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How to derive the quantum commutation in matrix mechanics?

  1. Mar 4, 2016 #1
    1. The problem statement, all variables and given/known data
    I would like to know how to derive the quantum commutation relations in matrix form,
    $$i \hbar \partial_t x(t)= [x(t),E]$$
    $$i \hbar \partial_t P(t)= [P(t),E]$$
    Where X(t), P(t) and E are the position, momentum and the energy of the particle, respectively.

    2. Relevant equations

    The solution of harmonic oscillator as expressed in Fourier expansion;
    $$X(t)=\sum X_{mn} e^{i \omega (m-n)t} $$
    $$P(t)=\sum P_{mn} e^{i \omega (m-n)t} $$
    differentiate X(t) relative to t would give P(t) assuming the mass of the particle=1.
    $$\partial_t X(t)=\sum X_{mn} i\omega(m-n) e^{i \omega (m-n)t} $$
    $$\partial_t X(t)=\sum X_{mn} i \frac{E_m - E_n}{\hbar} e^{i \omega (m-n)t} $$
    But how to arrive to the commutation formula from here?
    3. The attempt at a solution
    I tried retrospectively from the commutation relation to prove the equivalence with the original equation;
    $$i \hbar \partial_t x(t)= [x(t),E]$$
    So I started from;
    $$=[X(t) E - E X(t)]$$
    $$=X_{mn} E_m \delta_{mn} - E_m \delta_{mn} X_{mn}$$
    where $$E=E_{mn}=E_m \delta_{mn}$$
    But this did not give me;
    $$\partial_t X(t)=\sum X_{mn} i \frac{E_m - E_n}{\hbar} e^{i \omega (m-n)t} $$
     
    Last edited: Mar 4, 2016
  2. jcsd
  3. Mar 4, 2016 #2

    blue_leaf77

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    Those relations apply for a general operator actually, not only restricted to ##x(t)## and ##p(t)##.
    Note that
    $$x(t)=e^{iEt/\hbar}xe^{-iEt/\hbar}
    $$
    Now differentiate this with respect to ##t## (I assume ##E## is an operator).
     
  4. Mar 4, 2016 #3
    $$\partial_t x(t)=\partial(e^{iE_mt/\hbar}xe^{-iE_nt/\hbar})$$
    $$\partial_t x(t)=\frac{i(E_m-E_n)}{\hbar} (e^{iE_mt/\hbar}xe^{-iE_nt/\hbar})$$
    $$i\hbar \partial_t x(t)= (E_n-E_m) (e^{iE_mt/\hbar}xe^{-iE_nt/\hbar})$$
    I assumed E is a different energy level. I don`t know how to proceed if E is an operator because for your equation, the exponential function will reduce to 1.
     
    Last edited: Mar 4, 2016
  5. Mar 4, 2016 #4

    blue_leaf77

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    I am rather confused with what you actually want to do. I thought you wanted to prove
    which are true provided ##E## is the Hamiltonian operator. If ##E## in those equations is a value of certain energy, i.e. a number, then the commutation above should equal zero. This further implies that ##x(t)## is time independent, which contradicts the way it was written. So, what do you want to prove actually?
     
  6. Mar 4, 2016 #5
    I just want to derive those commutation from the original equation of Fourier expansion of x(t) and p(t). In those expansion, ##\omega(m-n)=\frac{E_m-E_n}{\hbar}##
    I know how to derive the commutation from your equation,
    $$\partial_t x(t)=\partial(e^{iEt/\hbar}xe^{-iEt/\hbar})$$
    $$\partial_t x(t)=\frac{i(E)}{\hbar}(e^{iEt/\hbar}xe^{-iEt/\hbar})-(e^{iEt/\hbar}xe^{-iEt/\hbar})\frac{iE}{\hbar}$$
    $$\partial_t x(t)=\frac{i(E)}{\hbar}x(t)-x(t)\frac{iE}{\hbar}$$
    $$\hbar\partial_t x(t)=i(E x(t)-x(t) E)$$
    $$i\hbar\partial_t x(t)=x(t)(E)-(E)x(t)$$
    $$i\hbar\partial_t x(t)=[x(t),(E)]$$ as desired.
    But I want to do the same from matrix algebra.
     
  7. Mar 4, 2016 #6

    blue_leaf77

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    What is ##E##? Is it an operator or a scalar?
     
  8. Mar 4, 2016 #7
    In your equation, E is an operator, but in ##E_n## and ##E_m## in Fourier expansion are numbers.
     
  9. Mar 4, 2016 #8

    blue_leaf77

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    The root of the trouble seems to be the equations:
    I honestly never encounter them. ##X_{mn}## is a matrix element of ##X## and then you basically sum them all along the rows and columns. Then somehow it turns out to be equal to the operator ##X(t)##. Where do you get these formula from?
     
  10. Mar 4, 2016 #9
    https://en.wikipedia.org/wiki/Matrix_mechanics see the section under Heisenberg reasoning. So, I summat over all m and n by mistake. In the same page, under the matrix basics, the commutation relation is shown without derivation, in matrix form. My question, how can we derive that?
    On the other hand, you gave $$x(t)=e^{iE/\hbar t}xe^{-iE/\hbar t}$$
    Is this a solution to the harmonic oscillation?
     
  11. Mar 4, 2016 #10

    blue_leaf77

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    Now you are referring to the commutation between x and p, which is a different thing from the equations you said you wanted to prove and ask how to derive this.
    So, how do the correct formula look like? I am really curious with these Fouriert transform equation for ##X(t)## and ##P(t)##.
    No, that's just how an operator in the Heisenberg picture is defined. But wait, in your original problem statement, harmonic oscillator problem was not mentioned at all. Why are you considering harmonic oscillator?
     
  12. Mar 4, 2016 #11
    This is almost the same. the equations I want to prove using matrix algebra are:
    $$i\hbar\partial_tx(t)=[x(t),E]$$
    and
    $$i\hbar\partial_tp(t)=[p(t),E]$$
    Now consider a harmonic oscillator
    $$H=\frac{P^2}{2}+x$$ assuming f(x)=-1, Also;
    $$\partial_tp(t)=f(x)$$
    plug this into the second equation,
    $$i\hbar=[p(t),E]$$
    $$=p(\frac{p^2}{2}+x)-(\frac{p^2}{2}+x)p$$
    $$=px-xp=[x,p]$$
    So the key to derive the commutation ##[x,p]## is to derive the first two equations which is my question. But it is ok also if you have another way to derive the relation without considering harmonic oscillator. But please let me know who Heisenberg definition of operator is first derived? (in matrix algebra not wave mechanics).
    It should look like:
    $$x_{mn}(t)=\sum x_{mn}(0)e^{i\omega_{mn}t}$$
    I think that was the main essence behind the matrix mechanics, where ##\omega_{mn}## is replacing ##(m-n)\omega## as in classical mechanics. So long as the quantum theory demands introduction of ##\omega_{mn}## as a matrix element, the Fourier coefficient must be also a matrix element ##x_{mn}(0)##
     
    Last edited: Mar 4, 2016
  13. Mar 4, 2016 #12

    blue_leaf77

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    No, that's not a Hamiltonian of harmonic oscillator.
    Just an intermezzo though, you assumption f(x)=-1 is actually true, it's not just an assumption. Using Baker-Hausdorff formula, you can prove that
    $$
    p(t)=e^{it(p^2/2+ax)}pe^{-it(p^2/2+ax)} = p-at
    $$
    so that ##\partial_tp(t)=-a##.
    That cannot be right because ##mn## in the left hand side is a fixed indices while ##mn## in the right hand side is the indices for summation. If you are interested in knowing the matrix element of ##x(t)##, it's just ##x_{mn}(t)=x_{mn}(0)e^{i(\omega_m-\omega_n)t}##, no summation is needed.
    The commutation between x and p is the so-called fundamental commutation relation in quantum mechanics. I believe this is a principle which was not invented through derivation, rather it actually comes from the analogy with the Poisson bracket between x and p in classical mechanics.
    From what I comprehend about the current situation, you want to prove
    $$
    i \hbar \partial_t x(t)= [x(t),E]$$
    using matrix algebra, which failed because you started with a wrong expression for ##x_{mn}(t)##. Now that I have told you that ##x_{mn}(t)=x_{mn}(0)e^{i(\omega_m-\omega_n)t}##, you should be able to complete the derivation by showing that
    $$
    \langle m |[x(t),E]| n\rangle
    $$
    does becomes ##x_{mn}(0)e^{i(\omega_m-\omega_n)t}##. Or is it not the matrix algebra that you desired?
     
    Last edited: Mar 4, 2016
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