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Beginner to Braket State Vectors

  1. Jun 14, 2008 #1
    I'm trying to understand what Braket state vectors are... and I'm not sure if I completely understand what's going on. Could someone clear this up? Here's what I (think) I understand now.

    A ket is pretty much a state vector (written with different notation). A state vector is pretty much a vector corresponding to a particular observable. It has everything you need to know about the system. If you want to find (statistical) information regarding a particle, suppose the total energy, then you have to solve the eigenvalue equation

    [tex]\hat{H} \psi = E \psi[/tex]

    where [tex]\hat{H}[/tex] represents the operator regarding the observable and [tex]E[/tex] represents the eigenvalue. Here, [tex]E[/tex] is the mean value observed for the observable (in this case, the mean energy observed).

    What I am completely missing is the concept of adding together state vectors... how exactly does this work? And most importantly, what does this mean? What can we do with them? I'm really sorry if I don't understand some of what seem to be the easier concepts, but I learned everything that I know from online sources only. :uhh:

    In addition, I am sort of unsure as to how the entire complex number idea works with braket state vectors... am I understanding correctly that ket state vectors (and bra vectors) are in some N-dimensional (N isn't necessarily finite o.0?) complex vector space? If this is the case, what is the use of having bra vector if we already have ket vectors, which are just the conjugate transposes of bra vectors? I guess that they are used for inner and outer products, but could someone give me an actual example or a link to an introduction example where these are actually used to solve problems?
     
    Last edited: Jun 14, 2008
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  3. Jun 14, 2008 #2
    Actually its exactly the state vector (not pretty much).
    A state vector can correspond to more than one observable.
    E is not a mean value. Its a particular value. The entire spectrum of possible eigenvalues for that vector determine the mean value.
    They can be used to find anything you want to about a quantum mechanical system. E.g. they can be used to determine the uncertainty of a given observable as well as the mean value.
    Complex numbers come into play when one is determining things like the mean value of the uncertainty of an observable. The vector itself doesn't have a real or complex value. Only things like the norm of a vector or the inner product of a two vectors.
    bra vectors are required in order to form the inner product of two vectors. The inner product represents things that are actually measured.

    Pete
     
  4. Jun 14, 2008 #3
    Ah okay, I think I understand now. So let's then say that we have two states, |A> and |B>, the former of which corresponds to a photon propagating in the x-direction, while the latter which corresponds to a photon propagating in the z-direction. Let's also say that |C> = |A> + |B>. Then does this mean that, after |C> is normalized, then |C> pretty much means the photon which propagates in the x and z directions (with a 45 degree angle)? Or does this mean that there is a probability that the photon propagates in the x and z directions and in fact propagates in only one direction (assuming I'm a Copenhagen-ist)? Also, what would <A|, <B|, and <C| signify physically?
     
  5. Jun 14, 2008 #4

    Fredrik

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    You need the scalar product to calculate probabilities. If the system is in state [itex]|\alpha\rangle[/itex], and [itex]|n\rangle[/itex] is an eigenvector of an observable H with eigenvalue [itex]E_n[/itex], then the probability that a measurement of H will yield the result [itex]E_n[/itex] and put the system into the state [itex]|n\rangle[/itex] is [itex]|(|n\rangle,|\alpha\rangle)|^2[/itex], where I'm using the notation (x,y) for the scalar product of two vectors x,y.

    (Some other common notations are <x,y>, (x|y) and <x|y>. The last one may be the most common, but it's very confusing to use it while explaining the bra-ket notation).

    A ket is just a vector in a Hilbert space H. A bra is a member of the dual space H*, i.e. it's a linear function from H into the set of complex numbers. The bra [itex]\langle\beta|[/itex] is the member of H* that's defined by

    [tex]\langle\beta|(|\alpha\rangle)=(|\beta\rangle,|\alpha\rangle)[/tex]

    There's a theorem that guarantees that there exists exactly one such bra for each ket. The left hand side isn't usually written like that. The convention is to drop the parentheses and one of the |s, and write it as [itex]\langle\beta|\alpha\rangle[/itex]. So the conventional way to write the probability mentioned above is [itex]|\langle n|\alpha\rangle|^2[/itex].

    We can define the sum of two bras in an obvious way, and the product of a complex number and a bra in an equally obvious way. These definitions turn H* into a vector space.

    Note that we don't really need the dual space in quantum mechanics. We're just using it because it simplifies the notation sometimes.

    Here's an example of the same equation written without bras and then with bra-ket notation. It's not meant as an example of why the notation is useful. It's just an example of how it's used.

    [tex](|\alpha\rangle,A|\beta\rangle)=(A^\dagger|\alpha\rangle,\beta\rangle)[/tex]

    [tex]\langle\alpha|(A|\beta\rangle)=(\langle\alpha|A)|\beta\rangle[/tex]
     
  6. Jun 14, 2008 #5

    Fredrik

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    It means that right now it's neither propagating left nor right. It's in a state that doesn't correspond to a possible result of any measurement of its direction. It also means that there's a 50% chance that a measurement will find it propagating right, and a 50% chance that a measurement will find it propagating left. After the measurement, the photon is either in state |A> or |B>, depending on the result.

    The same things as |A>, |B> and |C>.
     
  7. Jun 14, 2008 #6

    G01

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    This is an example of a system being in a superposition of eigenstates.

    The thing is, if you try to measure the observable (in this case direction) you will always obtain one of your eigenvalues. i.e. If you try to find out which way this particle is going, your never going to measure it going in the +45 degree direction. You'll either find it going up or right, not at +45 degrees, since this isn't a "direction eigenstate" in this hypothetical made up example.

    The photon's wave function will "collapse" into one of the eigenstates, either |A> or |B>, when measured, thus giving the direction associated with the one it collapsed into. The probability of the system collapsing into one state vs. the other is given by the square of the coefficient in front of each eigenvector. Here, when properly normalized we'll have:

    [tex]|C>=\frac{1}{\sqrt{2}}|A> + \frac{1}{\sqrt{2}}|B>[/tex]

    Thus, here we have a probability of 1/2, or 50%, of the system being found in either "eigendirection" (Has a nice ring to it:smile:)

    In this next case we have a probability of 1/10 for the A direction, and 9\10 for the B direction:

    [tex]|C>=\frac{1}{\sqrt{10}}|A> + \frac{3}{\sqrt{10}}|B>[/tex]

    So, this hypothetical state doesn't really describe a photon going in the +45 degree direction. You second explanation is the better one.
     
  8. Jun 15, 2008 #7
    This sounds like an incomplete question to me. Did you mean to write more?

    |C> is a state which is the superposition of eigenstates and as such does not represent a state which is propagating in either direction.
    Yes. There is a probability that the photon propagates in the x and z directions .
    No. Nothing can be said about what direction it is going until a measurement is made.
    Each represents the quantum state that the letter refers to. E.g. <A| represents |A>. <A| is said to be a dual vector to |A>. In relativity a vector might be represented by [itex]V^\mu[/B] while there is a dual vector is represented by [itex]V_\mu[/itex] and is known as a 1-form. The quantity <A| is used to form an inner product which is formed by justipositioning a bra and a ket as (<A|)(|B>) = <A|B>. If |A> is the state of a system and Q is a phyical observable (which is an operator) then <A|Q|A>2 is the probability density if measuring q, an eigenvalue of Q.

    Pete
     
  9. Jun 15, 2008 #8
    Hmm... actually, as far as writing more for the first question, I mean the last question to be: Then does this mean that, after |C> is normalized, then |C> pretty much means the photon which propagates in the x OR z directions (with a 45 degree angle)? (as opposed to an and)

    And can't we actually determine some information regarding which way the photon goes (x or z directions) before making the measurement? Particularly, I mean the probabilities that it goes in a particular direction. Hmm... and if a state is a mixture of states, say

    [tex]\psi = a_1 \psi_a + b_1 \psi_b[/tex]

    where [tex]\psi_x[/tex] represents a state with eigenvalue [tex]x[/tex] (with an operator [tex]\hat{O}[/tex]), then the expected average for this operator is

    [tex]|a_1|^2 a + |b_1|^2 b[/tex],

    implying that the probability that [tex]\psi_a[/tex] takes over is [tex]|a_1|^2[/tex] while the probability that [tex]\psi_b[/tex] takes over is [tex]|b_1|^2[/tex] assuming that [tex]\psi[/tex] is normalized and [tex]|a_1|^2 + |b_1|^2 = 1[/tex]. Of course, once the measurement is made, [tex]\psi[/tex] collapses onto one of the states. Is this correct?
     
  10. Jun 15, 2008 #9
    I sent you a PM Domnu. It doesn't have to do with this thread but I wanted you to know incase you don't notice it.
     
  11. Jun 15, 2008 #10
    I've been thinking about this in the last few days. It occured to me when I first read this post that there is no reason to assume that different kets would correspond to different directions of propagation. This doesn't make sense to me. Here is my reasoning - I've never heard of different ket vectors representing directions of propagation. It seems to me that the momentum would determine the direction of propagation and since the operators of all components of momentum mutually commute then the momentum, and thus direction, would be well defined. What would it mean to have eigenkets corresponding to different directions? I don't think there is any such thing.

    Any thoughts?

    Pete
     
  12. Jun 15, 2008 #11
    Okay, the entire thing dealing with directions is just an example. What (I think) kets actually represent are quantum states of ANYTHING. So, a ket could represent literally anything involving some sort of quantum motion. It doesn't only have to be direction, etc. It could, for example, just be a state in which a particle moves at say 0.5c. Now, on a different note, since subatomic particles generally travel close to the speed of light, shouldn't relativistic calculations be taken into account? Is this only in quantum field theory?
     
  13. Jun 16, 2008 #12
    Yes. I understood that. I was simply curious about it. I don't mean to take this thread in a different direction.
    A ket represents the state of a quantum system. So if the state has a physical existance then it has a ket representation.
    Yes.
    When the particles are relativistic then one needs to use relativistic quatum mechanics. This is different than field theory.

    Pete
     
  14. Jun 16, 2008 #13
    I just came across a comment in a quantum mechanics text that I used in grad school. The text is Quantum Mechanics, Cohen-Tannoudji, Diu and Laloe, Volume I, page 15 in which the authors write
    Field theory is a quantum theory of fields. It's about the second quantinization, i.e. quantizing the field itself. I think that quantum electrodynamics (QED) is a quantum and relativistic theory of radiation which is a bit different ... I think. If Reilly shows up I'm sure he can answer this more precisely.

    Pete
     
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