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- Thread starter LagrangeEuler
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- #2

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To explain this we need more context, i.e., which concrete example are you discussing?

- #3

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But they are matrices in any other representation, e.g. momentum representation. And even in coordinate representation they can be viewed as matrices, but in this representation calculations can be simplified by viewing them as ordinary numbers.##x## and ##y## are not matrices in coordinate representation.

- #4

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For instance 2d linear harmonic oscillator with this perturbation.To explain this we need more context, i.e., which concrete example are you discussing?

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A 1d quantum system "lives" in a Hilbert space ##\mathcal{H}_{x}## that is spanned by linear combination of base kets of the form ##\left|x\right\rangle##. You can also have another independent degree of freedom living in a Hilbert space ##\mathcal{H}_{y} ## with base kets of the form ##\left|y\right\rangle ##. Then, you can combine both degrees of freedom into a Hilbert space ##\mathcal{H}_{xy}=\mathcal{H}_{x}\otimes\mathcal{H}_{y}##, that is how the Hilbert space of a 2d quantum system is build. The base kets are of the form ##\left|xy\right\rangle \equiv\left|x\right\rangle \otimes\left|y\right\rangle ##

So ##V'=\alpha xy## is

See Cohen-Tannoudji, quantum mechanics, vol 1, page 160.

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