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Help with Questions for a beginner

  1. Jan 29, 2005 #1
    I am brand new to Qunatum Mechanics, and I have just been reading through my first text assignment. I have really been blown away by some of the difficulty in this stuff. I had some questions, and I was hoping that here might be a good place to get them answered...

    1) There is a proof that asks me to show the relation between the uncertainty in position and the uncertainty in the energy. I know the actual inequlaity and how to show the steps via the generalized uncertainty principle. My question is, how come there is uncertainty in the energy? I thought that there are specific and discrete states of total energy. Therefore why isn't the uncertainty in the Hamiltonian zero?

    2) I am having some difficulty in understanding how a commutator functions. I understand its physical significance, but I do not understand how to find [A, B] for any given operators.

    Again, I am very new to this stuff, and I am sure that my questions may seem dumb to some people. However, if you could please help me out with understanding this initial stuff, I would really appreciate it!

    I am glad that there is a forum like this available with so many knowledgeable people to help us tyros.
     
  2. jcsd
  3. Jan 29, 2005 #2

    dextercioby

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    What???Where did u pick that one ??To my knowledge,there ain't such thing... :uhh:

    The definition is simple.The fact that u are /are not used to working with the Poisson bracket is quite important...

    Compute
    [tex] [\hat{A}\hat{B},\hat{C}\hat{D}]_{-} [/tex]

    Apply the definition and its properties...

    Daniel.
     
  4. Jan 29, 2005 #3
    When you make a measurement of the energy of a system, the result must be one of the discrete energy eigenvalues and the system will collapse into the corresponding eigenstate. However, in between measurements the system can evolve out of that state so that the energy is no longer certain until another measurement is made.

    An operator which doesn't commute with the Hamiltonian will have a different set of eigenstates associated with it. This means that the energy eigenstates are superpositions of the other operator's eigenstates. So if you make a measurement with this operator the system will collapse into one of its eigenstates which, rather than having a specific, exactly-known energy, will be in a superposition of energy eigenstates and therefore the energy will not be known exactly.

    Hope that helps!

    Mike
     
  5. Jan 29, 2005 #4
    Uncertainties

    I think I understand what you are saying. It is the measurement that causes the wave function to "collapse" into the eigenstate. That really clears things up.

    Oh and also, in regard to the question I posed about the uncertainties. What I should have said is that my text told me to prove an energy-position uncertainty using a Schwartz inequality. The point of the exercise is to determine that this relation is essentially useless for a stationary state. I was curious as to how there was an uncertainty in energy, which was now answered for me.
     
  6. Jan 30, 2005 #5
    You're right in a certain way, when the system is in an eigenstate of the Hamiltonian, it's energy is well defined and it has no uncertainty ([tex]\Delta E=0[/tex]); such a state it's stationary, i.e. it doesn't evolve, and so, one could say that, in a sense, [tex]\Delta t=\infty[/tex]. But a system can also be in a linear combination of stationary states (that's a very important fact of QM) and when that happens, there's no definite value of the energy and there is some uncertainty [tex]\Delta E[/tex] in it, and so the state evolves in time with a charateristic time [tex]\Delta t[/tex]; the time-energy uncertainty relation just tell you how those 2 quantities are related: [tex]\Delta E\Delta t\gtrsim \hbar [/tex].
    Just to be clear, if someone says "the system can only be on states of definite energies" that's not true, the system may have "specific and discrete states of total energy" but that doens't mean the system can't be in a linear combitation of states with different energies, and when that happens the energy is, of course, uncertain. Do I explain myself? (this is a crucial and important point to understand QM).
    One more comment on that, in the Hydrogen atom, for example, one calculates the the eigenstates and eigenenergies of the Hamiltonian, those states should be stationary, but we find experimentally that they are not, they're unstable Why is that so?
    The problem is that you're treating the hydrogen atom as if it were totally isolated while it's actually in constant interaction with external electromagnetic fields and that causes the states to be unstable; however, since that interaction is weak, it's a good aproximation no completly neglect it's presence except when you're precisely interested on the inestabilty, which is noramally studied phenomelogically using precisely the energy-time uncertainty relation.
    Finally, there's a way to relate the postion-momentum u.r. with the energy-time u.r. for a free one-dimensional wave packet, is that what you were asked to do? do you need help with that?

    .
    Well, once I had the same doubts...
    It all comes to reduce any conmutation relation to some fundamental conmutators that you usually postulate (for example [tex][r_i,p_j]=i\hbar \delta_{ij}[/tex]), by using the different properties of conmutators (such as the one dextercioby asked you to compute); you can usually do that since any given operator should usually be function of you're canonical variables (like [tex]\vec{r}[/tex] and [tex]\vec{p}[/tex]) (the same thing happens in classical mechanics: for you to calculate the Poisson bracket of any dynamical quantities, you have to have them as a function of your canonical variables (again, usually, [tex]\vec{r}[/tex] and [tex]\vec{p}[/tex]))
    If you don't have an operator as a function of your canonical variables, then you usually postulate its conmutators (as you do with the spin operators, for example).
    To get an idea of what I mean, try to compute the orbital angular momentum uncertainty relation [tex][L_i,L_j]=i\hbar \epsilon_{ijk}L_k[/tex] from it's definition ([tex]L_i=\epsilon_{ijk}r_j p_k[/tex]) and from the fundamental conmutator [tex][r_i,p_j]=i\hbar \delta_{ij}[/tex]. Just ask if you need any help with that.
     
  7. Jan 30, 2005 #6
    Uncertainties

    I do now understand what is meant by the uncertainty in energy. The above explanation was very clear, and I thank you for that. Let me describe exactly what I was asked to do for this particular problem:

    I was given that (delta x *delta E) >= (h/2m) <p>, and I was asked to prove that. Basically I just manipulated the generalized uncertainty to prove that inequality. After that though, I was asked why this inequality is not useful for stationary states. That is what I need help with. I don't understand why it is not useful- I think that some of the qualitative details are still escaping me.

    As for the commutators, using the fundmaental commutator [x,p]= ih, I understand that part of what you said. Still, I am unsure of how to attack the angular momentum problem that you gave me. Maybe if you could show me a few steps on getting started? dextercioby is right- I am not used to working with the brackets.

    Thanks for the help!
     
  8. Jan 30, 2005 #7
    I'm not sure what your prof may be looking for, but here's my take on it...

    The more commonly encountered uncertainty relation involving energy is [tex]\Delta E \Delta t \ge \hbar/2 [/tex]. In an eigenstate of the Hamiltonian (a stationary state) we will have [tex]\Delta E = 0[/tex]so in order to satisfy Heisenberg, [tex]\Delta t\to \infty[/tex]. What this tells us is that the uncertainty in energy remains [tex]\Delta E=0[/tex] on the order of the time interval [tex]\Delta t[/tex], which is essentially infinite - the very definition of a stationary state, i.e. one that won't change for all time.

    If we now look at the position-energy uncertainty relation for a stationary state, we see that [tex]\Delta x\to\infty[/tex], which basically says that the particle could be anywhere in space. To me this does not seem particularly useful since it really doesn't tell us very much about the system.

    I'd be interested to know what the answer ends up being if this isn't it.
     
  9. Jan 30, 2005 #8
    Uncertainties

    That makes so much sense that I can't believe I didn't think of it. Thanks for getting me thinking on the right track.

    Oh, and by the way, how do get all the mathematical symbols in the posts?
     
  10. Jan 30, 2005 #9
    No worries, glad it made sense :)

    The math is created with TeX code - for info, check out this post: https://www.physicsforums.com/showthread.php?t=8997
     
  11. Feb 1, 2005 #10
    I've also a question to commutators (uncertainty).

    commutator with momentum and position: [p,X]
    Then: [p,X]=pX-Xp and so (k is psi)
    -i*h*k'*X-X*(-i*h*k')
    now we can:
    -i*h (k'X-Xk')

    Now the question:
    Why the hell is (k'X-Xk') not 0? Solution is (k'X-Xk')=1, but why?
     
  12. Feb 1, 2005 #11

    Galileo

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    If you want to know what a commutator does, then it's usually a good rule to apply it to a test function. After all the operations are done, you can discard the test function and see what the net effect of the commutator is.

    For example: Calculate [x,p]. Take a test function f.

    [tex][x,p]f=(xp-px)f=xpf-pxf=\frac{\hbar}{i}xf'-\frac{\hbar}{i}(xf)'=\frac{\hbar}{i}(xf'-(f+xf'))=(i\hbar) f[/tex]

    Therefore [itex][x,p]=i\hbar[/itex].
     
  13. Feb 1, 2005 #12
    Now I got it, thank you Galileo.

    But why is:
    (xf')=(f+xf') in your formula or did I look something over?
     
  14. Feb 1, 2005 #13

    Galileo

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    The notation was a bit shorthand, but I was taking the derivative of xf. So it said (xf)'=xf'+f by the product rule.

    In nicer notation it would've been:

    [tex]\frac{\partial}{\partial x} (xf)=f+x\frac{\partial f}{\partial x}[/tex]
     
  15. Feb 2, 2005 #14
    But if you use the product rule, shouldn't x be a function?
    x is an operator. Shouldn't it be:
    (xf)' = xf' (factor rule)
     
  16. Feb 2, 2005 #15
    The position operator [itex]\hat{x}[/itex] is the function x ! So it's no fancy derivative operator or anything, the position operation just involves multiplying by the position coordinate.
     
  17. Feb 2, 2005 #16
    aaa, X is a function. something like: f(x)=x (1 dimensional)
     
  18. Feb 2, 2005 #17
    Well, x is just the position coordinate. It plays a bit of a double role in nonrelativistic quantummechanics as an operator and a position coordinate: [itex]\hat{x}f=xf[/itex]!
     
  19. Feb 2, 2005 #18

    Galileo

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    Well, the operator X applied to a function f, gives the function xf: it multiplies the function by x.
    So if I denote the operator by X, derivation wrt x by D and the independent variable by x, then step-by-step:

    [tex][X,D]f=(XD-DX)f=(XD)f-(DX)f=X(Df)-D(Xf)=[/tex]
    [tex]Xf'-D(xf)=xf'-(xf)'=xf'-xf'-f=-f[/tex]

    So [X,D]=-1.
     
  20. Feb 2, 2005 #19
    Isn't the product rule: (xf)' = x'f+xf'? Or is x'f=f?
     
  21. Feb 2, 2005 #20
    Not 'or', and! That's why you have to be careful with this shorthand notation , you might forget to what variable you're differentiating (x!):

    [tex](xf)' = x'f+xf' = \frac{dx}{dx}f+ x \frac{df}{dx}=f+x \frac{df}{dx}[/tex] as [tex]\frac{dx}{dx} = 1[/tex]
     
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