Timeline for Positional probability density for combined spin and position states
Current License: CC BY-SA 3.0
19 events
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| Jun 11, 2020 at 9:33 | history | edited | CommunityBot |
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| May 29, 2017 at 21:22 | comment | added | user100411 | @milo The probability density in this sense is just the expectation value of the projection operator. | |
| Jan 6, 2017 at 21:29 | comment | added | milo | 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 $θ\ and\ ϕ$. | |
| Jan 6, 2017 at 16:47 | comment | added | Alex | $$\frac{1}{3}|R_{21}|^2\int|Y_{1}^{0}|^2 \sin^2 \theta d \theta d \phi + \frac{2}{3}|R_{21}|^2\int|Y_{1}^{1}|^2\sin^2 \theta d \theta d \phi = |R_{21}|^2$$ then since we are only considering the case where we are spin up (which has a probability of $\frac{1}{3}$), we have a final probability density of $$\frac{1}{3}|R_{21}|^2$$ Is there a more rigorous way of finding this or is the reasoning fine? Thanks a lot. | |
| Jan 6, 2017 at 16:47 | comment | added | Alex | Hi, one more thing. If we now measure both the $z$ component of the spin and the distance from the origin, what is the probability density for finding the particle with spin up and at radius $r$? My reasoning is that we start with the probability density for finding the particle at $(r, \theta, \phi)$ as above (ignoring spin for now), we have $$\frac{1}{3}|R_{21}|^2|Y_{1}^{0}|^2 + \frac{2}{3}|R_{21}|^2|Y_{1}^{1}|^2$$ Hence since we are only interested in the probability density at radius $r$, we integrate over $\theta$ and $\phi$, thus we have the probability density at $r$ given by | |
| Dec 23, 2016 at 21:14 | history | bounty ended | Alex | ||
| Dec 23, 2016 at 15:07 | comment | added | Alex | Yes that is something that I have seen in "Introduction to Quantum Mechanics" by Griffiths, just not in bra-ket notation, but rather using matrices. | |
| Dec 23, 2016 at 15:04 | comment | added | milo | Basically, the representation of $|1⟩$ in the z-eigenbasis is $|1⟩=1/\sqrt{2}(|χ+⟩+|χ−⟩)$. You can get it by doing: $|1⟩=(|χ+⟩⟨χ+|+|χ−⟩⟨χ−|)|1⟩$ | |
| Dec 23, 2016 at 14:59 | comment | added | milo | That was just from memory :) But you can look here: en.wikipedia.org/wiki/… | |
| Dec 23, 2016 at 14:49 | vote | accept | Alex | ||
| Dec 23, 2016 at 14:49 | comment | added | Alex | Yeah I understand thanks for the discussion and answer. Out of interest, how did you get that $\langle \chi_{+}| + \langle \chi_{-}| = \sqrt{2}\langle 1 |$ (positive $x$ direction spin state)? | |
| Dec 23, 2016 at 7:25 | comment | added | milo | Yes, that is correct. Only note that at that stage you are already dealing with "numbers" and not kets or operators, so the tensorial product should be replaced with a regular product. | |
| Dec 23, 2016 at 5:55 | comment | added | Alex | Using the projection operator in your answer I calculated and obtained the following: $\frac{1}{3}|R_{21}|^2|Y_{1}^{0}|^2 \otimes 1 + \frac{\sqrt{2}}{3}|R_{21}|^2Y_{1}^{0*}Y_{1}^{1} \otimes 0 +\frac{\sqrt{2}}{3}|R_{21}|^2Y_{1}^{1*}Y_{1}^{0} \otimes 0 + \frac{2}{3}|R_{21}|^2|Y_{1}^{1}|^2\otimes 1$, is this correct then, and then does this simplify to the desired result $$\frac{1}{3}|R_{21}|^2|Y_{1}^{0}|^2 + \frac{2}{3}|R_{21}|^2|Y_{1}^{1}|^2?$$ | |
| Dec 23, 2016 at 5:55 | comment | added | Alex | Thanks for the extra detail, I think I get what you are saying, it's as in the single dimensional case where the probability density is given as $|\psi(x)|^2 = \psi^*(x)\psi(x) = \langle \psi| x \rangle \langle x| \psi \rangle$. | |
| Dec 23, 2016 at 0:25 | history | edited | milo | CC BY-SA 3.0 |
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| Dec 23, 2016 at 0:02 | comment | added | milo | 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. | |
| Dec 22, 2016 at 23:54 | comment | added | milo | 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. | |
| Dec 22, 2016 at 21:21 | comment | added | Alex | 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 $|(| \mathbf{r} \rangle \otimes | \chi_{+} \rangle \langle \mathbf{r}| \otimes \langle \chi_{+} | + | \mathbf{r} \rangle \otimes | \chi_{-} \rangle \langle \mathbf{r} | \otimes \langle \chi_{-} | ) | \psi \rangle |^2$? | |
| Dec 22, 2016 at 19:04 | history | answered | milo | CC BY-SA 3.0 |