Understanding tensor product and direct sum

[dwd40physics]

TL;DR Summary

Understanding tensor product and direct sum of vector spaces in general and applied to total angular momentum.

Hi, I'm struggling with understanding the idea of tensor product and direct sum beyond the very basics. I know that direct sum of 2 vectors basically stacks one on top of another - I don't understand more than this . For tensor product I know that for a product of 2 matrices A and B the tensor product essential means each element in A gets multiplied by the matrix B.

What I don't understand in more formalism what the direct sum and tensor product do for: 1. vectors 2. matrices.

For example for the direct sum of e1 = (1,0) and f2 =(0,1,0) would be (1,0|0,1,0) -
how would we form (1,0,0,0,0) from the basis elements e1 and f2 ? - how does this make sense ?

In QM we covered the idea of total angular momentum where we saw that the total angular momentum of two angular momenta
J1and J2 is: JT = 1xJ2 + J1x1 - where x is a tensor product.

How do I make sense of this relation ?

 

[andrewkirk]

For example for the direct sum of e1 = (1,0) and f2 =(0,1,0) would be (1,0|0,1,0) -
how would we form (1,0,0,0,0) from the basis elements e1 and f2 ? - how does this make sense ?

$$(1,0,0,0,0)= 1\times e_2\oplus 0 \times f_2= 1\times (1,0)\oplus 0 \times(0,1,0)$$
where $\oplus$ is the direct sum operator and we use $\times$ here to denote left-multiplication of the vector by a scalar.

In QM we covered the idea of total angular momentum where we saw that the total angular momentum of two angular momenta
J1and J2 is: JT = 1xJ2 + J1x1 - where x is a tensor product.

How do I make sense of this relation ?

For this one you'll need to provide more context and set out the formulas more precisely - perhaps with a picture from your text. We can't work out from this what was intended.

 

[gentzen]

Do you understand why going from $\phi(x)$ for a single particle to $\phi(x_1,x_2)$ for two particles (with $x,x_1,x_2\in\mathbb R^3$) translates to a tensor product of the Hilbert space with itself (for two particles)? If yes, then a simple way to translate this understanding to spin is to replace $x,x_1,x_2\in\mathbb R^3$ by $\sigma,\sigma_1,\sigma_2\in\{\uparrow,\downarrow\}$. So $\phi(\sigma)$ are simply the two complex numbers $\phi(\uparrow)$ and $\phi(\downarrow)$, and $\phi(\sigma_1,\sigma_2)$ then naturally gives you four complex numbers. And if you have $n$-spin qubits, then by the same logic you get $2^n$ complex numbers.

 

[dwd40physics]

I don not understand the two particle case - do you have some recommended reading for where I can go to find out more. Thanks

 

[PeroK]

I don not understand the two particle case - do you have some recommended reading for where I can go to find out more. Thanks

Have you tried these:

https://en.wikipedia.org/wiki/Tensor_product

https://www.math3ma.com/blog/the-tensor-product-demystified

In terms of the physics: if we have a system of two spin $\frac 1 2$ particles, then we find (experimentally, say) that the following measurement outcomes are possible:

$S^2 = 0, \ S_z = 0$
$S^2 = 1, \ S_z = 0 \ \text{or} \ \pm 1$

And this is what you get if you use the tensor product of two spin 1/2 Hilbert spaces and calculate the eigenvalues etc.

 

[gentzen]

I don not understand the two particle case - do you have some recommended reading for where I can go to find out more. Thanks

If you tell me what exactly you don't understand about the two particle case, then maybe I could suggest something specific.

Are you familiar with the (quantum) 1D harmonic oscillator? Can you see how a 2D harmonic oscillator corresponds to two 1D harmonic oscillator? (And a 3D to three 1D...?)

Or if you have two events, whose indiviual probabilities would be given by $p_1(e_1)$ and $p_2(e_2)$. Do you understand why their joint probability is given by a function $p_{12}(e_1,e_2)$? (Of course, if they are independent, then you have simply $p_{12}(e_1,e_2)=p_1(e_1)p_2(e_2)$. But in general $p_1(e_1)=\sum_{e_2}p_{12}(e_1,e_2)$ and $p_2(e_2)=\sum_{e_1}p_{12}(e_1,e_2)$ is all you can say.)

 

[dwd40physics]

$$(1,0,0,0,0)= 1\times e_2\oplus 0 \times f_2= 1\times (1,0)\oplus 0 \times(0,1,0)$$
where $\oplus$ is the direct sum operator and we use $\times$ here to denote left-multiplication of the vector by a scalar.

For this one you'll need to provide more context and set out the formulas more precisely - perhaps with a picture from your text. We can't work out from this what was intended.

Here's a screenshot - don't understand going from 3.322 to 3.323. (Sakurai)

 

[dwd40physics]

If you tell me what exactly you don't understand about the two particle case, then maybe I could suggest something specific.

Are you familiar with the (quantum) 1D harmonic oscillator? Can you see how a 2D harmonic oscillator corresponds to two 1D harmonic oscillator? (And a 3D to three 1D...?)

Or if you have two events, whose indiviual probabilities would be given by $p_1(e_1)$ and $p_2(e_2)$. Do you understand why their joint probability is given by a function $p_{12}(e_1,e_2)$? (Of course, if they are independent, then you have simply $p_{12}(e_1,e_2)=p_1(e_1)p_2(e_2)$. But in general $p_1(e_1)=\sum_{e_2}p_{12}(e_1,e_2)$ and $p_2(e_2)=\sum_{e_1}p_{12}(e_1,e_2)$ is all you can say.)

I have come across the basics of two particle systems e.g. parity operator exchanging particles 1 and 2 but not much more depth than that not the tensor product

 

[dwd40physics]

$$(1,0,0,0,0)= 1\times e_2\oplus 0 \times f_2= 1\times (1,0)\oplus 0 \times(0,1,0)$$
where $\oplus$ is the direct sum operator and we use $\times$ here to denote left-multiplication of the vector by a scalar.

I don't understand how you get this ? What happens to the basis of R^2 and R^3 when you direct sum, how do we construct the basis of R^5 from R^3 and R^2 ?

 

[PeroK]

Here's a screenshot - don't understand going from 3.322 to 3.323. (Sakurai)
View attachment 321987

That's another example of what I tried to communicate above. We know that there is some sort of "addition" of angular momentum, but the experimental results (if nothing else) lead us to use the tensor product of Hilbert spaces, kets and operators; rather than the simple direct sum.

Sakurai doesn't give a justification in that particular extract you quote, but he may do elsewhere.

 

[gentzen]

Here's a screenshot - don't understand going from 3.322 to 3.323. (Sakurai)

With my suggestion, you would read $\mathbf{J = L + S}$ as $\mathbf{J}(\sigma_1,\sigma_2,\sigma'_1,\sigma'_2) = (\mathbf{L + S})(\sigma_1,\sigma_2,\sigma'_1,\sigma'_2)$ and then translate $\mathbf{J} = \mathbf{L}\otimes 1 + 1 \otimes \mathbf{S}$ as $\mathbf{J}(\sigma_1,\sigma_2,\sigma'_1,\sigma'_2) = \mathbf{L}(\sigma_1,\sigma'_1)\cdot \delta_{\sigma_2,\sigma'_2} +\delta_{\sigma_1,\sigma'_1} \cdot \mathbf{S}(\sigma_2,\sigma'_2)$. ($\delta$ here is a simple Kronecker delta.) If you want, you can even "define" $1(\sigma_1,\sigma'_1):=\delta_{\sigma_1,\sigma'_1}$ and $1(\sigma_2,\sigma'_2):=\delta_{\sigma_2,\sigma'_2}$, and then write it as $\mathbf{J}(\sigma_1,\sigma_2,\sigma'_1,\sigma'_2) = \mathbf{L}(\sigma_1,\sigma'_1)\cdot 1(\sigma_2,\sigma'_2) + 1(\sigma_1,\sigma'_1) \cdot \mathbf{S}(\sigma_2,\sigma'_2)$.

Of course, you also have to specify how such an operator (~"matrix") transforms a wavefunction (~"vector"). Say for example as $(\mathbf{J}\phi)(\sigma_1,\sigma_2):=\sum_{\sigma'_1,\sigma'_2}\mathbf{J}(\sigma_1,\sigma_2,\sigma'_1,\sigma'_2)\phi(\sigma'_1,\sigma'_2)$.

 

[andrewkirk]

I don't understand how you get this ? What happens to the basis of R^2 and R^3 when you direct sum, how do we construct the basis of R^5 from R^3 and R^2 ?

For any two vector spaces $V_a, V_b$ with corresponding bases $B_a,B_b$ and zero vectors $\vec 0_a, \vec 0_b$, the set of vectors
$$\{\vec u \oplus \vec 0_b\ |\ \vec u\in V_a\} \cup
\{\vec 0_a \oplus \vec v \ |\ \vec v\in V_b\} $$forms a basis for $V_a\oplus V_b$.

In our case, with $V_a=\mathbb R^3, V_b=\mathbb R^2$, we choose bases $B_a = \{(1,0,0), (0,1,0), (0,0,1)\}, B_b=\{(1,0), (0,1)\}$
Then a basis for $\mathbb R^5 = \mathbb R^3\oplus\mathbb R^2$ is
$$\{b\oplus (0,0)\ |\ b\in B_3\}\cup
\{(0,0,0)\oplus b \ |\ b\in B_2\}$$which has elements
$$(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1)$$with the first three coming from the first of the two sets in the union and the last two coming from the second of the two sets.