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I'm looking to improve my understanding of quantum superposition. Consider the two-slit setup, where $|A\rangle$ is the state of the electron at the source, $|z\rangle$ is the state at some point on the detector, and $|S_{1}\rangle$ and $|S_{2}\rangle$ are the states at the two slits. As I understand it, the probability of finding the electron in $|z\rangle$ is:

$\langle z|A\rangle= \langle S_{1}|A\rangle\langle z|S_{1}\rangle + \langle S_{2}|A\rangle\langle z|S_{2}\rangle$

which yields

$|A\rangle= \langle S_{1}|A\rangle|S_{1}\rangle + \langle S_{2}|A\rangle|S_{2}\rangle$

Does this imply that $|A\rangle$ is a superposition of states $|S_{1}\rangle$ and $|S_{2}\rangle$? I'm confused as to what this would mean -- given the electron is in $|A\rangle$, there is a chance upon measuring it to find it in $|S_{1}\rangle$ or $|S_{2}\rangle$?

[I found this at: http://physics.mq.edu.au/~jcresser/Phys301/Chapters/Chapter7.pdf ]

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Maybe a way for you to be not confused is to imagine a time dependence. For instance, let suppose three times $t_i, t, t_f$ with $t_i < t < t_f$. One may suppose that the particle is in the initial state $|A\rangle$ at time $t_i$, is in the final state $|z\rangle$ at time $t_f$, and, at the intermediary time $t$ is in one of the $2$ states $|S_1\rangle$ or $|S_2\rangle$. The law of composition of amplitudes say that :

$\langle z, t_f|A,t_i\rangle =\langle z, t_f|S_1,t\rangle\langle S_1, t|A,t_i\rangle + \langle z, t_f|S_2,t\rangle\langle S_2, t|A,t_i\rangle$

This is true for all $z$, so we have :

$|A,t_i\rangle = |S_1,t\rangle\langle S_1, t|A,t_i\rangle + |S_2,t\rangle\langle S_2, t|A,t_i\rangle$

Now, you may interpret this equation as follows : Given that the particle is in the state $|A\rangle$ at time $t_i$, the probability amplitude to find the particle in the state $|S_1\rangle$, at time $t$, is : $\langle S_1, t|A,t_i\rangle$

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  • $\begingroup$ So then any state is considered a superposition of all states in which the system could be found at a certain (later) time? $\endgroup$ – sschlassa Dec 17 '13 at 1:33
  • $\begingroup$ This is correct. However, my argument was only to maybe help you to have a different view of the problem (more rigour would be necessary, in fact). The textbook you gave in reference is very well written and totally correct, so you don't need time dependence at all, for instance, to write $|A\rangle= \langle S_{1}|A\rangle|S_{1}\rangle + \langle S_{2}|A\rangle|S_{2}\rangle$. So I do not want to confuse you. It is better to follow the formalism used in your reference. $\endgroup$ – Trimok Dec 17 '13 at 9:46
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Let's say you've prepared an electron n the state: $$|\psi\rangle= \alpha|0\rangle + \beta|1\rangle$$ And you want to measure it in the state: $$|\phi\rangle= \alpha'|0\rangle + \beta'|1\rangle$$ Then the probability of the electron collapsing into the state $\phi$ upon measurement is: $$|\langle\phi|\psi\rangle|^2$$ This means that it is possible for an electron prepared $\psi$ to collapse into any $\phi$. To answer your question, $\psi$ is indeed in a superposition of $|0\rangle$ and $|1\rangle$. It can collapse into either of them upon measurement, and $|\phi\rangle = |0\rangle + 0|1\rangle$ and $|\phi\rangle = 0|0\rangle + |1\rangle$ respectively.

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