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Operator and measurement

  1. Jun 5, 2009 #1
    Mathematically, I understand the following eigen equation: A|a> = a |a>, where A is an operator, |a> is the eigenstate, and a is the eigenvalue. In terms of mathematics, it is nothing more than a linear transformation.

    However, physically, what does the equation mean? Is it equivalent to the following?
    (1) If we measure observable A, and if the state of the system immediately before the measurement is an eigenstate |a>, then the measurement will not change the state of the system.
    (2) In addition, the measurement will record a value of a for the observable A.

    In general, does A |a> = |b> mean that a measurement of the observable A for a system initially in state |a> sends the system into state |b> ?

    If we accept the above physical interpretation, I run into trouble in the following example. Consider the spin of an electron. Suppose initially the state is in the z+ direction: [tex]|a>=|\sigma_z+>[/tex]= (1 0)[tex]^{T}[/tex]. Note that the Pauli matrix for a measurement in the x direction is
    [tex]$\sigma_x = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) = A$[/tex]

    If we measurement the spin in the x direction, then the resulting state will be
    [tex]$\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) \left( \begin{array}{c} 1 \\ 0 \end{array}\right) = \left( \begin{array}{c} 0 \\ 1 \end{array}\right) = |\sigma_z->=|b>$[/tex]
    So we will observe [tex]|b>=|\sigma_z->[/tex] with certainty? But we are supposed to observe it with probability 1/2.

    Where am I wrong? Thanks.
     
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  3. Jun 5, 2009 #2

    malawi_glenn

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    You have mixed the bases, the (0,1) and (1,0) are not eigen kets to the sigma_x matrix. Thus you have not gain any eigenvalue to the operation. The points (1) and (2) are correct.

    So you have to find the spin-z up ket (1,0) in terms of eigenstates to the sigma_x operator, then after that you will perform the eigenvalue equation and you will obtain the desired result.

    To find the eigenstates to sigma_x, just do as in ordinary linear algebra, and then normalize them. Then you find out how to expand (1,0) in terms of those two new basis (the x-basis) and then operate with sigma_x on that linear combination.

    GL HF
     
  4. Jun 5, 2009 #3
    Hi malawi_glenn,

    Thanks for your explanation. I understand what you said. What bothers me is: what does the equation [tex]\sigma_x |\sigma_z+>=|\sigma_z->[/tex]
    mean? According to Sakurai's book, applying any operator to a ket produces a new ket. Does the equation mean that after a measurement [tex]\sigma_x[/tex], the state [tex]|\sigma_z +>[/tex] is transformed into [tex]|\sigma_z ->[/tex]?

    Not only that, the dimension is wrong too. To see this, consider [tex]S_x, |S_x\pm>, |S_z\pm>[/tex] instead.
    1. We know
    [tex]S_x |S_x+>=\frac{\hbar}{2} |S_x+>[/tex]
    so is
    [tex]\frac{\hbar}{2} |S_x+>[/tex]
    a new ket? But the dimension is not right since we have an extra term [tex]\hbar/2[/tex].
    2. We know
    [tex]S_x |S_z+>=S_x (\frac{1}{\sqrt{2}}|S_x+> + \frac{1}{\sqrt{2}}|S_x->)=\frac{\hbar}{2} \frac{1}{\sqrt{2}}|S_x+> - \frac{\hbar}{2} \frac{1}{\sqrt{2}}|S_x->=\frac{\hbar}{2} |S_z->[/tex]
    so is [tex]\frac{\hbar}{2} |S_z->[/tex] the new state?
     
  5. Jun 5, 2009 #4

    malawi_glenn

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    The constant n front of the ket doesn't matter since we are working with rays in Hilbert Space.

    You can of course solve this easy since the operator has units h-bar as well.. so the units on the LHS equals the units on RHS.
     
  6. Jun 5, 2009 #5

    Fredrik

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    It definitely doesn't mean that. One thing it does mean is that [tex]|\sigma_z +>[/tex] isn't an eigenvector of [itex]\sigma_x[/itex]. If it had been an eigenvector, let's say with eigenvalue s, then that would have meant that you will get the result s with certainty (and leave the system in the same state it was in before the measurement).

    An ideal measurement that doesn't destroy the system always leaves it in an eigenstate of the observable that's being measured. For example, if you measure [itex]\sigma_x[/itex] when the state is [tex]|\sigma_z +>[/tex], then the particle will be left in the state [tex]|\sigma_x +>[/tex] (if the result is +1/2) or the state [tex]|\sigma_x ->[/tex] (if the result is -1/2). It's always one of those two.
     
  7. Jun 5, 2009 #6
    Got it. But what exactly does
    [tex]S_x |S_z+>=\frac{\hbar}{2}|S_z->[/tex]
    mean?

    Thanks.
     
  8. Jun 5, 2009 #7

    Fredrik

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    What do you mean? It's an equation, so it tells us the relationship between the mathematical quantities that appear in it. I told you about the most important thing that's implied by this particular relationship. If that's not the answer you're looking for, then what are you looking for? A complete list of all the things it implies?
     
  9. Jun 6, 2009 #8
    Fredrik and malawi_glenn,

    I seem to understand a little better. I think the equation
    [tex]S_x |S_z+>=\frac{\hbar}{2}|S_z->[/tex]
    itself has nothing to do with any measurement. [tex]S_x[/tex] represents a unitary transformation only.

    To measure an observable, we need a projective operator of that observable. For example, to measure [tex]S_x +[/tex] of a system of state [tex]|\phi>[/tex], we first apply the projective operator [tex]|S_x+><S_x+|[/tex] to [tex]|\phi>[/tex], i.e.,
    [tex]|S_x+><S_x+| \phi > [/tex]
    and then normalize the resulting state vector.

    Please let me know whether my understanding is right or wrong. Thanks.
     
  10. Jun 6, 2009 #9

    malawi_glenn

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    yes that is better! That is the procedure I told you about in post #2.

    For a measurement, we need the eigenvalue relation. and

    [tex]
    S_x |S_z+>=\frac{\hbar}{2}|S_z->
    [/tex]

    is not an eigenvalue relation, but operate with S_x on |S_x+><S_x+| \phi > will result in an eigenvalue relation.
     
  11. Jun 6, 2009 #10

    Fredrik

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    [itex]S_x[/itex] is hermitian (i.e. self-adjoint), but not unitary. ([itex]\sigma_x[/itex] happens to be unitary, but that's more of a coincidence than anything else). If you meant that it represents a unitary time evolution, then I would disagree with that, because a spin-component operator doesn't define a Hamiltonian by itself, and it's the Hamiltonian that defines the time evolution.

    Now you're describing how to calculate a state that 50% of the time will be the state after the measurement. The other 50% of the time, you'd have to use the other projection operator. There's actually a neat way to express the time evolution of the state during a measurement, using density operators to represent states instead of vectors.

    [tex]\rho\rightarrow\sum_a P_a\rho P_a[/tex]

    where [itex]\rho=|\psi\rangle\langle\psi|[/itex] represents the state before the measurement, and [itex]P_a=|a\rangle\langle a|[/itex]. (I have no idea why the summation symbol looks so weird).
     
  12. Jun 6, 2009 #11
    Yes, you are right. An operator is usually Hermitian, but not necessarily unitary. The time evolution operator is unitary, being a sufficient condition for the probability preservation (=1).

    I am trying to figure out why the above relation is true. I think after a measurement of eigenstate [itex] a [/itex], the new density matrix will be [itex] |a \rangle \langle a|[/itex].

    The density matrix [itex]\rho'=\sum_a P_a\rho P_a[/itex] is for a mixed ensemble for which we measure the observable corresponding to [itex]|a\rangle[/itex] according to the proabability composition of [itex]|\psi \rangle[/itex]. The derivation is as follows.
    [tex] \begin{array}{lll}\rho' & = & \sum_a \mbox{Prob(new state is in eigenstate } a \mbox{)} |a \rangle \langle a| \nonumber \\
    &=& \sum_a (\langle a| \psi \rangle \langle \psi | a \rangle) |a \rangle \langle a| \\
    &=& \sum_a |a \rangle (\langle a| \psi \rangle \langle \psi | a \rangle) \langle a| \\
    &=& \sum_a |a \rangle \langle a| (|\psi \rangle \langle \psi |)| a \rangle \langle a| \\
    & = & \sum_a P_a \rho P_a
    \end{array}[/tex]

    Please let me know whether I am right or wrong. Thanks.
     
  13. Jun 6, 2009 #12

    Fredrik

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    Yes, if you know that the result of the measurement was a, then [itex]|a\rangle\langle a|[/itex] is the correct density matrix. The expression I used represents the situation when a measurement has been performed and we don't know what the result is. I should have been more clear about that.

    Your derivation is fine.

    Note that there are two seemingly different situations in which the density operator can be used: 1. You're dealing with an ensemble of systems. 2. You're dealing with a single system in a specific but unknown state. I said "seemingly", because the second situation is actually a special case of the first. This is because saying that the probability of obtaining the result a is P(a) really means that if you make the same measurement on a large number of identically prepared systems you will obtain the result a with frequency P(a). This set of identically prepared systems is an ensemble too.
     
  14. Jun 6, 2009 #13
    Fredrik and malawi_glenn,

    I am trying to reply to and comment on a few posts at once. Please let me know whether my understanding is right or wrong. Thanks.

    I am not sure if I understand situation 2 right. I think by "a single system in a specific but unknown state", the single system is in a pure state, and therefore density matrix is not particularly useful. How about this substitution: We have a single system in a specific but unknown state, and we want to describe the new state after a measurement is made.

    The new state will have a proability distribution over all possible eignestates. To determine the distribution, we need to carry out the same measurement for many identically prepared systems. The result is that we will end up with an ensemble of systems, and they are in different eigenstates. So we need the density matrix to describe them.

    We can't measure an eigenstate. When we make a measurement, we throw the system into one of a number of eigenstates. So "I think after a measurement of eigenstate [itex] a [/itex]" should be "If we observe an eigenvalue [itex]a[/itex] after a measurement".

    The question is why and when do we encounter the above equation? One answer is that the above equation may be part of the calculation of a projection measurement such as in the Stern-Gerlach experiment (keeping the [itex]S_x+[/itex] beam after passing the atoms through an uneven magnetic field in the x+ or x- axis).
    [tex]|S_x+\rangle \langle S_x+| = \frac{\sigma_x + 1}{2} = \frac{S_x}{\hbar} + \frac{1}{2}[/tex]
    Applying [itex]|S_x+\rangle \langle S_x+| [/itex] to a state [itex]|S_z+\rangle[/itex] to be projected will have the term
    [tex]S_x |S_z+>=\frac{\hbar}{2}|S_z->[/tex]

    Are there other situations we use [itex]S_x |S_z+>[/itex]?
     
  15. Jun 6, 2009 #14

    Fredrik

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    That's not what I meant. I suppose I could have been more clear. If we subscribe to one of the interpretations that say that state vectors represent objective properties of physical systems, then we have to say that the system is in a pure state after a measurement. But if we don't know the result of the measurement, we would still have to use a density operator, because if we repeat the experiment many times we would be dealing with an eigenstate |a> some fraction of the time and another eigenstate |a'> some fraction of the time. The systems that are involved in all those experiments can be thought of as part of one ensemble.

    If we don't make the unnecessary (and in my opinion completely unjustified) assumption that states represent objective properties of physical systems, then it makes more sense to say that the state after the measurement (with a result that's unknown to us) is the density operator.

    If state vectors don't represent objective properties of physical systems, then QM is just an algorithm that tells us the probabilities of possible results of experiments, given the results of other experiments. This is equivalent to saying that state vectors represent objective properties of ensembles of identically prepared systems (the many copies of the system that are used when we test the theory's predictions). So the fact that a density operator represents statistical properties of an ensemble doesn't mean that it's weird to say that the state of a single system is a density operator. State vectors aren't really that different. They too are representations of statistical properties of ensembles.
     
  16. Jun 21, 2009 #15
    Fredrik, Thanks for the explanation, and I am with you now.
     
  17. Jun 21, 2009 #16

    Hurkyl

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    So what exactly do you mean by this? There are couple aspects:

    (1) What is an "objective property of a physical system"? Normally I would consider this obvious, but I have seen people who appear to use the phrase in a very limited way, somewhat synonymous with "local realism".

    (2) Is this your overarching view on the philosophy of science (e.g. you would say that Special Relativity doesn't represent objective properties of systems either)? Or are you asserting that QM is less qualified to speak about objective properties of systems than other theories like SR or classical mechanics?
     
  18. Jun 21, 2009 #17

    Fredrik

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    I only have time for a relatively short answer right now, so I'll focus on (2).

    It's possible to have two classical theories that make the same predictions about the results of experiments, but describe things in different ways. (See below for an example). I would say that the mere possibility of the existence of two different theories that make the same predictions is a good reason not to think of a theory (even a classical theory) as "an approximate description of our universe". It makes a lot more sense in my opinion to think of a classical theory as "an exact description of a fictional universe in which the results of experiments are approximately the same as in ours".

    I don't mean that it's absolutely horrible to say that SR describes our universe (approximately). I just consider it a mild abuse of the appropriate terminology, not much worse than calling [itex]\mathbb R^n[/itex] a vector space. In other words, I think it's OK.

    However, I believe that a strong case can be made for the idea that QM can't be a description of any universe, even a fictional one. I'm a little reluctant to say that in such an assertive way because I know you'll make me try to prove it (as you should). It won't be today though. You can start by finding the other thread where we talked about this. I remember that I tried to argue for this view there too, but right now I don't even remember the whole argument I used there.

    _____________

    The example: I believe that the alternative version of GR that was mentioned in a fairly recent post gives us a good example. Spacetime is flat in this theory, but instead of spacetime curvature, we have a deformation of our measuring devices (along with the rest of the matter in the universe I presume). This theory makes the exact same predictions as standard GR. The details of this theory aren't important, and I don't know them anyway. I just wanted an example to make things more concrete.
     
    Last edited: Jun 22, 2009
  19. Jun 22, 2009 #18

    Fredrik

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    I found the other thread. This is the only post that's relevant. It doesn't prove my claim, but at least it's a relevant observation. The argument in sections 9.1-9.3 of Ballentine is stronger I think. You should probably check it out. Ballentine argues that a state vector isn't "a complete and exhaustive description of an individual system". The alternative is that it "describes the statistical properties of an ensemble of similarly prepared systems".
     
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