I A basis of a composite system

Lebnm

If I have a composite system, like a two particle system, for exemple, I can construct my Hilbert space as the tensor product of the hilbert spaces of these particles, and, if $\{|A;m \rangle \}$ and $\{|B;n \rangle \}$ are basis in these hilbert spaces, a basis in the total hilbert space is $\{|A;m \rangle \otimes |B;n \rangle \}$. But, to it make sense, shuoldn't the hilbert spaces of the subsystems represent non-interacting systems? For exemple, one of the most know exemples of composite system is the Hydrogen atom, where the hamiltonian is $$H = \frac{p_{p}^{2}}{2m_{p}} + \frac{p_{e}^{2}}{2m_{e}} + V(r),$$where $r$ is the relative position of the proton and the electron and $V$ is the Coulomb potential. However, we can write this in the rest of C.M. $$H = \frac{P^{2}}{2M} + \frac{p^{2}}{2\mu} + V(r).$$This hamiltonian is equivalent to a system with a free particle and a particle subject to an external potential, but these particles do not interact with each other. So, in this case, I can write a basis of the total system using the tensor products $|P \rangle \otimes |n, l, m \rangle$.

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Truecrimson

I'm not sure I understand your question, but not every state in $U \otimes V$ is of the form $|u\rangle \otimes |v\rangle, |u\rangle \in U, |v\rangle \in V$. Only your basis vectors are, but the choice of bases is arbitrary and has no physical meaning.

• vanhees71

Lebnm

I have a doubt about the definition of the kets of a composite system. If I understood correctly, when we have a composite system, that is, a system made of any number of subsystems (like two particle or a particle with spin), the Hilbert space $H$ of the total system will be the tensor product of the hilbert spaces of the subsystems $\bigotimes _{i=1}^{N} H^{(i)}$, and a basis in $H$ wil be a tensor product of the vectores of the basis of $H^{(i)}$. For exemple, consider two non-identical spin 1/2 particles. In this case, a basis in $H$ is $$\{ |+,+ \rangle , |+,- \rangle , |-,+ \rangle , |-,- \rangle \}.$$ If the system is in the state $|+,- \rangle \equiv |+ \rangle \otimes |- \rangle$, particle 1 have spin +1/2 and particle 2 spin -1/2. This make sense to me because the spin of particle one dosen't affect the spin of particle two. But, if I have two particles that interact, to me is strange construct the basis using states of isoleted particles. For exemple, if $|a \rangle$ is a state of $H^{(a)}$ and $|b \rangle$ a state of $H^{(b)}$, a state of the composite system can be $|a \rangle \otimes |b \rangle$, but it's strange, because $|a \rangle$ is one state of system $A$ isolated. The interaction with system $B$ will change the states of $A$, and vice versa, so we can't use neither the basis $\{|a \rangle \}$ nor $\{ |b \rangle \}$ to construct a basis in $H$.

vanhees71

Gold Member
Well, now you have to distinguish one-particle, two-particle, ... $N$-particle operators making up your Hamiltonian. For your example of a two-spin system, your Hamiltonian may look like this
$$\hat{H}=\hat{H}_{1}^{(1)} \otimes \hat{1} + \hat{1} \otimes \hat{H}_{2}^{(1)} + \hat{H}^{(2)}.$$
Here $\hat{H}_1^{(1)}$ and $\hat{H}_2^{(1)}$ are operators defined in $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively. Acting with them on product states, keeps them as product states, and thus they do not describe any interaction, while $\hat{H}^{(2)}$ is a genuine two-body operator which maps a product state in a superposition of product states and thus describes interactions.

Lebnm

It was not exactly my question. Let me consider another exemple: A hydrogen atom. The hamiltonian is $$\hat{H} = \frac{\hat{P}_{p}}{2 m_{p}} + \frac{\hat{P}_{e}}{2 m_{e}} + V(\hat{r})$$If I take $\mathcal{H}$ to be $\mathcal{H}_{p} \otimes \mathcal{H}_{e}$, where $\mathcal{H}_{e(p)}$ is the hilbert space of a isolated electron (proton), a basis in $\mathcal{H}$ will be $\{ |P_{p} \rangle \otimes |P_{e} \rangle \}$, but this is obviously incorrect. On the other hand, I can write the hamiltonian as $$\hat{H} = \frac{\hat{P}}{2 M} + \frac{\hat{p}}{2 m} + V(\hat{r}),$$ where $P$ and $M$ are the momentum and mass of C.M., $p$ is the relative momentum and $m$ the reduced mass. This hamiltonian is the hamiltonian of a system composed by a free particle of mass $M$ and momentum $P$ and another particle of momentum $p$ and mass $m$ subject to a central potential $V$, but these two particle don't interact between themselves. In this case, I can write $\mathcal{H} = \mathcal{H}_{1} \otimes \mathcal{H}_{2}$, and a basis in $\mathcal{H}$ is $\{ |P \rangle \otimes |n,l,m \rangle \}$, and we know this is correct.
So, I can conclude that, in a composite system, I can write the Hilbert space as $\mathcal{H} = \mathcal{H}_{1} \otimes \mathcal{H}_{2} \otimes ...$, but I have to choose the hilbert spaces $\mathcal{H}_{i}$ such that the total hamiltonian can be written as $\hat{H} = \sum _{i} \hat{H}_{i}$, where $\hat{H}_{i}$ is the hamiltonian associated with $\mathcal{H}_{i}$. In this case, a basis in $\mathcal{H}$ is $\{ |a_{1} \rangle \otimes |b_{2} \rangle \otimes ... \}$, where $\{|a_{i} \rangle \}$ is a basis in $\mathcal{H}_{i}$.
Is this correct?

stevendaryl

Staff Emeritus
It was not exactly my question. Let me consider another exemple: A hydrogen atom. The hamiltonian is $$\hat{H} = \frac{\hat{P}_{p}}{2 m_{p}} + \frac{\hat{P}_{e}}{2 m_{e}} + V(\hat{r})$$If I take $\mathcal{H}$ to be $\mathcal{H}_{p} \otimes \mathcal{H}_{e}$, where $\mathcal{H}_{e(p)}$ is the hilbert space of a isolated electron (proton), a basis in $\mathcal{H}$ will be $\{ |P_{p} \rangle \otimes |P_{e} \rangle \}$, but this is obviously incorrect. On the other hand, I can write the hamiltonian as $$\hat{H} = \frac{\hat{P}}{2 M} + \frac{\hat{p}}{2 m} + V(\hat{r}),$$ where $P$ and $M$ are the momentum and mass of C.M., $p$ is the relative momentum and $m$ the reduced mass. This hamiltonian is the hamiltonian of a system composed by a free particle of mass $M$ and momentum $P$ and another particle of momentum $p$ and mass $m$ subject to a central potential $V$, but these two particle don't interact between themselves. In this case, I can write $\mathcal{H} = \mathcal{H}_{1} \otimes \mathcal{H}_{2}$, and a basis in $\mathcal{H}$ is $\{ |P \rangle \otimes |n,l,m \rangle \}$, and we know this is correct.
So, I can conclude that, in a composite system, I can write the Hilbert space as $\mathcal{H} = \mathcal{H}_{1} \otimes \mathcal{H}_{2} \otimes ...$, but I have to choose the hilbert spaces $\mathcal{H}_{i}$ such that the total hamiltonian can be written as $\hat{H} = \sum _{i} \hat{H}_{i}$, where $\hat{H}_{i}$ is the hamiltonian associated with $\mathcal{H}_{i}$. In this case, a basis in $\mathcal{H}$ is $\{ |a_{1} \rangle \otimes |b_{2} \rangle \otimes ... \}$, where $\{|a_{i} \rangle \}$ is a basis in $\mathcal{H}_{i}$.
Is this correct?
If I understand your question, if for each subsystem $i$ you have a complete basis $|a_{in}\rangle$, then you can (usually) come up with a complete basis for the composite system to be all those states of the form $|a_{1n_1}\rangle \otimes |a_{2n_2}\rangle \otimes ...$. But there can be other complete bases for the composite system that don't have that form.

You give the example: For the electron-proton system, a complete basis can be written as $|\overrightarrow{P},\overrightarrow{p}\rangle$, where $\overrightarrow{P}$ is the momentum of the center of mass, and $\overrightarrow{p}$ is the momentum associated with the reduced mass. This is not a product of the one-particle basis.

Lebnm

If we have a system composed by two interacting particles, it's writeen in many textbooks (like "Quantum Physics", Le Bellac) that we can write the total Hilbert space $\mathcal{H}$ as the tensor product of he Hilbert spaces of the particles isoleted ($\mathcal{H}_{1}$ and $\mathcal{H}_{2}$), and, if $\{|a_{1} \rangle \}$ is a basis in $\mathcal{H}_{1}$ and $\{ | b _{2} \rangle \}$ a basis in $\mathcal{H}_{2}$, a basis in $\mathcal{H}$ is $\{ |a_{1} \rangle \otimes |b_{2} \rangle \}$. But it dosen't make sense to me if the systems are interacting. For exemple, if we consider a Hydrogen atom, and choose $\mathcal{H}_{1(2)}$ as the Hilbert space of the elecron (proton) isoleted, the electron (positron) isolated is only a free particle, so a basis in $\mathcal{H}_{1(2)}$ is simply formed by the eigenkets of its momentum operator : $\{ | p_{e(p)} \rangle \}$. So, a basis to the Hydrogen atom should be $\{ | p_{e} \rangle \otimes |p_{p} \rangle \}$, but this is obviously wrong, I can't get the wavefunctions of H-atom from this ket, right?
So, I think I should choose the Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ such that they represent independent systems. In the H atom, I could write the hamiltonian in the rest of the center of mass and take $\mathcal{H}_{CM}$ and $\mathcal{H}_{rel}$, and a basis would be $\{ | P_{CM} \rangle \otimes |n, l, m \rangle \}$.

PeterDonis

Mentor
it dosen't make sense to me if the systems are interacting
The two particles interacting doesn't change the Hilbert space, it changes the Hamiltonian. You can still use the same basis built from free particle eigenstates with an interacting Hamiltonian; it will just make the Hamiltonian look more complicated, whereas choosing a different basis for the two-particle Hilbert space makes an interacting Hamiltonian look simpler. That's basically what you are discovering.

Truecrimson

a basis to the Hydrogen atom should be $\{ | p_{e} \rangle \otimes |p_{p} \rangle \}$, but this is obviously wrong, I can't get the wavefunctions of H-atom from this ket, right?
Yes, you can.

You have made a few threads to ask basically the same question. The answer will still be the same: a choice of bases for a vector space is arbitrary and has no physical meaning, and as such cannot effect physically relevant notions such as the locality of a Hamiltonian or the energy levels. The confusion is similar to asking "is the state $|\psi \rangle$ a superposition?" The question has no meaning because it depends on the choice of bases; every state is a superposition of some other states.

To reiterate maybe in different words, the tensor product $\mathcal{H}_1 \otimes \mathcal{H}_2$ is still a vector space. The set of all unitary transformations (equivalently, Hamiltonians) on $\mathcal{H}_1 \otimes \mathcal{H}_2$ includes interacting ones (as well as non-interacting ones). So the tensor product construction is sufficient to treat any interaction between multiple systems.

For 2 two-level systems, you can show that any product state (like $|+,+\rangle$, using your notations from the other thread) can be written as a superposition of four entangled states (the "Bell basis") $$\{ |+,+\rangle + |-,-\rangle, |+,+\rangle - |-,-\rangle, |+,-\rangle + |-,+\rangle, |+,-\rangle - |-,+\rangle \},$$
themselves I wrote in the form of superpositions of product states.

vanhees71

Gold Member
Indeed, you can in principle describe the hydrogen energy eigenfunctions in terms of the product basis, but it's awfully complicated, because due to the interaction the energy eigenfunctions are entangled states, and it's way easier to use the appropriate standard basis, which is a product state of the center-mass motion and relative motion. A very illuminating AJP paper on this usually not much discussed issue is

• Truecrimson

DrClaude

Mentor
You have made a few threads to ask basically the same question.
Indeed. These threads have now been merged into this one.

• Truecrimson

vanhees71

Gold Member
The hydrogen Hamiltonian is indeed precisely of the form I wrote. You just used the usual shorthand-notation not noting the identity operators for the single-body piece of the Hamiltonian. Of course, the entire system lives on the product space $\mathcal{H}_{\text{p}} \otimes \mathcal{H}_{\text{e}}$.

It's however extremely unwise to use the naive product basis to solve the energy eigenvalue problem. As @stevendaryl told you, it's much better to start by finding complete sets of compatible observables, including the Hamiltonian to find the basis where the diagonalization of $\hat{H}$ is easy. Since the observables in the complete set of compatible observables must commute with the Hamiltonian and among each others, you just have to look for the conserved quantities of your dynamical problem.

Here, it's indeed easy to find such a complete set of compatible observables by just thinking about the classical problem: It's a closed system of two Newtonian particles with a conservative central force. Thus the total momentum (center-of-mass momentum) is conserved, providing three compatible conserved observables (the three components of center-of-mass momentum).

The next step is to introduce new variables to express you Hamiltonian. You want the center-of-momentum components,
$$\vec{P}=\vec{p}_{\text{e}}+\vec{p}_{\text{p}}$$
Then it's good to introduce the relative position
$$\vec{r}=\vec{r}_{\text{e}}-\vec{r}_{\text{p}}$$
and it's canonical momentum $\vec{p}$.

Expressing the Hamiltonian in these coordinates, you'll find after some simple algebra (using Heaviside-Lorentz units)
$$\hat{H}=\frac{\vec{P}^2}{2M} + \frac{\vec{p}^2}{2 \mu} -\frac{e^2}{4 \pi r}.$$
Here $M=m_{\text{e}}+m_{\text{p}}$ and $\mu=m_{\text{e}} m_{\text{p}}/(m_{\text{e}} + m_{\text{p}})$.

Since $\vec{P}$ commutes with $\vec{p}$ and $r$ you can use $P$ eigenfunctions to diagonalize the Hamiltonian, i.e., you split your Hilbert space $\mathcal{H}=\mathcal{H}_{\text{p}} \otimes \mathcal{H}_{\text{e}} = \mathcal{H}_{\text{CM}} \otimes \mathcal{H}_{\text{rel}}$.
Now you only need to solve for the effective one-body problem with the Hamiltonian
$$H_{\text{rel}}=\frac{\vec{p}^2}{2 \mu} -\frac{e^2}{4 \pi r}.$$
Since everything is radial symmetric, you can use $H_{\text{rel}}$, $\vec{\ell}^2$, and $\ell_z$ as a complete set of energy eigenvectors, where $\vec{ell}=\vec{r} \times \vec{p}$ is the orbital angular momentum of the relative motion.

So the appropriate basis to solve for the energy eigenvalue problem is $|\vec{P},E_{\text{rel}},\ell,m \rangle$. The eigenspectra are $\vec{P} \in \mathbb{R}^3$, $\ell \in \{0,1,2,\ldots\}$ and (for each given $\ell$) $m \in \{-\ell,-\ell+1,\ldots,\ell \}$. The Eigenvalues of $\vec{\ell}^2$ are $\hbar^2 \ell(\ell+1)$ and that of $\ell_z$ are $\hbar m$.

• weirdoguy

Lebnm

Ok, I understood. Thank you guys!

"A basis of a composite system"

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