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Understanding the Commutator

  1. Jul 9, 2012 #1
    Hi!

    In my textbook the explanation of the expectation value in general covering any observable [tex] Q [/tex] is:

    [tex] \overline{Q} = \int \Psi^\ast (x,t) \hat{Q}\Psi(x,t) dx [/tex]

    Then they define the commutator as:

    [tex] [\hat{A},\hat{B}] = \hat{A}\hat{B}-\hat{B}\hat{A}[/tex]

    Now for position and momentum they give directly it as:

    [tex] [\hat{x},\hat{p}] = i\hbar[/tex]

    Now my question is: how is that possible? I know what the operators of the observables for both position and momentum are, but I do not understand why the commutator can be derived trough the general form of EXPECTATION value?? Expectation value is a value but the commutator is an operator on operators??
    Can anyone help me with this confusion? Im new to quantum physics. Thanx
     
  2. jcsd
  3. Jul 9, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Elekko! Welcome to PF! :smile:

    (write "itex" rather than "tex", and it won't keep starting a new line :wink:)
    Are you asking how the LHS is an operator, but the RHS is only a number?

    It's because the RHS is really [itex]i\hbar I[/itex], where I is the identity operator (the operator that converts anything to itself) … we don't usually bother to write it. :smile:
     
  4. Jul 9, 2012 #3

    Fredrik

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    I don't understand your questions. You may want to clarify them a bit. The words "through the general form of expectation value" are especially confusing. Have you seen a derivation of a commutator that involves an expectation value? What do you mean by "operator on operators"? It's just an operator. The "product" AB is defined by (AB)f=A(Bf) where f is a square-integrable function from ##\mathbb R^3## to ##\mathbb C##. So the "product" AB is just the composition ##A\circ B##. The sum of two operators is defined by (A+B)f=Af+Bf. So [A,B]f=A(Bf)-B(Af).
     
  5. Jul 9, 2012 #4
    The commutator is not defined from the expectation value, the expectation value necessarily depends on the particular wavefunction whereas the commutator is defined as a property of a pair of operators, regardless of any particular wavefunction.
     
  6. Jul 9, 2012 #5
    Well, in a sense the commutator itself is indeed an operator on operators, since the input is two operators and it gives a new operator. More exactly if [itex]\mathcal B[/itex] is the space of operators, then the commutator is a function [itex]\mathcal B \times \mathcal B \to \mathcal B : (\hat A, \hat B) \mapsto \hat A \hat B - \hat B \hat A[/itex].

    As for the OP's initial question, it seems he's under the impression that the identity [itex][\hat x , \hat p ] = i\hbar[/itex] was derived using expectation values; he probably thought so because the book seemed to imply it by introducing commutators after introducing expectation values and immediately stating "one can see that [itex][\hat x , \hat p ] = i\hbar[/itex]". I'm not sure if I interpreted the OP correctly though, but if so: no worries, the identity has nothing to do with expectation values. To get the "[itex] [\hat x , \hat p ] = i\hbar[/itex]" identity, one writes [itex] [\hat x , \hat p ] [/itex] out in terms of the actual operators (i.e. derivatives) and then applying that to a general [itex]\psi[/itex] one can prove, using some algebra, that this is equal to [itex]i\hbar \psi[/itex].
     
  7. Jul 10, 2012 #6
    Re: Welcome to PF!

    Hi all,

    I appreciate for all your reply!

    So this means that in the case of momentum and position, the "identity" operator is: [itex]\Psi[itex] ???
     
  8. Jul 10, 2012 #7

    tiny-tim

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    Hi Elekko! :smile:

    (remember to close your latex with a / :wink:)
    nooo :confused:

    the identity operator is I

    the effect of I on ψ is ψ​
     
  9. Jul 10, 2012 #8
    So here, it means that when we derive the commutator for momentum and position to [itex]i\hbar \psi[/itex] the [itex]\psi[/itex] is always applied and it can be seen as and identity operator as tiny-tim mentioned?
     
  10. Jul 10, 2012 #9
    Wow you're quick in replying.
    So let me see if I have got it right:

    the expectation value [itex] \overline{Q} = \int \Psi^\ast (x,t) \hat{Q}\Psi(x,t) dx [/itex] has nothing to do with to derive the commutator.
    The commutator of momentum and position is simply [itex]i\hbar[/itex] by using the observables in the definition of commutator. And [itex]\psi[/itex] can be applied in case of showing it in general?
     
  11. Jul 10, 2012 #10
    Post #8 doesn't seem correct, (though post #9 does)

    To find the commutator, evaluate [AB-BA](phi). If you get zero, then the operators commute, otherwise you will get X(phi) where X is the commutator operator, and you can read off what X is. It is always an operator.

    For position and momentum,

    [x,p](phi) = iħ*phi ≡ X(phi)

    Note the brackets are still used on the right, because the commutator is not simply a number "iħ", it is an operator, equivalent to "multiply argument by iħ".

    In this sense, [x,p] is not strictly iħ, it is iħ*, but the multiplication sign is often omitted.
     
  12. Jul 10, 2012 #11

    tiny-tim

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    See what MikeyW :smile: says.
    I have no idea what you mean. :redface:

    ψ is any wavefunction

    any operator equation A = B means "Aψ = Bψ for any ψ"​
     
  13. Jul 10, 2012 #12
    I think I got it now guys. Thank you!
    Tiny-tim, nevermind. It seems that I don't understand why we even do have to apply [itex]\psi[/itex] on the commutator to begin with. Its just simply xp-px where x = x, p = -ihd/dx
     
  14. Jul 10, 2012 #13

    Fredrik

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    Right, but unless you have xp-px act on a function, how are you going to prove that [x,p]=iħ?
     
    Last edited: Jul 10, 2012
  15. Jul 10, 2012 #14
    Thanks!
    I got it now!
    In case of quantum mechanics we use the waveFUNCTION [itex] \psi [/itex] in order to prove the commutator.
    I appreciate for all your help guys!
     
  16. Jul 10, 2012 #15

    vanhees71

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    To make things really clear: The commutation relation between position and momentum opertor (in the same direction) follows from the fact that, by definition and as in classical mechanics, the (canonical) momentum of a system is defined by it's operation as a generator of spacial translations.

    In the position-space representation, which sometimes is called "wave mechanics", but in fact it's just a representation of quantum theory in the position eigenbasis, the operators thus read
    [tex]\hat{x} \psi(x)=x \psi(x), \quad \hat{p} \psi(x)=-\mathrm{i} \hbar \mathrm{d}_x \psi(x).[/tex]
    The commutation relations are found by applying these operators in different orders
    [tex](\hat{x} \hat{p}-\hat{p} \hat{x})\psi(x)=-\mathrm{i} \hbar [x \mathrm{d}_x \psi(x) - \mathrm{d}_x (x \psi(x))]=\mathrm{i} \hbar \psi(x).[/tex]
    This means that for any wave function the commutator acts as by simply multiplying the wave function by [itex]\mathrm{i} \hbar[/itex], and this is written in operator form as
    [tex][\hat{x},\hat{p}]=\mathrm{i} \hbar \mathbb{1}.[/tex]
     
  17. Jul 11, 2012 #16

    Fredrik

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    I suspect that most people who are new to this don't fully understand how results like ##\hat p\hat x\psi(x)=\frac{d}{dx}(x\psi(x))## follow from the definitions, so I will explain that. I will only consider the 1-dimensional case here, so our wavefunctions are functions from ##\mathbb R## into ##\mathbb C##. First note that when ##\psi## denotes such a function, ##\psi(x)## denotes a complex number in its range. So ##\psi(x)## never denotes a function. It's always just a number. This means that an operator like ##\hat x## or ##\hat p## acts on ##\psi##, never on ##\psi(x)##.

    ##\hat x## is defined by specifying a set of functions that will be its domain, and then specifying ##\hat x\psi## for all ##\psi## in that set. Since ##\hat x\psi## is a function with domain ##\mathbb R##, the way to do this is to specify ##\hat x\psi(x)## for all ##x\in\mathbb R##. (Note that ##\hat x\psi(x)## means ##(\hat x\psi)(x)##. ##\hat x## acts on ##\psi## and ##\hat x\psi## acts on ##x##).

    The domain of ##\hat x## is the set of all square-integrable ##\psi:\mathbb R\to\mathbb C## such that the map ##x\mapsto x\psi(x)## is square-integrable. For each ##\psi## in that set, we define
    $$\hat x\psi(y)=y\psi(y),$$ for all ##y\in\mathbb R.## (Note that it never matters what variable symbol is used in a "for all" statement. I chose to use y instead of x because I wanted to make it clear that we don't have to use the same symbol that's used in the notation for the position operator).

    ##\hat p## can be defined without even mentioning a variable (like x) that represents a real number. The domain of ##\hat p## is the set of all square-integrable ##\psi:\mathbb R\to\mathbb C## such that ##\psi'## is square-integrable. For all such ##\psi##, we define
    $$\hat p\psi=-i\psi'.$$ I like to use units such that ##\hbar=1##. That's why I don't include ##\hbar## explicitly on the right-hand side.
    The "product" ##\hat p\hat x## is defined as follows: For all ##\psi:\mathbb R\to\mathbb C## in the domain of ##\hat x##, such that ##\hat x\psi## is in the domain of ##\hat p##, we define
    $$(\hat p\hat x)\psi=\hat p(\hat x\psi).$$ This definition is the reason why we don't need to insert parentheses when we write things like ##\hat p\hat x\psi##.

    The other product ##\hat x\hat p## is defined similarly. The domain of the commutator is the intersection of the domains of ##\hat p\hat x## and ##\hat x\hat p##. Note that this means that the identity operator that we don't write out on the right-hand side of ##[\hat x,\hat p]=i\hbar## is the identity operator on that set, not the identity operator on the full set of square-integrable functions.

    Now, let's use the definitions to evaluate ##\hat p\hat x\psi(x)##, where ##\psi## is an arbitrary member of the domain of ##[x,p]##.
    $$\hat p\hat x\psi(x)=\hat p(\hat x\psi)(x)=-i(\hat x\psi)'(x)=-i\frac{d}{dx}(x\psi(x)).$$ The notation on the right means "-i times the derivative of the map ##y\mapsto y\psi(y)## evaluated at x". The definition of ##\hat x## tells us that that map is precisely what we denote by ##\hat x\psi##, so the right-hand side above means ##-i(\hat x\psi)'(x)##.

    The rest of the calculation that proves the identity ##[x,p]=i## goes like this:
    $$
    \begin{align}&=-i\big(\psi(x)+x\psi'(x)\big)=-i\big(\psi(x)+\hat x(\psi')(x)\big)=-i\big(\psi(x)+\hat x(i\hat p\psi)(x)\big)\\
    &=-i\psi(x)+\hat x\hat p\psi(x)=(-i+\hat x\hat p)\psi(x).
    \end{align}$$ I used that ##\hat x## is linear here. I haven't proved that, but it's easy to do, using the definition of ##\hat x##.
     
  18. Jul 11, 2012 #17
    That's a perfect explanation.
    Thank you very much!
     
  19. Jul 11, 2012 #18

    Fredrik

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    You're welcome. Thanks for letting me know that you read it and understood it.

    Here's a slightly different version that doesn't even use the d/dx notation. I will not include all the information about domains and stuff this time. Here I denotes the function that takes x to x. To avoid confusion, the identity operator will be denoted by 1 instead of I. We define two operators, Q and D.

    Qf=If (This means that for all real numbers x, Qf(x)=(If)(x)=I(x)f(x)=xf(x), so the Q we have defined here is the position operator).
    Df=f' (We define the momentum operator by P=-iD, so D=iP).

    \begin{align}
    DQf(x) &=(If)'(x)=I'(x)f(x)+I(x)f'(x) =f(x)+xf'(x)=f(x)+QDf(x) =(f+QDf)(x) \\
    DQf &=f+QDf=(1+QD)f\\
    DQ &= 1+QD\\
    \left[Q,P\right] &= \left[Q,-iD\right]=-i\left[Q,D\right]=-i(-1)=i.
    \end{align}
     
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