Graduate Tensor product in Cartesian coordinates

[LagrangeEuler]

I am confused. Why sometimes perturbation $V'=\alpha xy$ we can write as $V'=\alpha x \otimes y$. I am confused because $\otimes$ is a tensor product and $x$ and $y$ are not matrices in coordinate representation. Can someone explain this?

 

[vanhees71]

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

 

[Demystifier]

$x$ and $y$ are not matrices in coordinate representation.

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.

 

[LagrangeEuler]

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

For instance 2d linear harmonic oscillator with this perturbation.

 

[andresB]

I'm confused by your confusion.

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 always a shorthand for $V'=\alpha x \otimes y$, because that's how the operators acting on the Hilbert space of 2d quantum systems are build, by definition.

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