Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider a system of N quantum mechanical particles described on Hilbert spaces $\mathcal{H}_1,...,\mathcal{H}_N$ and with Hamiltonians $H_1,...,H_N$. The Hamiltonian operator $H_1$ acts on the Hilbert space $\mathcal{H}_1$ etc. These particles are non-interacting if the Hamiltonian of the system can be written as the sum of the Hamiltonians of each particle: $$H=H_1+...+H_N$$

Why the Hilbert space of the combined system of these non-interacting particles is considered to be a tensor product of the Hilbert spaces of each particle?

$$\mathcal{H}=\mathcal{H}_1\otimes...\otimes \mathcal{H}_N$$

For interacting particles is it a tensor product too?

share|improve this question
add comment

migrated from math.stackexchange.com Dec 6 '11 at 16:48

This question came from our site for people studying math at any level and professionals in related fields.

3 Answers

The existing answer does not directly address the question, « ¿Why? »

For simplicity, consider two distinguishable particles each of whose wave functions would be described by $\cal H_i = L^2(R)$. In order to describe the states of them both considered as a system, we must have at least all the following possibilities:

$\psi_1$, the wave function describing the first particle (say a pion) and $\psi_2$, (say, a deuteron). So the state space we want must have at least all ordered pairs $(\psi_1,\psi_2)$ in it.

But on the other hand, it must be a Hilbert space. It is a fundamental principle of Quantum Mechanics that physical states can be superposed and you get a new, possible, physical state. So the state space we want must have all formal linear combinations of these ordered pairs in it.

Nevertheless, some of these formal combinations will be physically the same state, for example $(\psi_1 + \phi_1, \psi_2)$ must be the same state of the combined system as $(\psi_1,\psi_2) + (\phi_1,\psi_2)$. Grunging along systematically like this you wind up with the generators and relations for the tensor product of $\cal H_1$ and $\cal H_2$.

Now, there is also a more intuitive way to see this. If $\{ v_i\}$ is an orthonormal basis for the space of the electron, say by energy levels, then since the other particle could be in any state $\psi_2$ independently of what the electron is doing, we have a separate copy of its state space for each $i$. And by superoposition, we must be able to take linear combinations of these different copies since the electron might not have been in an energy eigentstate, but might have been in some superoposition. So we get $$\sum_i L^2{R_2^3}$$ where I put the subscript 2 to show it is the spatial position of the deuteron. But again, this is just the tensor product.

The third way to see this is the most intuitive and uses the functorial properties of the tensor product. For every wave function $\psi_1(x,y,z)$ for the electron, it describes the probability amplitude for the electron to be found at the point $(x,y,z)$. Since the particles are distinguishable, each of these possibilities can be associated with a possible state of the deuteron. So to describe the combined system, we should replace $\psi_1$ by a function which, to each point $(x,y,z)$, attaches a possible state of the deuteron, with the probability amplitude of the electron's being at that point. But once again, this is the tensor product since $L^2(V) \otimes L^2(W*)$ is equal to $L^2(L(W,V))$, where the second $L$ is the usual notation for linear maps from one Banach space to another.

share|improve this answer
    
@Jo I've have no problem with your solution, but I have no idea what the word "Grunging" means. It's hard enough trying to remember all the definitions for quantum physics without resorting to looking up obscure Indie bands from Seattle. This comment is meant in jest, but really what the heck is grunging? –  metzgeer Feb 5 '12 at 1:37
    
I just made it up: it means working out the tedious details as you go, backtracking when you make a mistake, and piling up eraser shreds all over the tables and chairs.... –  joseph f. johnson Feb 5 '12 at 1:47
add comment

In classical mechanics, the phase space of a single particle consists of pairs $(q,p)$ of the position $q$ and momentum $p$ of the particle. In quantum mechanics, this is replaced by dropping either position or momentum (depending on the representation used), and working with wave functions that are functions $\psi(q)$ in the position representation or $\psi(p)$ in the momentum representation. The question has nothing to do with whether or not the particles are interacting.

In classical mechanics, $N$-particle phase space consists of the list $(q_1,p_1,\dots,q_N,p_N)$ of the individual positions $q_j$ and momenta $p_j$ of the particles. In quantum mechanics, this is again replaced by dropping either the positions or the momenta, and working with wave functions that are functions $\psi(q_1,\dots,q_N)$ in the position representation or $\psi(p_1,\dots,p_N)$ in the momentum representation. Abstractly, the space of functions $\psi(q_1,\dots,q_N)$ is (ignoring topological details) just the tensor product of $N$ spaces of single-particle wave functions $\psi(q_j)$.

Thus the tensor product is the natural formulation of multiparticle quantum mechanics.

share|improve this answer
add comment

If the particles never did interact in the past and won't ever interact in the future, then it indeed makes not much sense to model the system as a tensor product. Keep in mind that it does not matter how long ago the interaction happened as long as the particles are not otherwise disturbed - that's one of the points of the EPR 'paradoxon'.

However, care must be taken when dealing with identical particles: They are excitations of the same underlying field and should always be considered as having interacted.

In fact, the state space of identical particles is not the tensor product, but its symmetrization (in case of bosons) or anti-symmetrization (in case of fermions), ie only entangled states are considered.

These are just my thoughts and by no means authorative - anyone who deals with such issues on a more hands-on basis probably has a better grasp of the underlying issuess...

Considering the unexplained downvote, a more technical answer to the problem:

In non-relativistic Quantum mechanics, the evolution of a system in governed by inital state and Schrödinger's equation.

If we start out in an un-entagled state and the Hamiltonian does not contain interactions between the different sub-systems, we will stay in an unentangled state and we can consider the sytems independantly.

More formally, let our inital state be

$$\psi = \psi_1\otimes...\otimes\psi_n$$

and the Hamiltonian given by

$$H = \tilde H_1 + ... + \tilde H_n$$ $$= H_1\otimes1\otimes...\otimes1 + 1\otimes H_2\otimes1\otimes....\otimes1 + ... + 1\otimes...\otimes1\otimes H_n$$

Then the state after time $\Delta t$ is given by

$$ \mathrm{e}^{-iH\Delta t}\psi = \mathrm{e}^{-i\tilde H_1\Delta t}\cdot...\cdot\mathrm{e}^{-i\tilde H_n\Delta t}\psi_1\otimes...\otimes\psi_n = (\mathrm{e}^{-iH_1\Delta t}\psi_1)\otimes...\otimes(\mathrm{e}^{-iH_n\Delta t}\psi_n) $$

where the first equality follows from $[\tilde H_i,\tilde H_j] = 0$.

share|improve this answer
    
please don't downvote without leaving a comment - we're all here to learn, but one can only do that if you actually tell us what's wrong with my answer... –  Christoph Dec 6 '11 at 17:34
    
I didn't do the downvote, but I think there would be objection to the first paragraph - taking the tensor prod. of single particle Hilbert spaces to form the multiparticle version is often included amongst the postulates of QM. Also re. second para: identical particles wouldn't be considered as interacting if there's no interaction Hamiltonian. –  twistor59 Dec 6 '11 at 17:57
    
@twistor59: see my edit - if we start in an unentagled state and the Hamiltonian doesn't contain an interaction term we stay unentangled and we can consider the systems seperately; otherwise, if you take QM seriosly, you couldn't do any physics at all as there actually only is a single wavefunction $\psi$ describing the whole universe –  Christoph Dec 6 '11 at 18:00
    
@twistor59: you're correct that a better wording for the identical particle case would have been entangled, as that's what the formalism actually does... –  Christoph Dec 6 '11 at 18:03
    
Entangled particles are non-interacting. –  Andyk Dec 6 '11 at 18:14
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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