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Removed my last comment which is obviously wrong.
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Aharon Brodutch
Aharon Brodutch
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It is easy to construct any operator (in continuous variables) using the set of operators $$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $$\{a^k (a^\dagger)^\ell\}.$$

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,\ell}$ such that $$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

It is easy to construct any operator (in continuous variables) using the set of operators $$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $$\{a^k (a^\dagger)^\ell\}.$$

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,\ell}$ such that $$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

It is easy to construct any operator (in continuous variables) using the set of operators $$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $$\{a^k (a^\dagger)^\ell\}.$$

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,\ell}$ such that $$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)

added 29 characters in body; edited tags; edited title
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Qmechanic ♦
Qmechanic ♦
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Do the ladder operators $a$ and $a^\dagger$ form a complete algebra basis?

It is easy to construct any operator (in continuous variables) using the set of operatorsoperators {$|l\rangle\langle m |$},$$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $\hat M=\sum_{l,m}\alpha_{l,m}|l\rangle\langle m |$$$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{l,m}$$\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $\{a^k (a^\dagger)^l\}$.$$\{a^k (a^\dagger)^\ell\}.$$

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,l}$$\alpha_{k,\ell}$ such that $|0\rangle\langle 0|=\sum_{k,l}\alpha_{k,l}a^k (a^\dagger)^l$.$$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

Do the ladder operators $a$ and $a^\dagger$ form a complete basis?

It is easy to construct any operator (in continuous variables) using the set of operators {$|l\rangle\langle m |$}, where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $\hat M=\sum_{l,m}\alpha_{l,m}|l\rangle\langle m |$ where $\alpha_{l,m}$ are complex coefficients. My question is, can we do the same thing with the set $\{a^k (a^\dagger)^l\}$.

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,l}$ such that $|0\rangle\langle 0|=\sum_{k,l}\alpha_{k,l}a^k (a^\dagger)^l$. (here $|0\rangle$ is the vacuum and I take $a^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

Do the ladder operators $a$ and $a^\dagger$ form a complete algebra basis?

It is easy to construct any operator (in continuous variables) using the set of operators $$\{|\ell\rangle\langle m |\},$$ where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $$\hat M=\sum_{\ell,m}\alpha_{\ell,m}|\ell\rangle\langle m |$$ where $\alpha_{\ell,m}$ are complex coefficients. My question is, can we do the same thing with the set $$\{a^k (a^\dagger)^\ell\}.$$

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,\ell}$ such that $$|0\rangle\langle 0|=\sum_{k,\ell}\alpha_{k,\ell}a^k (a^\dagger)^\ell.$$ (here $|0\rangle$ is the vacuum and I take $a^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

typo corrected
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Aharon Brodutch
Aharon Brodutch
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It is easy to construct any operator (in continuous variables) using the set of operators {$|l\rangle\langle m |$}, where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $\hat M=\sum_{l,m}\alpha_{l,m}|l\rangle\langle m |$ where $\alpha_{l,m}$ are complex coefficients. My question is, can we do the same thing with the set $\{a^k (a^\dagger)^l\}$.

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,l}$ such that $|0\rangle\langle 0|=\sum_{k,l}\alpha_{k,l}a^k (a^\dagger)^l$. (here $|0\rangle$ is the vacuum and I take $\alpha^0=I$$a^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

It is easy to construct any operator (in continuous variables) using the set of operators {$|l\rangle\langle m |$}, where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $\hat M=\sum_{l,m}\alpha_{l,m}|l\rangle\langle m |$ where $\alpha_{l,m}$ are complex coefficients. My question is, can we do the same thing with the set $\{a^k (a^\dagger)^l\}$.

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,l}$ such that $|0\rangle\langle 0|=\sum_{k,l}\alpha_{k,l}a^k (a^\dagger)^l$. (here $|0\rangle$ is the vacuum and I take $\alpha^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

It is easy to construct any operator (in continuous variables) using the set of operators {$|l\rangle\langle m |$}, where $l$ and $m$ are integers and the operators are represented in the Fock basis, i.e any operator $\hat M$ can be written as $\hat M=\sum_{l,m}\alpha_{l,m}|l\rangle\langle m |$ where $\alpha_{l,m}$ are complex coefficients. My question is, can we do the same thing with the set $\{a^k (a^\dagger)^l\}$.

Actually, this boils down to a single example which would be sufficient. Can we find coefficients $\alpha_{k,l}$ such that $|0\rangle\langle 0|=\sum_{k,l}\alpha_{k,l}a^k (a^\dagger)^l$. (here $|0\rangle$ is the vacuum and I take $a^0=I$)

My feeling is that an impossibility proof would involve the fact that $|0\rangle\langle 0|$ is rank 1, but I am not sure.

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Aharon Brodutch
Aharon Brodutch
  • 93
  • 4