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Basic quantum probability question

  1. Jul 27, 2011 #1
    Probability = |<x|x>|2

    but if |x> = e-ix

    then prob = e+ixe-ix = 1

    but that is not |e-ix|2 right?

    |e-ix|2 becomes e-2ix???

    thanks!

    edit: i was watching a lecture on quantum mechanics and the lecturer wrote

    Px = |a e-iEt|2 = a*a
     
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  3. Jul 27, 2011 #2

    Fredrik

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    The absolute value of a complex number z is [itex]|z|=\sqrt{z^*z}[/itex], so for all real numbers x, [tex]|e^{-ix}|^2=e^{-ix}e^{ix}=e^{-ix+ix}=e^0=1.[/tex]
     
  4. Jul 27, 2011 #3

    vanhees71

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    I also hope that the lecturer mentioned that the plane wave (momentum eigenstate) is not representing a state in quantum theory. Only square integrable functions represent states!
     
  5. Jul 27, 2011 #4
    wow thank you i never knew that the absolute of a complex number is that.

    i thought its just changing it to +
     
  6. Jul 27, 2011 #5
    1) are you talking about -ih(bar)d/dx? the momentum operator?

    2) also i thought a state means like if light passes through a polarizer, then if its polarized in the horizontal position, that means its a state |x> ? , if vertical means |y> ?

    3) anyway, the lecturer wrote something like this
    Px = |a e-iEt|2 = a*a
    where Px is probability

    4) btw, what do you mean by square integrable functions? you mean this?
    http://en.wikipedia.org/wiki/Square-integrable_function

    so whats the catch? is it got to do with normalizing?
     
  7. Jul 27, 2011 #6

    Fredrik

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    He's talking about the eigenfunctions of that operator. They are the functions [itex]u_p[/itex] defined by [itex]u_p(x)=e^{-ipx/\hbar}[/itex] for all x. The catch is that they are not square integrable. If you plug one of the [itex]u_p[/itex] functions into the formula for the norm of a square integrable function, the result is [itex]\infty[/itex].

    If these functions should be called "states" or not is to some extent a matter of taste.

    Are you still having difficulties with (3)? The absolute value of a complex number z=x+iy where x and y are real is [itex]|z|=\sqrt{z^*z}=\sqrt{x^2+y^2}[/itex]. What does that tell you about |wz| where w is another complex number?

    By the way, the function <,> defined on that Wikipedia page is a semi-inner product, not an inner product. The article got that detail wrong.
     
  8. Jul 27, 2011 #7
    oh, so is there anything i should do to make an eigenfunction square integrable? or theres nothing i can do about it?

    i understand (3),
    but with regards to the |wz|, do you mean i have to squareroot [(wz)*(wz)] ? which means i complex conjugate the whole of wz this time?

    with regards to semi-inner product,
    a site said this
    "Hence, a semi-inner product on a vector space is just like an inner product, but for which <v|v> can be zero (even if v=0 )."

    i thought as long as v=v it is 1, if its not = to each other, its 0?
    so in a semi-inner product, if <0|0>, then it is 0 and not 1? this is the catch?
     
  9. Jul 27, 2011 #8

    Fredrik

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    Yes, that's what the definition of the absolute value says. You really need to do the following (easy) exercise. Write w=a+ib, z=c+id, where a,b,c,d are real. Define the absolute value by [itex]|z|^2=\sqrt{c^2+d^2}[/itex] and the complex conjugate by z*=c-id. Show that (wz)*=w*z* and that |wz|=|w||z|. (This is the kind of stuff you should have done within the first hour after you learned about complex numbers).

    Both inner products and semi-inner products are linear in one of the variables and antilinear in the other. So if either of x or y is =0, then <x,y>=0. (For example, if y=0, we have <x,y>=<x,0y>=0<x,y>=0). The difference is that inner products satisfy the condition if <v,v>=0, then v=0. The function <,> defined in that article is a semi-inner product, because for example, if f is the function defined by [tex]f(x)=\begin{cases}0 &\text{if}\ x\neq 0\\ 1 &\text{if}\ x=0,\end{cases}[/tex] we have <f,f>=0.

    u≠v obviously doesn't imply <u,v>=0. That would actually mean that <v,v>=(1/2)<v,2v>=0 for all v, so that we actually have <u,v>=0 for all u,v.

    You are very careless with your mathematical statements. I assume that you understand that it's not possible for v to not be =v.
     
    Last edited: Jul 27, 2011
  10. Jul 27, 2011 #9

    jambaugh

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    [itex] 1^2 = 1[/itex]
     
  11. Jul 27, 2011 #10
    [(a+ib)(c+id)]*
    =ac-bd-iad-ibc
    but how do i go from here to (a-ib)(c-id)? long division?
    or just expand (a-ib)(c-id) = ac-bd-iad-ibc ?

    so for |wz| = sqrt[(wz)*(wz)] = sqrt(w*z*wz) = sqrt(w*wz*z)= sqrt(w*w)sqrt(z*z)=|w||z| ?






    oh so for normal inner product <v|v> = 1
    but if <v|v> = 0, then it is a semi-inner product and v can only be = 0? so the inner product of something with itself can only be 0 if it is 0 itself? <0|0> = 0, nothing else?thats the catch to semi-inner products?




    i thought <u|v> = 0 if u≠v ? or is that the kronecker delta?
    oh if u is orthogonal to v then <u|v> is 0 right?

    also, i don't really understand this part
    <v,v>=(1/2)<v,2v>=0 for all v, so that we actually have <u,v>=0 for all u,v.

    why will having half of the inner product of v,2v be equal to 0 which equal <v|v> ? is this showing orthogonality?
     
  12. Jul 27, 2011 #11
    haha i see
     
  13. Jul 28, 2011 #12

    jtbell

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  14. Jul 28, 2011 #13
    |e-ix|2 is not equivalent to (e-ix)2, right?
     
  15. Jul 28, 2011 #14

    jtbell

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    Right.

    [tex]|e^{ix}|^2 = (e^{ix})(e^{ix})^* = (e^{ix})(e^{-ix}) = e^{ix-ix} = e^0 = 1[/tex]

    [tex](e^{ix})^2 = (e^{ix})(e^{ix}) = e^{ix+ix} = e^{2ix}[/tex]
     
  16. Jul 28, 2011 #15
    Then OP is not quite true...
     
  17. Jul 28, 2011 #16

    Fredrik

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    This is the easiest way to do it.

    This is correct, but to justify the fourth equality, you need to prove either that z*z is real for all complex z, or that sqrt(wz)=(sqrt w)(sqrt z) for all complex w,z. The first option is the easiest: z*z=(c-id)(c+id)=c^2+d^2.

    For all non-zero v, yes.

    No. If it's an inner product, this equality implies v=0. If it's a semi-inner product, it doesn't.

    Yes.

    Only if you have already specified that u and v are members of the same orthonormal subset of a Hilbert space. It's certainly not true for arbitrary u,v.

    Yes. The definition of "orthogonal" says that u and v are said to be orthogonal if <u,v>=0.

    I was showing you a consequence of your claim that for all u,v, we have <u,v>=0. Both inner products and semi-inner product are linear in the second variable, so for any complex c, we have <v,v>=(1/c)<v,cv>. Choose c≠1. Now your claim implies that the second factor on the right is 0. So we find that for all v, we have <v,v>=0. Since you had already assumed that <u,v>=0 whenever u≠v, the conclusion is that for all u,v, we have <u,v>=0. This would be a completely useless semi-inner product. If it's an inner product, it even implies that the only vector in the inner product space is the zero vector.
     
  18. Jul 28, 2011 #17

    vanhees71

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    It cannot be stressed often enough. That's why I put it here again and in bold letters:

    A wave function represents a state in quantum mechanics if and ONLY if it is square integrable and thus normalizable to 1!

    Momentum "eigenstates", i.e., plane waves are NOT square integrable and thus do not represent a state. They belong to another space, namely the dual space of the dense subspace in Hilbert space, where the momentum operator is defined, i.e., it is to be considered as a generalized function (or distribution).

    The same is true for any "eigenstate" for a "eigenvalue" of an operator in the continuous part of its spectrum.

    E.g., for the generalized eigenvector of the position operator to the "eigenvalue", [itex]x_0[/itex] one has

    [tex]u_{x_0}(x)=\langle x|x_0 \rangle = \delta(x-x_0),[/tex]

    where I consider a one-dimensional problem for simplicity's sake.

    The generalized eigenvector of the momentum operator in the position representation is

    [tex]u_{p_0}(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p).[/tex]

    I've set [itex]\hbar=1[/itex] for simplicity.

    Generalized eigenstates are normalized to a [itex]\delta[/itex] distribution. Indeed, one has

    [tex]\langle p'|p \rangle=\int \mathrm{d} x \langle p'|x \rangle \langle x|p \rangle = \int \mathrm{d} x \frac{1}{2 \pi} \exp[\mathrm{i} x(p-p')]=\delta(p-p').[/tex]

    For "real states", represented ONLY by square integrable wave functions in the position representation, for any two observables the uncertainty relation holds. For position and momentum it reads

    [tex]\delta x \delta p \geq 1/2,[/tex]

    and this shows that there doesn't exist any real state, for which either position or momentum are precisely determined. You can make the uncertainty in one of the observables arbitrarily small at the expense of the accuracy the other is known, but you can never make the uncertainty of either variable really 0! Thus, there cannot exist true eigenstates for position or momentum, because for those the uncertainty would vanish!
     
  19. Jul 28, 2011 #18

    Fredrik

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    vanhees71, did you perhaps post in the wrong thread by accident? Your reply would make more sense as a reply to post 25 in this thread.
     
  20. Jul 29, 2011 #19

    vanhees71

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    Maybe, but also in this thread I've seen some strange ideas like

    [tex]\langle x| x \rangle =1 \quad (\mathrm{WRONG}).[/tex]

    Generalized position eigenstates are distributions and not normalizable to 1 and it doesn't make any sense to take this scalar product. The correct expression is

    [tex]\langle x' | x\rangle =\delta(x'-x),[/tex]

    but that I tried to explain in my previous posting.
     
  21. Jul 30, 2011 #20
    thx fredrik i understand now i hope

    with regards to vanhees,




    erm does it mean that if i can normalize something, it means it is square integrable? like adding a 1/sqrt2 in front of the expression?



    do you mean that the momentum operator is not a state? i watched a quantum lecture on youtube, it says operators are observables? states are wavefunctions of the system? something like that?


    ah i have seen the lecturer talk about this, is this about the eigen vectors of position operator as delta functions while eigen vectors of momentum being cos and sine waves?

    i got lost here, are you trying to say that states are normalized to a delta function and not to 1? what is p prime, p', x' etc?

    also it reminds me of what fredrik said earlier
    is this the formula for the norm of a square integrable function?

    ??? so the momentum eigenstate e-ipx/h is 'false' ??
     
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