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Frederik vom Ende
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Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.

Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.

Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.
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Frederik vom Ende
  • 3.5k
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Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.

Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.

Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.
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Frederik vom Ende
  • 3.5k
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Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.

Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.

Quantum states

Quantum states: general properties

State transformations

... from the perspective of operator theory

  • The bounded operators on a Hilbert space $\mathcal H$ may be larger than $\mathcal H\otimes\mathcal H^*$. The reason for this is that $\mathcal H\otimes\mathcal H^*\simeq\mathcal B^2(\mathcal H)$ with the latter being the Hilbert-Schmidt operators, cf. this phys.SE answer; but in infinite dimensions there are bounded operators which are not Hilbert-Schmidt (a simple example here is the identity operator).
  • Taking the positive part commutes with conjugation with a state.
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added clarification that the LU-LC conjecture also applies to stabilizer states.
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Added counterexample: the LU-LC conjecture is false
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added counterexample in state transformations section
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Post Made Community Wiki by Frederik vom Ende