Problem 4.28b on page 159 of David Griffith’s “Introduction to Quantum Mechanics” says: An electron is in the spin state X = A[3i ; 4] . find < Sx >, < Sy >, and < Sz > < Sx > = 0 < Sy > = - (12/25) hbar < Sz > = - (7/50) hbar Could someone help me understand what this problem is telling me? Since < Sx > = 0 does this mean the spin must be in the yz plane? How would the spin get into this state--a prior measurement? Any interpretation of what this is all about would be greatly appreciated.
Also is there a way to determine what measurement would guarantee a result of +hbar/2 (spin up)? Oh, and A = 1/5.
yeah, that s what i got. seems correct to me. No!! this says that if you measure the x-component of the spin, you will get an average value of 0. but this spin state is not an eigenstate of Sx, so i won t say anything like "the spin is in the yz plane". a more accurate way to say it is that this state doesn t have a definite value of Sx. it is in some superposition of x-component spin up and spin down. yeah, this is not an eigenstate of spin along any of the coordinate axes, but it is an eigenstate of spin along some axis. you could prepare this state by measuring the spin along that axis, and then all the spins would collapse to either spin up or spin down along that axis. i suppose we could calculate what that axis is.
yes: measure the spin along the axis for which this spin state is an eigenstate, and you are guaranteed a spin up measurement.
Thanks lethe but here’s what’s confusing me. If you run an ensemble of electrons through the Stern Gerlach apparatus and measure the z component you’ll put them into an eigenstate of Sz. (Am I using the correct teminology here?) If you then measure the x component you’re equally likely to get a spin up or spin down result which would seem to imply that < Sx > = 0. So it would seem that if < Sx > = 0 the original spin state would have to have been perpendicular to the x axis or in the yz plane. Supposedly spin is one of the easiest things to understand but it’s confusing as hell to me. Can you determine this from the expectation values? James R That is just my clumsy way of writing a column vector.
yes, this is correct. actually..... i think maybe you are correct. i balked before because i don t like you to say things like "the spin is such and such". much better to talk about what the results of a spin measurement will be, or what spin state the particle is in (which is not a definite spin along any of the axes in question) but i guess the point you were getting at, which is correct, is that if <Sx>=0, and if we find the axis for which this spin state is an eigenstate, then that axis will be in the yz plane. if that axis had any component parallel to the x-axis, then there would be some nonzero expectation value for Sx. saying it the above way is less offensive then saying "the spin of the particle is in the yz plane". as you know, if you measure Sx once, you will definitely get a nonzero result. i don t know who told you that spin was easy, but i think spin is one of the deepest rabbit holes in quantum mechanics. i feel it takes years to understand spin. i guess what you mean is (maybe Griffiths says something along these lines) the spin Hilbert space is finite dimensional, so it is a lot simpler than the L<sup>2</sup>(R) Hilbert space. but that only makes some aspects of the mathematics easier. it doesn t make the issue of measurement in quantum mechanics any easier (which is what you are talking about in this thread), and it belies the fact that there is some really deep mathematics behind the meaning of spin (especially half integer spin) can i determine which axis this spin state is an eigenstate from the expectation values alone? is that what you are asking? hmmm.... lemme think.... certainly we can determine the axis from knowing the spin state, and i m pretty sure that we can determine the spin state from the expectation values. lemme see: the spin operator for an arbitrary axis is S*n where n is a normal vector. um... i m gonna try it out on paper. i ll get back to you in a minute
yes, you can determine the spin state from the expectation values (up to an overall phase), and from there, you can determine the axis for which the spin state is an eigenstate. give it a try. if you want to see the calculation, i will post it.
Yes, now I see what you are saying. Yes, he says it's so simple that many authors start out with it. Thanks. That makes me feel a lot less stupid.
I would very much like to see it. Like many others here I feel I learn something new everytime you write a post. (Plus my brain hurts from trying to understand spin and spin math)
alright suppose we know the expectation values of S: <S<sub>x</sub>>=A <S<sub>y</sub>>=B <S<sub>z</sub>>=C we want to solve for the spin state χ=[χ<sub>1</sub> χ<sub>2</sub>]<sup>T</sup> we know that χ is normalized, so that χ<sub>1</sub>*χ<sub>1</sub> + χ<sub>2</sub>*χ<sub>2</sub> = 1 χ<sub>1</sub>*χ<sub>2</sub> + χ<sub>2</sub>*χ<sub>1</sub> = A χ<sub>1</sub>*χ<sub>2</sub> - χ<sub>2</sub>*χ<sub>1</sub> = iB χ<sub>1</sub>*χ<sub>1</sub> - χ<sub>2</sub>*χ<sub>2</sub> = C using the normalization, and the last equation, we see that |χ<sub>1</sub>| = √((1+C)/2) and |χ<sub>2</sub>| = √((1-C)/2) now i will switch to polar coordinates: χ<sub>1</sub> = r<sub>1</sub>e<sup>iθ<sub>1</sub></sup> and likewise for χ<sub>2</sub> the first equation becomes cosh (θ<sub>2</sub> - θ<sub>1</sub>) = A/√(1-C<sup>2</sup>) and the second becomes sinh(θ<sub>2</sub> - θ<sub>1</sub>) = iB/√(1-C<sup>2</sup>) so now we have: χ = e<sup>iθ<sub>1</sub></sup>[√((1+C)/2)e<sup>i cosh<sup>-1</sup>A/√(1-C<sup>2</sup>)</sup> √((1-C)/2)e<sup>i cosh<sup>-1</sup>A/√(1-C<sup>2</sup>)</sup>]<sup>T</sup> and there it is.
Conservation of angular momentum ? Or S<sup>2</sup> ? I think you can prove something along the lines of A<sup>2</sup> + B<sup>2</sup> + C<sup>2</sup> = 1 but I would have to think on it. Bye! Crisp
yeah, something like this should work, right? we know that <S<sup>2</sup>> = 3/4 for a spin-1/2 particle. and <S<sup>2</sup>> = <S<sub>x</sub><sup>2</sup>> + <S<sub>y</sub><sup>2</sup>> + <S<sub>z</sub><sup>2</sup>> this is not quite the same thing as <S<sub>x</sub>><sup>2</sup> + <S<sub>y</sub>><sup>2</sup> + <S<sub>z</sub>><sup>2</sup> but perhaps with a little work, we can make them equal. i will think about it as well. i should point out, Sacroiliac, that i have a little more work to do to finish the calculation you asked for. you asked to find the axis for which a given spin state is an eigenstate, starting from only the expectation values. i have calculated the spin state starting only with the expectation values, but i have not yet calculated the axis of rotation. i ll get to that in a bit as well.
Indeed, you get the variance <S<sub>x</sub><sup>2</sup>> - <S<sub>x</sub>><sup>2</sup>. From this we know already that the three expectation values are dependent: the variance is constant (just like the expectation value) and we can relate all <S<sub>i</sub><sup>2</sup>> to the <S<sub>i</sub>><sup>2</sup> this way. And for the former we have a restriction. You can put it all in a formula if you like, but the fun part is over I guess Please Register or Log in to view the hidden image! Bye! Crisp
lethe I'd be happy just knowing how to find the axis for which the given spin state X =[3i 4]<sup>T</sup> is an eigenstate.
I think one way to do that would be to determine for what linear combination of Pauli matrices, the state X is an eigenstate.
Just 2 remarks 1: You have solved the equations: <σ<sub>x</sub>>=A <σ<sub>y</sub>>=B <σ<sub>z</sub>>=C but this is no problem' just take your results and replace A by 2A, etc. 2. From the equations cosh (θ<sub>2</sub> - θ<sub>1</sub>) = A/√(1-C<sup>2</sup>) and sinh(θ<sub>2</sub> - θ<sub>1</sub>) = iB/√(1-C<sup>2</sup>) Since cosh<sup>2</sup>x - sinh<sup>2</sup>x =1 (for any x) one finds that A, B and C must necessarily fulfill the relations: A<sup>2</sup> + B<sup>2</sup> + C<sup>2</sup> = 1.
i m not sure what you are saying here. why would i replace A by 2A? yes, very good. i actually saw that too, but i was really actually hoping for a physical justification why A, B and C are not independent. something along the lines of the argument Crisp was making. could we have known that they are not independent without actually calculating the the solution?
