In one dimension, given a particle in a quantum state $| \psi\rangle$, the probability density of position is given as $| \psi(x) |^2 = \psi^*(x) \psi(x) =\langle x | \psi \rangle\langle \psi | x \rangle $ where $| x \rangle$ is the base ket for the position operator. I have just learnt about tensor products role in describing states of multi-particles or singular particles where we consider different observable spaces simultaneously.
Consider the three dimensional problem: An electron in a hydrogen atom occupies the combined spin and position state given as $$R_{21} \bigg(\sqrt{\frac{1}{3}} Y_{1}^{0} \otimes \chi_{+} + \sqrt{\frac{2}{3}}Y_{1}^{1} \otimes \chi_{-} \bigg)$$
where $\chi_{+}$ and $\chi_{-}$ are the eigenstates of the z-component of spin (spin up and spin down). If you measure the position of the particle, what is the probability density of finding it at $r, \theta, \phi$?
The given answer suggests you use the scalar product defined for tensor products of Hilbert spaces (i.e. $\langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle = \langle \phi_1, \psi_1 \rangle_1 \langle \phi_2, \psi_2 \rangle_2 $ ) so we get $$\begin{align} |\psi|^2 &= |R_{21}|^2\bigg\langle \sqrt{\frac{1}{3} }Y_{1}^{0*} \otimes \chi_{+}^{\dagger} + \sqrt{\frac{2}{3}}Y_{1}^{1*} \otimes \chi_{-}^{\dagger}, \sqrt{\frac{1}{3} }Y_{1}^{0} \otimes \chi_{+} + \sqrt{\frac{2}{3}}Y_{1}^{1} \otimes \chi_{-} \bigg\rangle \end{align}\\
=|R_{21}|^2\bigg( \frac{1}{3}|Y_{1}^{0}|^2 \chi_{+}^{\dagger}\chi_{+} + \frac{\sqrt{2}}{3}Y_{1}^{1*}Y_{1}^{0} \chi_{-}^{\dagger}\chi_{+} + \frac{\sqrt{2}}{3}Y_{1}^{0*}Y_{1}^{1} \chi_{+}^{\dagger}\chi_{-} + \frac{2}{3}Y_{1}^{1}\chi_{-}^{\dagger}\chi_{-} \bigg) \\
= \frac{1}{3}|R_{21}|^{2} \big( |Y_{1}^{0}|^2 + 2|Y_{1}^{1}|^2 \big).$$
Question:
But if this is correct then what we have actually done is taken the scalar product $\langle \psi| \psi \rangle$ (which we usually want to be normalized to equal $1$) of a state $| \psi \rangle = R_{21} \bigg(\sqrt{\frac{1}{2}} Y_{1}^{0} \otimes \chi_{+} + \sqrt{\frac{2}{3}}Y_{1}^{1} \otimes \chi_{-} \bigg)$, which would usually give the summation of the probability coefficients rather than the probability density? What am I missing here? Why is this the way to get the probability density or is there a way which is more descriptive?
See my proposed answer below.
Thanks for any assistance.
Best Answer
I'll try to clear some of the confusion, first: $$ 1= ⟨ψ|ψ⟩ = \int ⟨ψ|x⟩⟨x|ψ⟩dx=\int|\psi(x)|^2dx $$ and $$\int |x⟩⟨x|dx=I$$
In your first question you compute $|\psi(r,\theta,\phi)|^2 $, which is fine, it's the density that you're looking for. It will only equal 1 after integration over $r, \theta\ and\ \phi$.
The mistake in the second derivation is that you assumed $⟨χ_+|+⟨χ_−|=I$, while in reality $⟨χ_+|+⟨χ_−|=\sqrt2⟨1|$ (Positive x-direction spin state). You were looking for: $|χ_+⟩⟨χ_+|+|χ_-⟩⟨χ_−|=I$.
Calculating $⟨ψ|(|r,\theta,\phi⟩⟨r,\theta,\phi|⊗I)|ψ⟩$ should do the trick.
bra-ket and projectors
I'll just add a really short summary here, to clarify further:
⟨ψ| is a bra (dual vector), |ψ⟩ is a ket (vector in Hilbert space).
$⟨ψ|\phi⟩$ is a complex number.
$|\phi⟩⟨ψ|$ is an operator, and $|ψ⟩⟨ψ|$ is a projector on state ψ.
Operator $P$ is a projector if $P^2=P$.
If you want the probability density at position x, you project on the eigenspace of the position operator, using $|x⟩⟨x|$.
If you want the probability density at $r,θ\ and\ ϕ$ you use the projection on the eigenspace of these operators. You don't care which spin you get, so you don't project on the spin space and just use the identity.
$⟨ψ|(|r,\theta,\phi⟩⟨r,\theta,\phi|⊗I)|ψ⟩$, or $⟨ψ|(|r,\theta,\phi⟩⟨r,\theta,\phi|⊗(|χ_+⟩⟨χ_+|+|χ_-⟩⟨χ_−|))|ψ⟩=⟨ψ|(|r,\theta,\phi⟩⟨r,\theta,\phi|⊗|χ_+⟩⟨χ_+|+|r,\theta,\phi⟩⟨r,\theta,\phi|⊗|χ_-⟩⟨χ_−|)|ψ⟩$, if you prefer, will give you what you're looking for. Notice that every open bra has its closing ket, which means you will get a number, and it is symmetric, so the the number is real and positive.