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If I had the wave function

$$\Psi\equiv\psi(r,\theta,\phi)\otimes\chi \in \mathscr{L}^2(\mathbb{R}^3)\otimes\mathbb{C}^{2S+1},$$

where $S$ is the spin of the state, is it correct to normalize the spin part of $\Psi$, namely $\chi$, regarding the spatial parameters $(r,\theta,\phi)$ as if they were fixed?

I mean: if $\psi\propto\sum Y_l^m (\theta,\phi)$, is it correct to say that the $Y_l^m(\theta,\phi)$'s are just numbers in $\mathbb{C}^{2S+1}$?

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2 Answers 2

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The norm squared of your wavefunction is

$$\left<\Psi\big|\Psi\right>=\left<\psi\big|\psi\right>\left<\chi\big|\chi\right>$$

and this should be $1$. In particular, you can normalize both $\psi$ and $\chi$ separately and your $\Psi$ will be also normalized.

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  • $\begingroup$ Yes, of course. But the norms on the right hand side of your equation are two different ones (different Hilbert spaces). My question was a little bit different. $\endgroup$ Dec 14, 2017 at 21:06
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    $\begingroup$ @VincenzoVentriglia Why does it matter? It would be helpful if you give a specific example. $\endgroup$
    – eranreches
    Dec 14, 2017 at 21:10
  • $\begingroup$ The point is: if I work just in $\mathbb{C}^{2S+1}$ without worrying about normalization in $\mathscr{L}^2(\mathbb{R}^3)$ (because I needn't entire wave function $\Psi$ to be normalized), is it legitimate to treat $(\theta,\phi)$, appearing in the spatial part $\psi$ of $\Psi$, like fixed parameters (i.e., numbers in $\mathbb{C}^{2S+1}$)? In this sense, spherical harmonics wouldn't be functions, but complex numbers, and this makes easier to calculate norms (in $\mathbb{C}^{2S+1}$). $\endgroup$ Dec 17, 2017 at 1:57
  • $\begingroup$ @VincenzoVentriglia If you work in $\mathbb{C}^{2S+1}$ then you can't have the $\mathscr{L}^{2}\left(\mathbb{R}^{3}\right)$ part of the wavefunction and your question is undefined. $\mathbb{C}^{2S+1}$ is a vector space over scalars from $\mathbb{C}$, not over functions from $\mathscr{L}^{2}(\mathbb{R}^{3})$. Take for example the following wavefunction $$\Psi\propto f(r)Y_{\ell}^{m}(\theta,\varphi)\left|\chi\right>$$ If you'll treat the $\mathscr{L}^{2}(\mathbb{R}^{3})$ part as a scalar, you'll end up getting $$\Psi=\left|\chi\right>$$ after the normalization, which is of course wrong. $\endgroup$
    – eranreches
    Dec 17, 2017 at 12:00
  • $\begingroup$ But I don't want to throw away the $\mathscr{L}^2 (\mathbb{R}^3)$ part of $\Psi$ at all. I just need to find probabilities for some spin measurements. $\endgroup$ Dec 17, 2017 at 13:29
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Actually, it seems you are asking a question about the definition of Hilbert space. A Hilbert space, as a vector space, is an inner product space -- that's the defining property of it. Therefore to define a Hilbert space, you have to define the inner product $\langle\psi_1|\psi_2\rangle$ of two vectors inside.

The inner product, by definition, should be a scalar. Below we restrict our discussion on wave-functions in quantum physics.

Firstly, for a simple Hilbert space structure $\mathcal{H}$, say, either $\psi$-space or $\chi$-space, when talking about wave-functions, it would be enough to define the scalar as a number. Therefore, when normalize vectors, i.e. wave-functions in $\mathcal{H}$, it is equivalent to require the result to be a number "$1$".

Now you are dealing with a tensor product space structure: $\mathcal{H}_0=\mathcal{H}_1\otimes\mathcal{H}_2$. The "vector" in $\mathcal{H}_0$ actually is defined as a tensor product of two simple vectors $\vec{v}_0 =\vec{v}_1\otimes\vec{v}_2$. Then the problem is what is the "scalar" in this $\mathcal{H}_0$? A more reasonable way to define a scalar is actually $a\otimes b$, where both $a, b\in\mathbb{C}$ are "daily-life" numbers. Therefore, more precisely, the normalization requirement now is modified as $\langle\vec{v}_0|\vec{v}_0\rangle = 1\otimes1\equiv \mathbb{1}$. That's why when you normalize it, you should do $\psi$ part and $\chi$ part separately.

Now you may ask: why don't I still use the old definition for scalar, as in a simple $\mathcal{H}$, i.e. just a "number"?

Formally (mathematically), definition of tensor product space should not include any operation mixing them, therefore it's not a good idea to mix two "$1$" in the $1\otimes1$ structure -- for sure you could define another operation if you want, but it has nothing to do with inner product and normalization procedure.

Practically (physically), you could consider the following condition: I fix the spin part as $\chi = |1\rangle$ by turning on some external coupling acting only on the spin, and then vary the potential energy affecting spatial part. In this case, apparently the normalization of spatial part should not affect spin part at all -- which has already been fixed by hand.

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    $\begingroup$ The point is: if I work just in $\mathbb{C}^{2S+1}$ without worrying about normalization in $\mathscr{L}^2(\mathbb{R}^3)$ (because I needn't entire wave function $\Psi$ to be normalized), is it legitimate to treat $(\theta,\phi)$, appearing in the spatial part $\psi$ of $\Psi$, like fixed parameters (i.e., numbers in $\mathbb{C}^{2S+1}$)? In this sense, spherical harmonics wouldn't be functions, but complex numbers, and this makes easier to calculate norms (in $\mathbb{C}^{2S+1}$). $\endgroup$ Dec 17, 2017 at 1:52
  • $\begingroup$ What I mean is $\psi$ and $\chi$ they belong to different Hilbert spaces if you don't define tensor product space; and if you define the tensor product space, there is still no relation between the computations in $\psi$-space and $\chi$-space. When you say you want to treat spherical harmonics as complex numbers in calculations, you are actually still thinking $\psi$ functions affects your calculation results in $\chi$-space (by multiplying a complex number) -- they don't, however. Simply because what I introduced above as the "scalar" $1\otimes 1$. $\endgroup$
    – Kite.Y
    Dec 17, 2017 at 18:43

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