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.

  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Thanks that was a typo, it should have been $1$, I edited it now. $\endgroup$
    – Alex
    Commented Dec 20, 2016 at 17:24
  • $\begingroup$ Are you aware that $R$ and the $Y$'s depend on position? $\endgroup$ Commented Dec 20, 2016 at 17:48
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Yes it is the spherical harmonic functions which depend on $\theta$ and $\phi$. $\endgroup$
    – Alex
    Commented Dec 20, 2016 at 17:52
  • $\begingroup$ Right, so when you are doing $\langle \psi | \psi \rangle$, you are actually not doing the integration over position part, so the thing you end up with depends on position and is in fact supposed to be the probability density. That is supposed to answer part of your question. $\endgroup$ Commented Dec 20, 2016 at 17:54
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Yes I understand that the $R_{21}Y$ part of the tensor products are already projected onto the position basis. But I am looking for the reasoning behind the method used in this case. What exactly are we doing in this method when you do the $\langle \psi | \psi \rangle$ to find the probability density. In the simple case I outlined at the start, it is clear, that you are finding the squared absolute value of the probability coefficients, this is the probability density for a continuous observable, that is clear. The reasoning here is not. $\endgroup$
    – Alex
    Commented Dec 20, 2016 at 18:03

3 Answers 3


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.

  • $\begingroup$ Are you sure you have the right expression to calculate the probability density at then end, did you maybe mean $|(|r,\theta, \phi \rangle \langle r , \theta, \phi| \otimes I)| \psi \rangle|^2$? So you don't agree with the suggestion in the comments above which advises to calculate $|(| \mathbf{r} \rangle \otimes | \chi_{+} \rangle \langle \mathbf{r}| \otimes \langle \chi_{+} | + | \mathbf{r} \rangle \otimes | \chi_{-} \rangle \langle \mathbf{r} | \otimes \langle \chi_{-} | ) | \psi \rangle |^2$? $\endgroup$
    – Alex
    Commented Dec 22, 2016 at 21:21
  • $\begingroup$ No, (|r,θ,ϕ⟩⟨r,θ,ϕ|⊗I)|ψ⟩ is a "ket" vector, you can't just take the absolute value of it and square it. You can, however, take the %L^2% norm of it, but it'll give you something else. I'll check the comments, but the equation you posted here makes no sense at all. $\endgroup$
    – milo
    Commented Dec 22, 2016 at 23:54
  • $\begingroup$ Ah, that's some real abuse of notation in the comments over there. I'll just add a bit about the bra-ket notation to my answer. It all really pretty clean and simple. $\endgroup$
    – milo
    Commented Dec 23, 2016 at 0:02
  • 1
    $\begingroup$ That was just from memory :) But you can look here: en.wikipedia.org/wiki/… $\endgroup$
    – milo
    Commented Dec 23, 2016 at 14:59
  • 1
    $\begingroup$ Yes, just use the appropriate projection operator, which in this case is: $|r⟩⟨r|⊗|χ+⟩⟨χ+|$. You will get: $⟨ψ|(|r⟩⟨r|⊗|χ+⟩⟨χ+|)|ψ⟩$ , which already includes the integral over $θ\ and\ ϕ$. $\endgroup$
    – milo
    Commented Jan 6, 2017 at 21:29

The example you give is a poor illustration of the idea you are thinking about because the spin up and spin down parts wouldn't have interfered with each other even if there had been no spin part to the have function. Try instead the same problem with $$R_{21} \bigg(\sqrt{\frac{1}{3}} Y_{1}^{0} \otimes \chi_{+} +\left( \sqrt{\frac{1}{3}} Y_{1}^{0}+\sqrt{\frac{1}{3}}Y_{1}^{1}\right) \otimes \chi_{-} \bigg)$$

Now you get $$\frac{1}{3}|R_{21}|^{2} \big( 2|Y_{1}^{0}|^2 + |Y_{1}^{1}|^2 \big).$$

If, on the other hand you ignored the spin part, then the formerly spin up part would interfere with the spin down part since they both have a $Y_{0}^{0}$ piece, and therefore do have overlap. Therefore you get $$\frac{1}{3}|R_{21}|^{2} \left( 4|Y_{1}^{0}|^2 + |Y_{1}^{1}|^2 \right)$$

Here we see the weighing of the $Y_{1}^{0}$ piece is greater because there was constructive interference. Destructive interference is also possible. If we had instead had $$R_{21} \bigg(\sqrt{\frac{1}{3}} Y_{1}^{0} \otimes \chi_{+} +\left(- \sqrt{\frac{1}{3}} Y_{1}^{0}+\sqrt{\frac{1}{3}}Y_{1}^{1}\right) \otimes \chi_{-} \bigg),$$ then the answer accounting for spin doesn't change, but when we ignore spin, the $Y_{1}^{0}$ pieces see each other and destructively interfere. Then the final answer we get after ignoring spin is $$\frac{1}{3}|R_{21}|^{2} |Y_{1}^{1}|^2 .$$

Notice that besides changing the relative densities at different points in space, ignoring spin also changes the overall normalization of the wave function. Thus it is very important to always correctly handle the spin.

  • $\begingroup$ Thanks for your answer, but I am particularly interested in the validity of the method suggested for reasons given and why it seems to differ from the usual way we find probability density? Also in the problem given, the spin does cancel some terms. $\endgroup$
    – Alex
    Commented Dec 20, 2016 at 17:30

I have tried using the advice as provided in the comments and using the scalar product associated with tensors defined as $\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 to find the probability density of finding the particle at $(r, \theta, \phi)$, I evaluated as follows: $$ \begin{align} &|(( \langle r | \otimes \langle \chi_{+}|) +( \langle r | \otimes \langle \chi_{-}))|\psi \rangle |^2\\ &=|( \langle r | \otimes \langle \chi_{+}|)| \psi \rangle + ( \langle r | \otimes \langle \chi_{-}|)| \psi \rangle|^2 \\& =\bigg|(\langle r | \otimes \langle \chi_{+}|)(\sqrt{\frac{1}{3}}R_{21}Y_{1}^{0} \otimes | \chi_{+} \rangle) + (\langle r | \otimes \langle \chi_{+}|)(\sqrt{\frac{2}{3}}R_{21}Y_{1}^{1} \otimes | \chi_{-} \rangle) + (\langle r | \otimes \langle \chi_{-}|)(\sqrt{\frac{1}{3}}R_{21}Y_{1}^{0} \otimes | \chi_{+} \rangle) +|(\langle r | \otimes \langle \chi_{-}|)(\sqrt{\frac{2}{3}}R_{21}Y_{1}^{1} \otimes | \chi_{-} \rangle)\bigg|^2 \\ &=\bigg|\sqrt{\frac{1}{3}}R_{21}(r)Y_{1}^{0}(\theta, \phi) \otimes 1 + \sqrt{\frac{2}{3}}R_{21}(r)Y_{1}^{1}(\theta, \phi)\otimes 0 + \sqrt{\frac{1}{3}}R_{21}(r)Y_{1}^{0}(\theta, \phi) \otimes 0 + \sqrt{\frac{2}{3}}R_{21}(r)Y_{1}^{1}(\theta, \phi) \otimes 1\bigg|^2 \\& = \bigg|\sqrt{\frac{1}{3}}R_{21}(r)Y_{1}^{0}(\theta, \phi) \otimes 1 + \sqrt{\frac{2}{3}}R_{21}(r)Y_{1}^{1}(\theta, \phi) \otimes 1\bigg|^2 \\& = \bigg\langle \sqrt{\frac{1}{3}}R_{21}(r)Y_{1}^{0}(\theta, \phi) \otimes 1 + \sqrt{\frac{2}{3}}R_{21}(r)Y_{1}^{1}(\theta,\phi)\otimes 1, \sqrt{\frac{1}{3}}R_{21}(r)Y_{1}^{0}(\theta, \phi)\otimes 1 + \sqrt{\frac{2}{3}}R_{21}(r)Y_{1}^{1}(\theta,\phi)\otimes 1 \bigg\rangle \\& = \frac{1}{3}|R_{21}(r)|^2|Y_{1}^{0}(\theta, \phi)|^2 + \frac{\sqrt{2}}{3}|R_{21}(r)|^2Y_{1}^{1}(\theta, \phi)Y_{0}^{*0}(\theta, \phi) + \frac{\sqrt{2}}{3}|R_{21}(r)|^2 Y_{1}^{0}(\theta,\phi)Y_{1}^{*1}(\theta, \phi) + \frac{2}{3}|R_{21}(r)|^2|Y_{1}^{1}(\theta, \phi)|^2 \end{align} $$

For some reason I get the extra two terms where as I should just get $$\frac{1}{3}|R_{21}(r)|^2|Y_{1}^{0}(\theta, \phi)|^2 + \frac{2}{3}|R_{21}(r)|^2|Y_{1}^{0}(\theta, \phi)|^2$$ Can anyone see where I went wrong?


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