2
$\begingroup$

I have read that in quantum mechanics, compound systems are constructed as tensor products.

But on page 177 of Griffith, for example, a two body wavefunction is introduced as Psi (x1,y1,z1,x2,y2,z2,t). (Six space dimensions plus time) This is clearly a direct sum, not a tensor product.

If this is correct, when do you use a direct sum and when do you use a tensor product?

Why isn’t the tensor product representing the two body system

Psi(x1x2,x1y2,x1z2,y1x2,y1y2,y1z2,z1x2,z1y2,z1z2,t) or something similar?
(nine tensor dimensions plus time). What am I misunderstanding?

And what are the tensor space dimensions?
Or how do these abstract dimensions relate to physical lab space dimensions?

|cite|improve this question
$\endgroup$
8
$\begingroup$

The composition law for quantum systems is always a tensor product. Your problem arises from a confusion over what the tensor product is applied to: you are trying to tensor product the spatial coordinates together, when it is in fact the basis vectors of the Hilbert space you should be tensoring together.

More formally, take two quantum systems A and B, with corresponding Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$. The state of the joint system (A & B) lives in the tensor product space $\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B$. This is a fancy-pants way of saying that the basis vectors of the combined Hilbert space look like $$|\phi\rangle = |A\rangle\otimes|B\rangle.$$ A general state can therefore be written as $$|\psi\rangle = \sum\limits_{jk} c_{jk} |A_j\rangle \otimes |B_k\rangle, $$ where the $\{|A_j\rangle\}$ and $\{|B_k\rangle\}$ span Hilbert spaces $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively, and $c_{jk}$ are complex numbers. As you would expect, if the two Hilbert spaces have dimension $d$ (if they are spanned by $d$ basis vectors each) then the combined Hilbert space has dimension $d^2$. A general $N$-body Hilbert space has dimension $d^N$.

If the two quantum systems are the position degrees of freedom of two particles, then the total wavefunction looks like $$\Psi = \sum\limits_{jk} c_{jk} \psi_j^{(1)}(x_1,y_1,z_1,t)\psi^{(2)}_k(x_2,y_2,z_2,t),$$ which clearly depends on 6 spatial coordinates plus time.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ But isn’t the space of wavefunctions on the config space, i.e. functions like $ \psi(x_1,y_1,z_1,x_2,y_2,z_2,t)$ bigger than your space of wavefunctions $\Psi = \sum\limits_{jk} c_{jk} \psi_j^{(1)}(x_1,y_1,z_1,t)\psi^{(2)}_k(x_2,y_2,z_2,t)$ ? $\endgroup$ – twistor59 Oct 26 '12 at 16:58
  • $\begingroup$ No, this form encapsulates every possible wavefunction. If it helps, I can always write any bipartite state of any kind in Dirac notation as $$|\Psi\rangle = \sum\limits_{j,k} c_{jk} |\psi^{(1)}_j\rangle \otimes |\psi^{(2)}_k\rangle. $$ If these states describe continuous degrees of freedom like positions, then I can project onto the basis kets as follows: $\endgroup$ – Mark Mitchison Oct 26 '12 at 17:43
  • 1
    $\begingroup$ No, I wasn't thinking of that. The question didn't specify that the particles were non-interacting. What if the w.f. depends on the separation distance between the particles? Can you represent it as a sum of products of wavefunctions of the "1" coordinates and wavefunctions of the "2" coordinates? $\endgroup$ – twistor59 Oct 26 '12 at 18:21
  • 2
    $\begingroup$ @Jim: Does this help? The tensor product $L^2(\mathbb{R}^3)\otimes L^2(\mathbb{R}^3)$ is isomorphic to $L^2(\mathbb{R}^3\times\mathbb{R}^3)$. And the cartesian product is $\mathbb{R}^3\times\mathbb{R}^3=\mathbb{R}^6$. $\endgroup$ – Raskolnikov Oct 26 '12 at 19:16
  • 1
    $\begingroup$ @JimGraber You are confusing the dimension of the configuration space of each particle, which is 3, with the dimension of the Hilbert space of each particle, which is infinite! There could in principle be an arbitrarily large number of $c_{jk}$, not necessarily 9 as you assume. I didn't talk about $d$ for the case of positions of particles because expressions like $\infty^2$ are confusing to everyone, especially me! $\endgroup$ – Mark Mitchison Oct 26 '12 at 19:36
2
$\begingroup$

A two-body state $|\Psi_{12}\rangle$ can be written as a tensor product $|\Psi_{12}\rangle = |\Psi_1\rangle \otimes |\Psi_2\rangle$ only when there are not correlations between both bodies: e.g., two separated non-interacting bodies. The same happens in classical mechanics, where the two-body state factorizes as $\rho_{12} = \rho_1 · \rho_2$ only in absence of correlations.

The Hilbert space is not the ordinary space and the quantum mechanical variables $(x_1, x_2)$ associated to the position basis $|x\rangle$ do not correspond to any point in ordinary space. Again the same happens in classical mechanics with the classical variables $(x_1, x_2)$ associated to the two-body configuration space.

The physical lab space dimensions are obtained in the one-particle limit.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Yes, I have read how Schrodinger himself struggled with this. I think I am beginning to understand it a little better now. $\endgroup$ – Jim Graber Oct 26 '12 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.