The things I'm pretty sure I understand: Let's say I have a single particle hamiltonian $H$ represented by a $2$x$2$ matrix, so it has two eigenstates $|\lambda_1\rangle$ and $|\lambda_2\rangle$. I can define two projectors onto these states $P_1=|\lambda_1\rangle\langle\lambda_1|$ and $P_2=|\lambda_2\rangle\langle\lambda_2|$. Now if we look at a 2-particle system, the Hamiltonian of particle 1 in the direct product space is $H_1=H\otimes I$ and the Hamiltonian of particle 2 is $H_2=I\otimes H$. If I understand correctly, the new space is spanned by, for example, the following 4 states: $|\lambda_i\rangle\otimes|\lambda_j\rangle$, for $i,j=1,2$. $H_1$ will have 4 eigenvectors which correspond to the energy eigenstates $H_1|n\rangle=E_n|n\rangle$ of particle 1 in the two particle system. I can then construct 4 projectors onto these states which are defined in a similar way as above, and I will call them $P'_n$ for $n=1,2,3,4$.

The question: I want to show that, for an arbitrary state in the two-particle vector space, $P'_nf(E_n)|\psi\rangle=P'_nf(H_1)|\psi\rangle$ where $f(E_n)$ is some function of the eigenvalues of the Hamiltonian and $|\psi\rangle$. I know how to do this for a one particle system, but I have no intuition for direct products yet and I can't work out how to reconcile the fact that $H$ has 2 states but $H_1$ has 4 in an attempt to express $P'_n$ in terms of $P_1$ and $P_2$. I can expand in terms of the two particle basis $|\psi\rangle=\sum_{i,j}c_ib_j|\lambda_i\rangle\otimes|\lambda_j\rangle$. Then apply the projector $|n\rangle\langle n|$ but at this point I'm out of ideas. Is there some way to write the eigenfunctions of $H_1$ in terms of the eigenfunctions of $H$?

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    $\begingroup$ just a quick point: $|\psi\rangle=\sum_{i,j}c_ib_j|\lambda_i\rangle\otimes|\lambda_j\rangle$ is not an arbitrary state but instead is the special case of a factorisable or unentangled state. The co-efficients here are the Kronecker product of the column vectors $c$ and $b$ and is wholly defined by $2 N$ complex values in the column vector. A general, entangled state has $|\psi\rangle=\sum_{i,j}C_{i\,j}|\lambda_i\rangle\otimes|\lambda_j\rangle$, i.e. you can't split the $N^2$ independent co-efficients into something of the form $b\otimes c$. $\endgroup$ Commented Dec 2, 2014 at 6:50
  • $\begingroup$ Thanks for pointing this out, I have't studied entanglement before so it's nice to see what it means in it's mathematical glory. $\endgroup$
    – Lachy
    Commented Dec 2, 2014 at 14:28
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    $\begingroup$ This is a common mistake in terminology: the direct product $\times$ is more akin to the direct sum $\oplus$; what you have is the tensor product $\otimes$. One important difference is that $\dim(A\oplus B)=\dim(A)+\dim(B)$, while $\dim(A\otimes B)=\dim(A)\dim(B)$. Unfortunately $2+2=2\cdot2$, so it's easier to get confused in this particular case. $\endgroup$
    – user10851
    Commented Dec 2, 2014 at 19:02
  • $\begingroup$ Okay so what am I actually dealing with here: a "tensor product space" or a "direct sum space"? I assume it's the tensor product space since $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$ and in this case it makes sense to think of $\text{dim}(\mathcal{H})=\text{dim}(\mathcal{H}_1)\text{dim}(\mathcal{H}_2)$ (from the analogy to the spin-1/2 case with two particles, where we think of the 4 resulting states as being ordered pairs of spin-ups and spin-downs). But the direct sum does come into the picture when we think of the "total" Hamiltonian $H=H_1\oplus H_2$, so I would like to be sure. $\endgroup$
    – Lachy
    Commented Dec 3, 2014 at 1:31

1 Answer 1


This is quite simple. Consider the operator $H$ on the Hilbert space $\mathscr{H}$, in your simple example it has a spectral resolution: $$H=\sum_{n}E_n \lvert n\rangle\langle n \rvert\; .$$ Each eigenvalue has multiplicity 1. Now the operators $H_1$ and $H_2$ on $\mathscr{H}\otimes\mathscr{H}$ have the same spectrum of $H$, but each eigenvalue has multiplicity $\mathrm{dim} [\mathscr{H}]$, and the orthonormal eigenvectors will be the tensor product of an eigenvector of $H$ and a vector of the orthonormal basis of $\mathscr{H}$ (in the respective order). The spectral resolutions are directly inherited by the one of $H$: $$H_1=\sum_{n}E_n \bigl(\lvert n\rangle\langle n \rvert\ \otimes 1\bigr)\;,\; H_2=\sum_{n}E_n \bigl(1\otimes\lvert n\rangle\langle n \rvert\ \bigr)\; .$$ Apply now the projection on a factorized vector $\psi_1\otimes\psi_2$ for example on $H_1$: $$\lvert \psi_1\otimes\psi_2\rangle\langle \psi_1\otimes\psi_2 \rvert H_1= \sum_n E_n \langle \psi_1,n\rangle \bigl( \lvert\psi_1\rangle\langle n\rvert\otimes\lvert\psi_2\rangle\langle\psi_2\rvert \bigr)\; .$$ Now suppose that $\psi_1=m$, where $m$ is an eigenvector of $H$, orthogonal to all the others (and normalized). Then the scalar products in the sum vanish apart from $n=m$. Thus you obtain: $$\tag{1}\lvert m\otimes\psi_2\rangle\langle m\otimes\psi_2 \rvert H_1=E_m \lvert m\otimes\psi_2\rangle\langle m\otimes\psi_2 \rvert \; .$$

Remark: Observe that I used more than once the fact that for factorized vectors $\lvert \psi_1\otimes\psi_2\rangle\langle \psi_1\otimes\psi_2 \rvert= \lvert\psi_1\rangle\langle \psi_1\rvert\otimes\lvert\psi_2\rangle\langle\psi_2\rvert$.

Equation (1)---with $\psi_2$ chosen as an element of the orthonormal basis of $\mathscr{H}$---is the relation on the projectors onto eigenvectors of $H_1$ you look for. This is extended to functions of $H_1$ since for any suitably regular function $f$, $f(H)=\sum_{n}f(E_n) \lvert n\rangle\langle n \rvert$.

  • $\begingroup$ I appreciate this response very much, thanks. It answers my question exactly, and very clearly, with a nice justification for each step. $\endgroup$
    – Lachy
    Commented Dec 2, 2014 at 16:36

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