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We know that the canonical commutation relation between the raising and lowering operator $\hat a_I$, $\hat a^\dagger_J$ should result in the identity $\delta_{IJ}$:

$$[\hat a_I, \hat a^\dagger_J] = \delta_{IJ}.$$

However, when I explicitly constructed these operators for the case of $D=2$ and $D=4$ (when the Fock space dimension for each particle is 2 and 4, respectively), I don't get $\delta_{IJ}$ but instead a traceless diagonal matrix. I suspect for other $D$, the same issue arises. Note that the case described here refers to bosonic particles whose Hilbert space dimension can be infinite, but for computational purposes, we work with truncated Hilbert spaces where $D$ is finite. Here we take $D=2$ and $D=4$.

  • For $D=2$, we have 2 states: $|n\rangle = |0\rangle, |1\rangle$ where $|0\rangle = \begin{pmatrix}1\\0\end{pmatrix}, \,\,|1\rangle = \begin{pmatrix}0\\ 1\end{pmatrix}$. Then the raising/lowering operators are $\hat a= \begin{pmatrix} 0 & 0 \\1 &0\end{pmatrix}, \,\, \hat a^\dagger = \begin{pmatrix} 0 & 1\\0 &0\end{pmatrix}$. It can be verified that: $\begin{equation}\hat a |0\rangle = 0, \hat a|1\rangle = |0\rangle\end{equation}$ and $\begin{equation}\hat a^\dagger |0\rangle = |1\rangle, \hat a|1\rangle = 0\end{equation}$. However, $[\hat a, \hat a^\dagger] = \begin{pmatrix} 1 & 0 \\ 0 &-1\end{pmatrix}$ and not the identity matrix $\mathbf{1}_2 = diag(1,1)$.

  • For $D=4$, there are 4 states: $|0\rangle = \begin{pmatrix} 1\\0 \\0 \\0\end{pmatrix}$, $|1\rangle = \begin{pmatrix} 0\\1 \\0 \\0\end{pmatrix}$, $|2\rangle = \begin{pmatrix} 0\\0 \\1 \\0\end{pmatrix}$, $|3\rangle = \begin{pmatrix} 0\\0 \\0 \\1\end{pmatrix}$. The raising/lowering operators are:

    $\hat a^\dagger = \sum^3_{i=0} \sqrt{i+1} |i+1\rangle \langle i|=\begin{pmatrix} 0 & 1 & 0 & 0\\0 &0 & \sqrt{2} & 0\\0& 0 & 0 &\sqrt{3} \\0& 0 & 0 & 0\end{pmatrix}$,

    $\hat a = \sum^3_{i=0} \sqrt{i+1} |i\rangle \langle i+1|=\begin{pmatrix} 0 & 0 & 0 & 0\\\sqrt{1} & 0 &0&0\\0& \sqrt{2}& 0 & 0 \\0& 0 & \sqrt{3} & 0\end{pmatrix}$,

The number operator can be constructed from the above raising and lowering operators to be $|n\rangle = \hat a\hat a^\dagger =\begin{pmatrix} 0 & 0 & 0 & 0\\0 &1 & 0 & 0\\0& 0 & 2 &0\\0& 0 & 0 & 3\end{pmatrix}$, which is correct.

Taking $\hat a$ and $\hat a^\dagger$ above to act on the states $|0\rangle, \ldots, |3\rangle$, we also get the correct behaviors. However, the commutation relation $[\hat a, \hat a^\dagger]$, which should result in a $4\times 4$ identity matrix $diag(1,1,1,1)$, turns out to be $diag(-1,-1,-1,3)$ instead.

Something must be wrong somewhere in the steps above, but I don't know where.

Qmechanic
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user195583
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2 Answers2

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In a finite dimensional Hilbert space, the relation ${\rm Tr} (AB) = {\rm Tr} (BA)$ holds for arbitrary linear operators $A,B$. In this case, the commutation relation $[a,a^\dagger]=1_n$ immediately leads to the contradiction ${\rm Tr }[a,a^\dagger]=0$ but ${\rm Tr}\, 1_n=n\ne 0$.

Hyperon
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As @Hyperon have already pointed in comments, one may not be able to build commutation relations in a finite-dimensional space... but one can build anti-commutation relations for 2-by-2 and 4-by-4 matrices - the obvious examples are Pauli and Dirac matrices.

Note that when speaking of Fermions or in another appropriate context canonical commutation relation may actually mean canonical anti-commutation relation.

Roger V.
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