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Quantum Mechanics - spin confused?

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider an electron, described initially by psi = 1/sqrt(10) with column vector (1 3). A measurement of the spin component is made along a certain axis, described by operator A, it has the eigenvalues +h/2 and - h/2 as possible outcomes, and the corresponding eigen states are;

    Psi(1) = 1/sqrt(10) column vector (1 3) and Psi(2) = 1/sqrt(10) column vector (3 -1)

    Explain why A returns the result h/2 with certainty.

    for the second part you must measure spin is the Z component


    2. Relevant equations

    Eigen equation A psi(1) = a(1) psi (1)

    3. The attempt at a solution

    Originally i thought the reason why it produced with certainty h/2 was because when you operate on a wavefunction, the eigenstate that is produced is the same as the initial wavefunction, thus the second eigenstate is different from that of the initial wave function and cannot be correct.

    Whats confusing me is if this is true, why does psi 1 still have an amplitude associated with spin down i.e the column vector (1 3) (the 3?)

    Could someone perhaps explain to me what it means to have two eigenstates corresponding to an operator?

    For part b, i get the same constant as my initial wavefunction however my column vector becomes (1 -3) as the eigenstates for spin are always either +h/2 or -h/2 - this again confuses me, as i thought the eigenstate had to be the same as the initial wavefunction (from the eigen equation)
     
  2. jcsd
  3. Feb 25, 2013 #2
    When you measure an observable of an arbitary system, for example it's spin, you will get a result which is an eigenvalue of this observable, that is to say it's operator representation. Thus the arbitary system collapses into it's appropriate eigenstate corresponding to the measured eigenvalue.

    So if the system initally consists of a mix of both states, spin up and spin down in this case, a measurement will force the system to collapse into either spin up, or spin down. There is no middle ground in quantum mechanics. You cannot get a value that would be h or h/3 when measuring the spin of a spin 1/2 fermion. Only it's eigenstates can be measured.

    Ofcourse measuring the spin of an ensemble of fermions could give you an average of 0 or h/3 or anything between -h/2 and h/2 for that matter. Doing this cascade of measurements on a system of fermions will let you deduce it's initial state.

    Also to help you understand this a little better, take the two states you are given. Confirm that they are orthogonal (linearly independent). If they are, you can easily rotate and scale those vectors to give you a more appropriate set of states. For the spin, you have 2 independent states so you can choose a simple basis (1 0) and (0 1) corresponding to spin up and spin down respectively for example.

    Please correct me if i'm wrong or something remains vaque.
     
  4. Feb 27, 2013 #3
    Thanks for the reply Thoros.

    They are indeed orthogonal, as they yield different eigenvalues.

    Does this mean i can take the eigenstates and multiply them with (1 0) and (0 1) (This will then tell me eigenstates of spin up and spin down respectively?)?

    What i think was confusing me initially, was when i operated on the initial wavefunction with the spinor(z) the wavefunction it returned was not the same as the initial wavefunction (due to the Sz operator having a -1 in it (as its a 2x2 matrix). However, i now believe this is just telling me that it is not an eigenstate of Sz. This is not surprising as the initial wavefunction is a eigenstate of the arbitrary operator A, so i suppose this means that they are not compatible.
     
  5. Feb 27, 2013 #4
    The (1 0) and (0 1) are the eigenstates of spin. Also multiplication of states is ambiguous unless explicitly saying what do you mean under multiplication. You can basically multiply them in two ways. Either to get a scalar using a scalar product, or a bigger object using a tensor product our outer product i think.

    The scalar product is the one you will be using 99% of the time. Basically you can construct any state vector using those 2 basis vectors (1 0) and (0 1). A scalar product with each of them will give you the amplitude for the state to collapse into that state.

    Exactly. Geometrically speaking, when an operator acts on a state-space element, that is a fancy word for your 2-dimensional state vector, it rotates your vector in that space. When a operator acts on its eigenvector (or eigenstate, same thing) it effectively leaves it unchanged. Normally, a change in scale is allowed, that is the length of the vector can be altered and even mirrored through the origin. For example, a unit vector pointing in the z direction is an eigenstate (eigenvector) of rotations about the z axis. (this is the best way for me to understand this atleast).
     
  6. Feb 28, 2013 #5
    You're completely right, i should have said scalar product. Right, so if i take the scalar product with my initial wavefunction and (0 1), this will give me the amplitude for my initial wavefunction to collapse into a spin down state? (take the modulus square and ill know the probability).

    I think i'm beginning to get a better feel for it, let me see if i follow your understanding...

    When i operate on a state vector, if it is an eigenstate of that operator it will be invariant, i.e wont change?
    If i operate on a state vector and it isn't an eigenstate of that operator it will rotate (and thus change), does this mean that this is not an allowed state for operator 'A', and in principle has a 0% chance of being an observed quantity (or its eigenvalue at least).

    I like how you think of the vector space etc, i think i find it easier to imagine what's happening, can you recommend any literature (particularly around these area's as i believe its the matrices, eigenvalues, operations etcetc which i may be weak at)? I enjoy QM, but it becomes very difficult without more knowledge in these particular mathematical area's, most the books i have already seem to just bomb through the maths and don't give you descriptive ways of thinking about it(like you have) or the opposite, tell you what you are dong in principle but not showing you the maths that goes with it. Hope this makes sense and i haven't rambled too much....

    Thanks again, you've been extremely helpful.

    Forgot to add: It still confuses me slightly, if you can always make a state vector (is this the same thing as an eigenstate?) by using (1 0) or (0 1), what is the point in having the 3 (2x2) matricies for spin(x), spin(y) and spin(z)? Or is just that these are not necessary if you already know the initial wavefunction?
     
  7. Feb 28, 2013 #6
    Yes, but beware. Since your state vector components are allowed to be complex. So a normal scalar product wont do, because it may become negative. This if fixed by the convention that you usually take the left column vector, transpose it to a row vector and take the complex conjugate of each entry (the transpose and complex conjugating together form what's called the Hermitian conjugate denoted by a dagger superscript). And you multiply them using the rule of multiplication for matrices.

    Yes, like when measuring spin, you can only get + 1/2 or - 1/2 (*hbar). Getting anything else has 0 probability. Getting 1/2 on the other hand can have a probability from 0 to 1.

    Well this is because you can measure the spin in every direction. I think i've been a little hasty on the part with (1 0) and (0 1) being the only possibilities. The values can be complex. The (1 0) and (0 1) correspond to the spin states in the z directions. You should find out the eigenvectors and eigenvalues for each of the pauli matrices on your own.
    Try the Wikipedia article on Pauli matrices - http://en.wikipedia.org/wiki/Pauli_matrices

    As for the literature part. My choice would be "Mathematical Physics, a modern introduction to it's foundations" by Sadri Hassani https://www.amazon.com/Mathematical-Physics-Sadri-Hassani/dp/0387985794/ref=la_B001HCWA08_1_1?ie=UTF8&qid=1362055719&sr=1-1
    It's relatively cheap but covers pretty much everything you need to know from simple algebra to general relativity.
     
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