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I have just started quantum mechanics with Shankar. In my understanding, quantum operators are linear operators in infinite-dimensional Hilbert spaces.

Shankar has repeatedly treated quantum operators as matrices.

I wonder how it is possible if the space of state vectors is infinite-dimensional.

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    $\begingroup$ You can do that by considering the states to be discrete. $\endgroup$ – my2cts Apr 15 at 10:56
  • $\begingroup$ We regularly get questions like this (I had them myself as a student). The answer is that a proper rigorous treatment requires a lot of functional analysis and operator theory. Physics texts often mention this in a preface (but who ever reads those :>) ). $\endgroup$ – Keith McClary Apr 15 at 18:36
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If the Hilbert space is infinite-dimensional, representing operators in terms of finite matrices is not possible. You can however use cumbersome infinite matrices which are not very useful... In any case some of the properties of matrices survive the passage to the (separable) infinite-dimensional case, at least dealing with bounded operators. Sometimes the space is infinite-dimensional because it is the Hilbert tensor product $H\otimes K$, where $K$ is finite-dimensional and $H$ is not. In those cases, operators acting in $K$, thus of the form $I\otimes A$, have the same properties as matrices ($A$). That is the case in quantum information where the qubits are referred to $K$, typically the polarization/spin space of an elementary particle and $H$ is infinite-dimensional, typically a $L^2$ space.

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  • $\begingroup$ I'd go beyond that and say most of the properties of matrices survive the passage. (After all, the difference rests in what measure you choose for the matrix-properties space ;D). $\endgroup$ – Emilio Pisanty Apr 15 at 12:27
  • $\begingroup$ Well, I think you are joking :) but there are some very important distinct features of operators in quantum theory which have no corresponding in matrix theory: the difference between Hermitian, symmetric and selfadjoint operator in particular. The spectral theorem applies only to selfadjoint operators whereas in matrix theory these three classes of matrices coincide and the spectral decomposition is true for the three types indistinctly. $\endgroup$ – Valter Moretti Apr 15 at 13:35

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