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States in Quantum Mechanics can be thought of as density operators, i.e., positive semi-definite, normalized trace class operators on a Hilbert Space $\mathcal{H}$. In the case $\mathcal{H}=\mathbb{C}^{2}$ we have that a generic state $\rho$ can be expressed as:

$$ \rho=\frac{1}{2}\left(\mathbb{I} + x^{1}\sigma_{1} + x^{2}\sigma_{2} + x^{3}\sigma_{3}\right) $$ where $\mathbb{I}$ is the identity matrix, the $\sigma_{i}$'s are the Pauli matrices and $||\,\vec{x}\,||\leq1$. This can be easily shown because $\mathcal{H}$ is finite-dimensional, and thus the set $\mathcal{B}(\mathcal{H})$ of linear operators is finite-dimensional.

I would like to know if there is a similar expression in the case in which $\mathcal{H}$ is infinite-dimensional, specifically, when $\mathcal{H}=\left(\mathcal{L}^{2}(\mathbb{R}^{n})\,;dl\right)$, where $dl$ is the Lebesgue measure.

EDIT

In order to be more specific, in the case $\mathcal{H}=\left(\mathcal{L}^{2}(\mathbb{R}^{n})\,;dl\right)$, I would like to find an expression for a generic $\rho$ suitable for the calculation of $tr(A\rho)$, where $A\in\mathcal{B}(\mathcal{H})$.

EDIT 2

In the setting I have in mind the operator $A$ is unknown. Moreover its definition depends on its action on states, in the sense that it is given as an assumption on the behaviour of $tr(A\rho)$, for example $tr(A\rho)>0$. Therefore, a generic expression for $\rho$ is needed in order to obtain informations on $A$.

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  • $\begingroup$ What would make an expression "similar"? In principle, you can pick any basis of $\mathcal B(\mathcal H)$ and expand $\rho$ in it. Which property do you want? $\endgroup$ – Norbert Schuch Feb 22 '15 at 15:20
  • $\begingroup$ @NorbertSchuch: The "similarity" I refer to is the possibility of writing $\rho$ in a way suitable for explicit calculations, just as is the case for $\rho$ when $\mathcal{H}=\mathbb{C}^{2}$, in which case the properties of Pauli matrices $\sigma_{i}$ make very useful the expression of $\rho$ in terms of $\sigma_{i}$. I know this is not quite specific but, at the moment, I do not know how to be more specific. $\endgroup$ – Ittiolo Feb 22 '15 at 15:25
  • $\begingroup$ What would be wrong e.g. with an expansion in a computational basis $\lvert i \rangle\langle j\rvert$? This is also quite useful for explicit calculations ... $\endgroup$ – Norbert Schuch Feb 22 '15 at 15:33
  • $\begingroup$ @NorbertSchuch: nothing would be wrong, but I am not able to produce such an expression for a generic $\rho$, i.e., without any assumptions on it rather than to be a quantum state. $\endgroup$ – Ittiolo Feb 22 '15 at 15:36
  • $\begingroup$ ??? Why not? Do you want a form where you can easily check whether $\rho\ge0$ and/or $\mathrm{tr}(\rho)=1$? It would be useful if you could specify which properties you like about this form. $\endgroup$ – Norbert Schuch Feb 22 '15 at 16:05
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The trace class operators form a Banach space. There is a concept of (countable) basis for Banach spaces that is called Schauder basis.

Not every Banach space has a Schauder basis, but it is true e.g. for the space of compact operators (the case of $\mathcal{K}(l^2)$ is given explicitly in the wikipedia article).

Since the trace class operators are imbedded in the compact operators you have also a Schauder basis for them (and that basis may be further specialized to converge in the trace norm I suppose).

Let $\{\rho_n\}_{n\in\mathbb{N}}$ be a Schauder basis for $\mathcal{L}^1(L^2(\mathbb{R}^d))$ (trace class operators on $L^2$), then for any $\rho\in\mathcal{L}^1$ there exist $\{\alpha_n\}$ complex numbers such that $$\rho=\sum_{n=0}^\infty \alpha_n\rho_n$$ and that sum converges in trace norm.

Also, if the operator $A$ is self-adjoint, then you may write it in the spectral decomposition $$A=\int_{\mathbb{R}}\lambda dP_\lambda$$ where $dP_\lambda$ is the projection valued measure. Then you may write $$Tr(A\rho)=\sum_n \alpha_n\int \lambda Tr(\rho_n dP_\lambda)\; .$$ If in addition the spectrum of $A$ is purely discrete (i.e. $A$ compact or with compact resolvent) you get $$Tr(A\rho)=\sum_{n,m=0}^\infty \alpha_n\lambda_m Tr(\rho_n \lvert\psi_m\rangle\langle\psi_m\rvert)$$ where $\psi_m$ is the eigenfunction and $\lambda_m$ is the corresponding eigenvalue.

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  • $\begingroup$ nice answer. I really thank you, especially for the expression of $Tr(A\rho)$ in terms of the PVM and the Schauder Basis. Can you give me some reference on Schauder Basis? $\endgroup$ – Ittiolo Feb 25 '15 at 9:51
  • $\begingroup$ I don't know much literature on these bases, however these free notes seem quite accurate. $\endgroup$ – yuggib Feb 25 '15 at 12:34
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In case $\mathcal{H}=L^{2}(\mathbb{R}^{n},d\mathbf{x})$ $\rho $ is a positive semi-definite trace class operator and can be expressed as $$ \rho =\sum_{k}\lambda _{k}|u_{k}><u_{k}| $$ where $\{u_{k}\}$ is a basis for $\mathcal{H}$ and $\lambda _{k}$ is non-negative with $$ \sum_{k}\lambda _{k}=1. $$

Addition in response to a question by Norbert Schuch.

This is common knowledge within the C*-algebra formulation of quantum mechanics. In the Hilbert space ($\mathcal{H}$) situation observables are bounded operators on $\mathcal{H}$ and states are positive linear functionals of the algebra of observables. One such class, the density operators, is contained in the trace class, the so-called pre-dual of the set of bounded linear operators $\mathcal{B}(\mathcal{H})$ as a Banach space. But, in the infinite-dimensional situation, there are also states in the dual of $\mathcal{B(H)}$ not contained in the trace class.

As an example consider a free particle in a box. Then the Hamiltonian $H$ has discrete spectrum bounded from below and the canonical equilibrium state is well-defined, $\sim \exp [-\beta H]$. But if the linear dimensions of the box tend to infinity $H$ becomes the kinetic energy operator $\mathbf{p}% ^{2}/(2m)$ which has purely continuous spectrum and $\exp [-\beta \mathbf{p}% ^{2}/(2m)]$ is no longer a trace class (density) operator. For a gas of free particles in a box we have a similar but more complicared situation in the thermodynamic limit. Also in certain scattering situations we encounter such a case.

I hope that this gives an idea of what is going on. In practice, for instance when only expectation values are involved, the introduction of such non-normal states can be avoided.

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    $\begingroup$ This is incorrect. E.g., for $\mathbb C^2$ your basis would only have two elements. $\endgroup$ – Norbert Schuch Feb 22 '15 at 15:19
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    $\begingroup$ This formula is actually true when $u_k$ is orthonormal set of $\rho$'s eigenvectors and $\lambda_k$ are the corresponding eigenvalues. This follows from the spectral decomposition of a compact bounded self-adjoint operator. Provided that it converges in the trace norm. $\endgroup$ – Apogee Feb 22 '15 at 17:20
  • $\begingroup$ @Apogee: Admitted. But I suspect the OP had a fixed basis in mind, not one which depends on $\rho$. Otherwise, choosing a basis which contains $\rho$ would make things even simpler! $\endgroup$ – Norbert Schuch Feb 22 '15 at 18:10
  • $\begingroup$ @Apogee: Yes, I should have mentioned that the basis is orthonormal. By the way, not all states can be represented as density operators. The latter are the so-called normal states. $\endgroup$ – Urgje Feb 22 '15 at 20:51
  • $\begingroup$ @Urgje "not all states can be represented as density operators" -- could you elaborate? $\endgroup$ – Norbert Schuch Feb 22 '15 at 22:36
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Every density matrix on $\mathbb C^2$ is of that form because $\rho$ must be Hermitian. This property is clearly satisfied because of the form and the algebra of Pauli matrices. This can then be generalised to higher dimensional Hilbert spaces by introducing a basis for Hermitian matrices which satisfy the same algebra. So the problem is then equivalent to whether one can find a basis $\{\sigma_i\}$ for the self-adjoint operators on $\mathcal H$ such that $$\{\sigma_i,\sigma_j\} = 2\delta_{ij}I,$$ i.e a real Clifford algebra, plus the trace-class condition.

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  • $\begingroup$ Why should $\rho$ be idempotent? $\endgroup$ – Norbert Schuch Feb 22 '15 at 15:18
  • $\begingroup$ My bad, $\rho$ need not be idempotent in general. I've fixed that $\endgroup$ – Phoenix87 Feb 22 '15 at 15:28
  • $\begingroup$ @Phoenix87: so, in the case $\mathcal{H}=\left(\mathcal{L}^{2}(\mathbb{R}^{n})\,;dl\right)$, one shold look for an infinite-dimensional real Clifford algebra? $\endgroup$ – Ittiolo Feb 22 '15 at 15:37
  • $\begingroup$ I believe so, although I have never seen this problem treated like this apart from the 2-dimensional case. $\endgroup$ – Phoenix87 Feb 22 '15 at 15:39
  • $\begingroup$ @Phoenix87 For powers of two, one can use tensor products of Paulis (and this done). $\endgroup$ – Norbert Schuch Feb 22 '15 at 16:07

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