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I The propagator and the Lagrangian

  1. Aug 6, 2017 #1
    I note the following:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\delta(\vec{x}_n-\vec{x}_0)e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}

    \end{split}

    \end{equation}


    I divide the time interval as follows:

    \begin{equation}

    \begin{split}

    t_n-t_0 &= \sum_{j=1}^{n} t_j - t_{j-1}

    \end{split}

    \end{equation}


    Such that:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\delta(\vec{x}_n-\vec{x}_0) e^{-i \frac{\mathcal{H}_n}{\hbar} \sum_{j=1}^{n} t_j - t_{j-1}}

    \\

    &=\delta(\vec{x}_n-\vec{x}_0) \prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \end{split}

    \end{equation}


    I note the following:

    \begin{equation}

    \begin{split}

    \delta(\vec{x}_n-\vec{x}_0)&=\int_{-\infty}^{\infty} d\vec{x}_{n-1} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{0})

    \\

    &=\int_{-\infty}^{\infty} d\vec{x}_{n-1}\int_{-\infty}^{\infty} d\vec{x}_{n-2} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{n-2})\delta(\vec{x}_{n-2}-\vec{x}_{0})

    \\

    &=\int_{-\infty}^{\infty} d\vec{x}_{n-1}\int_{-\infty}^{\infty} d\vec{x}_{n-2}\int_{-\infty}^{\infty} d\vec{x}_{n-3} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{n-2})\delta(\vec{x}_{n-2}-\vec{x}_{n-3})\delta(\vec{x}_{n-3}-\vec{x}_{0})

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1})

    \end{split}

    \end{equation}


    Such that:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1}) \prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1}) e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\delta(\vec{x}_{j}-\vec{x}_{j-1}) e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \end{split}

    \end{equation}


    Given:

    \begin{equation}

    \begin{split}

    \delta(\vec{x}_{j}-\vec{x}_{j-1})&=\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar} e^{i \frac{\vec{p}_j}{\hbar}(\vec{x}_{j}-\vec{x}_{j-1})}

    \end{split}

    \end{equation}


    Thus:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{i \frac{\vec{p}_j}{\hbar}(\vec{x}_{j}-\vec{x}_{j-1})} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{i (t_j - t_{j-1})\frac{\vec{p}_j}{\hbar}\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}-\mathcal{H}_n\right)}

    \end{split}

    \end{equation}


    Given ##\vec{v}_j=\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}##:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j \vec{v}_j-\mathcal{H}_n\right)}

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j \vec{v}_j\frac{m}{m}-\mathcal{H}_n\right)}

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\frac{\vec{p}_j^2}{m}-\mathcal{H}_n\right)}

    \end{split}

    \end{equation}


    Where:

    \begin{equation}

    \begin{split}

    \mathcal{H}_n&=T_n + V_n

    \\

    &=\frac{\vec{p}_n^2}{2 m}+ V_n

    \end{split}

    \end{equation}


    Such that:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n \right)}

    \end{split}

    \end{equation}


    However, ##\vec{p}_j \neq \vec{p}_n##. Therefore, where ##\mathcal{L}## denotes the Lagrangian:

    \begin{equation}

    \begin{split}

    \frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n &\neq \frac{\vec{p}_j^2}{2m}- V_n

    \\

    &\neq \mathcal{L}

    \end{split}

    \end{equation}


    What am I missing?
     
  2. jcsd
  3. Aug 7, 2017 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Hm, already (1) is wrong. You don't get a ##\delta## distribution times a unitary operator (in abstract Hilbert space), but on the left-hand side you write down the propgator, which is a c-number function, solving the Schrödinger equation with the initial condition that at ##t=t_0## it is a ##\delta## distribution.
     
  4. Aug 7, 2017 #3
    Two questions then:


    First, by time slicing:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\langle\vec{x}_n|\prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    |\vec{x}_{0}\rangle

    \end{split}

    \end{equation}


    Then:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|\prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})} |\vec{x}_{0}\rangle&=\langle\vec{x}_n| e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n - t_{n-1})}e^{-i \frac{\mathcal{H}_n}{\hbar} (t_{n-1} - t_{n-2})}

    \\

    &...e^{-i \frac{\mathcal{H}_n}{\hbar} (t_2 - t_{1})}e^{-i \frac{\mathcal{H}_n}{\hbar} (t_1 - t_{0})} |\vec{x}_{0}\rangle

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \langle\vec{x}_n| e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n - t_{n-1})}|\vec{x}_{l}\rangle\langle \vec{x}_{l} | e^{-i \frac{\mathcal{H}_n}{\hbar} (t_{n-1} - t_{n-2})} |\vec{x}_{l-1}\rangle

    \\

    &...\langle \vec{x}_{2} |e^{-i \frac{\mathcal{H}_n}{\hbar} (t_2 - t_{1})}|\vec{x}_{1}\rangle\langle \vec{x}_{1} |e^{-i \frac{\mathcal{H}_n}{\hbar} (t_1 - t_{0})} |\vec{x}_{0}\rangle

    \\

    &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\prod_{j=1}^{n} \langle\vec{x}_j| e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}|\vec{x}_{j-1}\rangle

    \end{split}

    \end{equation}


    Next:

    \begin{equation}

    \begin{split}

    \prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\prod_{j=1}^{n} \langle\vec{x}_j| e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}|\vec{x}_{j-1}\rangle&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\prod_{j=1}^{n}\left[\int_{-\infty}^{\infty}\frac{d\vec{p}_j}{2 \pi \hbar} \right] \langle\vec{x}_j|\vec{p}_j \rangle \langle \vec{p}_j| e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}|\vec{x}_{j-1}\rangle

    \end{split}

    \end{equation}


    Given:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_j|\vec{p}_j \rangle&=e^{\frac{i}{\hbar} \vec{p}_j \vec{x}_j}

    \end{split}

    \end{equation}


    And:

    \begin{equation}

    \begin{split}

    \langle \vec{p}_j| e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}|\vec{x}_{j-1}\rangle&=e^{-\frac{i}{\hbar} \vec{p}_j \vec{x}_0}e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \end{split}

    \end{equation}


    Thus:

    \begin{equation}

    \begin{split}

    \prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\prod_{j=1}^{n}\left[\int_{-\infty}^{\infty}\frac{d\vec{p}_j}{2 \pi \hbar} \right] \langle\vec{x}_j|\vec{p}_j \rangle \langle \vec{p}_j| e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}|\vec{x}_{j-1}\rangle

    \\

    =\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\prod_{j=1}^{n}\left[\int_{-\infty}^{\infty}\frac{d\vec{p}_j}{2 \pi \hbar} \right]e^{\frac{i}{\hbar} \vec{p}_j \vec{x}_j}e^{-i \vec{p}_j \vec{x}_0}e^{-\frac{i}{\hbar} \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \\

    =\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\prod_{j=1}^{n}\left[\int_{-\infty}^{\infty}\frac{d\vec{p}_j}{2 \pi \hbar} \right] e^{\frac{i}{\hbar} \vec{p}_j (\vec{x}_j- \vec{x}_{j-1})} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}

    \end{split}

    \end{equation}


    Where:

    \begin{equation}

    \begin{split}

    e^{\frac{i}{\hbar} \vec{p}_j (\vec{x}_j- \vec{x}_{j-1})} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}&=e^{\frac{i (t_j-t_{j-1})}{\hbar}\left(\vec{p}_j \frac{(\vec{x}_j- \vec{x}_{j-1})}{(t_j-t_{j-1})}-\mathcal{H}_n \right)}

    \end{split}

    \end{equation}


    Given ##\vec{v}_j=\frac{(\vec{x}_j- \vec{x}_{j-1})}{(t_j-t_{j-1})}##:

    \begin{equation}

    \begin{split}

    e^{\frac{i (t_j-t_{j-1})}{\hbar}\left(\vec{p}_j \frac{(\vec{x}_j- \vec{x}_{j-1})}{(t_j-t_{j-1})}-\mathcal{H}_n \right)}&=e^{\frac{i (t_j-t_{j-1})}{\hbar}\left(\vec{p}_j \vec{v}_j-\mathcal{H}_n \right)}

    \\

    &=e^{\frac{i (t_j-t_{j-1})}{\hbar}\left(\vec{p}_j \vec{v}_j \frac{m}{m}-\mathcal{H}_n \right)}

    \\

    &=e^{\frac{i (t_j-t_{j-1})}{\hbar}\left(\frac{\vec{p}_j^2}{m}-\mathcal{H}_n \right)}

    \end{split}

    \end{equation}


    Where:

    \begin{equation}

    \begin{split}

    \mathcal{H}_n&=T_n + V_n

    \\

    &=\frac{\vec{p}_n^2}{2 m}+ V_n

    \end{split}

    \end{equation}


    Such that:

    \begin{equation}

    \begin{split}

    \langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n \right)}

    \end{split}

    \end{equation}


    However, ##\vec{p}_j \neq \vec{p}_n##. Therefore, where ##\mathcal{L}## denotes the Lagrangian:

    \begin{equation}

    \begin{split}

    \frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n &\neq \frac{\vec{p}_j^2}{2m}- V_n

    \\

    &\neq \mathcal{L}

    \end{split}

    \end{equation}


    Again: What am I missing?


    That's the first question. The second question: how would you write the propagator in position, i.e., non-Dirac, notation?
     
  5. Aug 7, 2017 #4
    A quick correction in the math: In equations (16) and (17), ##\vec{x}_0## should be ##\vec{x}_{j-1}##.
     
  6. Aug 8, 2017 #5
    You start already from wrong assumptions. In this equation, the Hamiltonian should still be an operator. It doesn't know anything about times. The subscript just makes no sense at this point. So what you want is
    \begin{equation} \begin{split} \langle\vec{x}_n|\exp{\left( -i \frac{\hat{\mathcal{H}}}{\hbar} (t_n-t_0) \right)}|\vec{x}_{0}\rangle &=\langle\vec{x}_n| \left( \exp{\left( -i \frac{\hat{\mathcal{H}}}{\hbar} (\Delta t)\right) } \right)^n |\vec{x}_{0}\rangle, \quad \text{with } \Delta t = \frac{t_n - t_0} {n} . \end{split} \end{equation}

    Note, that there is no real reason to keep the subscripts in the time differences either, as you just slice up the operator in regular intervals anyway. Now, you make a very important approximation by splitting the Hamiltonian operator into kinetic and potential term, using the Baker-Campbell-Hausdorff formula and noting, that the commutator term is of order (Δ t)2.

    \begin{equation} \begin{split}
    \exp{\left( i \frac{\Delta t}{\hbar} \hat{H} \right)} \approx \exp{\left( i \frac{\Delta t}{\hbar} \hat{T} \right)} \exp{\left( i \frac{\Delta t}{\hbar} \hat{V} \right)} + \mathcal{O}((\Delta t)^2)
    \end{split} \end{equation}

    At this point, no time dependence has been introduced in the individual "time slices", this is all on the level of operators! Since we plan to take the limit n to infinity anyway, we drop the (Δ t)2 term, and have successfully splitted the potential and kinetic operators. NOW we introduce the partitions of unity in terms of momentum and position eigenstates at each time slice, such that the momentum exponential acts to the left on the momentum eigenstates, and the position exponential to the right on position eigenstates, ie.

    \begin{equation} \begin{split}
    <p_j| \exp{\left( i \frac{\Delta t}{\hbar} \hat{T} \right)} \exp{\left( i \frac{\Delta t}{\hbar} \hat{V} \right)} |x_{j-1}> = <p_j | x_{j-1}> \exp{\left( i \frac{\Delta t}{\hbar} \frac{p_j^2}{2m} \right)} \exp{\left( i \frac{\Delta t}{\hbar} V(x_{j-1}) \right)}
    \end{split} \end{equation}

    Now, you have an actual time dependence in the exponential, introduced by the eigenstates and eigenvalues inserted at each time slice. Again, this was only possible after we have properly splitted the operator exponential into two seperate operators, whose eigenstates we know! (If we knew the eigenstates of the Hamiltonian itself, we would not need to do this, but then again, we would probably also not need the path integral at all!)
     
  7. Aug 10, 2017 #6
    How would you write the equations in your explanation in coordinate basis, i.e., not in Dirac notation?
     
  8. Aug 11, 2017 #7
    Thank you for your response.


    The question I asked about writing the equations in non-Dirac notation was a serious one. Once you rewrite your equations in the coordinate basis, i.e., non-Dirac notation, then you should see why your answer is incorrect.


    As to why I use the subscript ##n## with the Hamiltonian:

    \begin{equation}

    \begin{split}

    \psi(\vec{x}_n,t_n) &= \int_{-\infty}^{\infty} d\vec{x}_0 \psi(\vec{x}_0,t_0) K(\vec{x}_n,t_n;\vec{x}_0,t_0)

    \\

    &=\int_{-\infty}^{\infty} d\vec{x}_0 \psi(\vec{x}_0,t_0) \langle \vec{x}|e^{-i \frac{\mathcal{H}_n}{\hbar}(t_n-t_0)}|\vec{x}_0\rangle

    \end{split}

    \end{equation}


    Where:

    \begin{equation}

    \begin{split}

    \psi(\vec{x}_n,t_n) &=e^{-i \frac{\mathcal{H}_n}{\hbar}(t_n-t_0)}\psi(\vec{x}_n,t_0)

    \end{split}

    \end{equation}
     
  9. Aug 13, 2017 #8
    Not trying to sound rude, but: Just no!
    Not only is my answer mathematically sound, it also aligns with the standard treatment of path integral quantization in the standard references such as Peskin&Schroeder, Srednicki or Weinberg Vol. 1. I think vanshees already gave a link to his excellent lecture notes in your last thread, which also cover this construction: http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

    I'm going to repeat the key points once more:
    The Baker-Campbell-Hausdorff formula reads
    $$\exp(A+B) = \exp(A) \exp(B) \exp(-\frac{1}{2}[A,B] +...),$$
    where ... stands for higher order nested commutators. Now, for the non-relativistic single particle path integral, ##A=-\frac{i \hbar \Delta t}{2 m} \nabla^2## and ##B=\frac{i \Delta t}{\hbar} V(\vec{x})##, and so it is easy to see that ##[A,B]## does not vanish, so there are no joint eigenstates, and in general we have no hope of cleanly splitting the operators. However, by construction it is proportional to ##(\Delta t)^2##, which means that it is suppressed if and only if ##\Delta t \rightarrow 0##.

    Your definition of the matrix elements of time evolution is simply wrong, not much to say about it. It should be
    $$K(\vec{x}_n,t_n;\vec{x}_0,t_0) = <\vec{x_n}|\exp(\frac{i (t_n-t_0)}{\hbar}\hat{H})|\vec{x}_0>,$$
    no invented subscripts! I am still not sure if your Hn is supposed to refer to the position xn or to the time tn. In the first case it is simply nonsense, the Hamiltonian doesn't know anything about which state it is acting on. And since position states are not eigenstates of the Hamiltonian, there is no replacement operator->eigenvalue going on!

    The second case also doesn't work, the path integral construction as discussed only holds for time independent Hamiltonians. For time dependent Hamiltonians, the time evolution is defined through the time ordering exponential:
    ##U(t_n,t_0) = \mathcal{T} \exp(\frac{i}{\hbar} \int_{t_0}^{t_n} dt \hat{H}(t)),##
    where the time ordering acts on each of the series expansion terms separately. To path integral quantize such a system you need the much more powerful Keldysh formalism, which is able to handle arbitrary time dependent Hamiltonians and arbitrary (pure and mixed) density matrices, see eg. Kamenev: https://arxiv.org/abs/cond-mat/0412296 , or vanshees lecture notes: http://th.physik.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf .

    But you shouldn't go into that before the standard time independent construction is perfectly clear. I can only emphasize: Read and understand the standard references i gave you above, and stick with their notation, don't make up your own.
     
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