Question regarding a two particle system

[armis]

From Griffiths' "Introduction to quantum mechanics"

"Suppose particle 1 is in the (one-particle) state $$\psi_{a}(r)$$, and particle 2 is in the state $$\psi_{b}(r)$$. In that case $$\psi(r_{1},r_{2})$$ is a simple product:

$$\psi(r_{1},r_{2}) = \psi_{a}(r_{1})\psi_{b}(r_{2})$$"

Why the wave function of a two particle system is simple product and not a linear combination or some other combination of the two states?

 

[Fredrik]

It's written as a product, but you should think of it as an ordered pair. Suppose that you had something described by a vector in a 3-dimensional vector space and something else described by a vector in a 2-dimensional vector space. If you wanted to describe both at the same time, you could use a vector in a 5-dimensional vector space constructed from two vector spaces you started with.

 

[armis]

Thanks
But I still don't get it :(. I guess I should review the FORMALISM chapter again and then return to your post

 

[armis]

I still don't understand what you mean. Could you explain it in more detail?

 

[Fredrik]

Unfortunately I don't know all the details, so I can't explain it as well as most things I find myself explaining in these forums. What I can tell you is this:

$\psi_a$ is a vector in a Hilbert space H_a and $\psi_b$ is a vector in a Hilbert space H_b. These two Hilbert spaces can be combined to form a new Hilbert space in several different ways. The method that's used to construct a space of many-particle states from two or more spaces of one-particle states is called a tensor product. $\psi$ is a vector in the tensor product space $H=H_a\otimes H_b$.

As you can see here, I've been having some difficulties understanding this concept myself. I think I'll make an new effort to understand this soon, because I hate not understanding concepts like this.

 

[Fredrik]

Most of the difficulties with this concept come from the fact that we're dealing with infinite-dimensional Hilbert spaces. If we were dealing with finite-dimensional spaces, it would be much easier. I think this would be a valid way to think about it: (I hope someone will let us know if I'm wrong).

Pretend that H_a is 2-dimensional and H_b 3-dimensional. Then we can write

$$\psi_a=\begin{pmatrix}a_1 \\ a_2\end{pmatrix}$$

$$\psi_b=\begin{pmatrix}b_1 \\ b_2\\ b_3\end{pmatrix}$$

and

$$\psi=\begin{pmatrix}a_1 \\ a_2\\b_1 \\ b_2\\ b_3\end{pmatrix}$$

An operator X on H_a would be a 2x2 matrix, and an operator Y on H_b would be a 3x3 matrix. X and Y can be combined into an operator on the tensor product space:

$$X=\begin{pmatrix}* & *\\ * & *\end{pmatrix}$$

$$Y=\begin{pmatrix}* & * & *\\ * & * & *\\ * & * & *\end{pmatrix}$$

$$X\otimes Y=\begin{pmatrix}* & * & 0 & 0 & 0\\ * & * & 0 & 0 & 0\\ 0 & 0 & * & * & *\\ 0 & 0 & * & * & *\\ 0 & 0 & * & * & *\end{pmatrix}$$

The asterisks represent the numbers that define X and Y. The block-diagonal form of $X\otimes Y$ guarantees that the a's and the b's won't be mixed when this operator acts on $\psi$. If the particles represented by H_a don't interact with the ones represented by H_b, the time evolution operator will be block-diagonal. If they interact, it won't be.

 

[Vanadium 50]

I think this is one of those places where it's easiest to start with the answer. Suppose particle 1 can be in 3 states and particle 2 can be in 4. How many states can the pair be in? That should tell you that we probably want to be multiplying our fundamental entities together.

 

[Fredrik]

Hm, if what George said here is right, then what I said in #6 is wrong. What he said implies that the tensor product space of a 2-dimensional space and a 3-dimensional space is 6-dimensional, but the space I described in #6 is 5-dimensional.

If, as in relativity, $V$ and $W$ are both finite-dimensional spaces, then $V \otimes W$ is (naturally) isomorphic to the vector space of bilinear maps from $V* \times W*$ to $\mathbb{R}$. For infinite-dimensional spaces $V \otimes W$ is isomorphic to a proper subspace of bilinear maps from $V* \times W*$ to $\mathbb{R}$. Therefore, this space of bilinear mapping is often taken to be the tensor product space.

If I had to make a bet right now, I'd bet that George got it right and I got it wrong.

 

[0xDEADBEEF]

Most of the difficulties with this concept come from the fact that we're dealing with infinite-dimensional Hilbert spaces. If we were dealing with finite-dimensional spaces, it would be much easier. I think this would be a valid way to think about it: (I hope someone will let us know if I'm wrong).

Pretend that H_a is 2-dimensional and H_b 3-dimensional. Then we can write

$$\psi_a=\begin{pmatrix}a_1 \\ a_2\end{pmatrix}$$

$$\psi_b=\begin{pmatrix}b_1 \\ b_2\\ b_3\end{pmatrix}$$

and

$$\psi=\begin{pmatrix}a_1 \\ a_2\\b_1 \\ b_2\\ b_3\end{pmatrix}$$

This is not the vector product of these two spaces. I am not a big fan of how these products are discussed but I'll try to give an insight with a bit of personal bastard notation.
The $$\otimes$$-product of two vectors is first and foremost some element we don't want to understand defined by two vectors, so if we write it in a tuple way it would look like this:
$$\psi_a \otimes \psi_b = \begin{pmatrix}\begin{pmatrix}a_1 \\ a_2\end{pmatrix}\\ \begin{pmatrix} b_1 \\ b_2\\ b_3\end{pmatrix}\end{pmatrix}$$
"How does this differ from $$\psi=\begin{pmatrix}a_1 \\ a_2\\b_1 \\ b_2\\ b_3\end{pmatrix}$$?",
you will ask. In the way how we define the scalar product (the numbers are simply different vectors, but there was no space for labels left):
$$\left< \psi_a^1 \otimes \psi_b^1, \psi_a^2 \otimes \psi_b^2 \right> = \left< \psi_a^1 , \psi_a^2 \right> \cdot \left< \psi_b^1, \psi_b^2 \right>$$
The way you defined psi implies an addition of the two products and not a multiplication. To make sure that we don't confuse things, we never write the product as this double tuple as I just did, but we leave it as $$\psi_a \otimes \psi_b$$. The product space has the following base vectors (I'll use the old element names as the base vector names I hope it doesn't confuse you $$a_1 = \begin{pmatrix}1\\ 0 \end{pmatrix}$$...):
$$\phi_1 = a_1\otimes b_1$$
$$\phi_2 = a_1\otimes b_2$$
$$\phi_3 = a_1\otimes b_3$$
$$\phi_4 = a_2\otimes b_1$$
$$\phi_5 = a_2\otimes b_2$$
$$\phi_6 = a_2\otimes b_3$$
So each combination of base states is a new base vector, with our new scalar product. You can understand how your idea brakes down when you consider a particle to be in this state:
$$\psi_{\mathrm{bad}}=\begin{pmatrix}1 \\ 0\\0 \\ 0\\ 0\end{pmatrix}$$
So we know the state of system a. What is the state of system b? Did it disappear? This is also why the product of two one dimensional states is not two dimensional. The only state that we have is one where both systems are in the only state allowed.
The nice thing is that we don't need to be confused as physicists at all. That's the job of mathematicians. We just imagine the vectors to be functions and everything automagically turns out right. So $$\psi_a \otimes \psi_2 = \psi(\mathbf{x}_{\mathrm{particle1}})\cdot \psi(\mathbf{x}_{\mathrm{particle2}})$$
And the scalar product is simply $$\int \int d\mathbf{x}_{\mathrm{particle1}}\,d\mathbf{x}_{\mathrm{particle2}}$$

 

[Fredrik]

Thanks for taking the time to explain. I actually realized this myself and came here to make a post about it, but you beat me to it. I realized that what I said (in #6) is wrong when I started thinking about how the map $\otimes$ that takes $(\psi_a,\psi_b)$ to $\psi=\psi_a\otimes\psi_b$ is supposed to be bilinear. My construction in #6 is very obviously not bilinear.

 

[armis]

Oh well, this is still over my head . But big thanks for the time and explanations. I'll certainly return to this thread once better armed with algebra

 

[malawi_glenn]

What if we look at the wavefunctions in the was as we look at probabilities?

Lets compare the situation $\psi _1 (x_1) \psi _2 (x_2) = \Psi (\vec{x})$ with the "two dice system".

The probability that you get, let's say 3 with dice number 1 and a 4 with dice number 2, is given by the product:

Probability_dice1 (x_1 = 3) * Probability_dice2(x_2 = 4) = Probability(x_1=3 ,x_2=4), Right?

Now, look at the 2 particle system $\Psi (\vec{x})$, the probability (density) that the particle number 1 is at position $x_1 = x_1'$ is $\psi _1 (x_1')$, and that particle 2 is at $x_2 = x_2^'$, is given by: $\psi _2 (x_2^')$. So the total probability for having particle1 at x_1' and particle2 at x_2' is simply:

$\psi _1 (x_1') \psi _2 (x_2') = \Psi (x_1', x_2')$