product and the tensor product?

16. Dec 2, 2007

### mathwonk

the cartesian product is an operation on sets, while the tensor product is an operation on modules over a ring. the cartesiaN PROduct is used in the definition of the tensor product.

i.e. the tensor product of two abelian groups A,B is a quotient of the free abelian group generated by their cartesian product. take a look in my free webnotes in algebra for math 845, the last part.

or search here for old posts on this oft discussed topic. i myself have written dozens of responses here on this question.

17. Dec 2, 2007

### mrandersdk

a lot, for example if you take two three dimensional vector spaces H_1 and H_2 then the castesian product of these is 6D, but the tensor product of them is 9D if i'm not mistaken.

18. Dec 4, 2007

### mathwonk

technically the cartesian product is only a set, not a vector space, hence has no linear dimension. but it is correct that the cartesian product has a natural structure as vector space, namely the "direct sum" vector space, and the dimension of that space is the sum of the dimensions of the factors.

and the dimension of the tensor product is indeed the product of the dimensions.