Integral Notations in Quantum Mechanics












2












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I've been learning about Quantum Dynamics, time evolution operators, etc. I am confused about the notation used in integrals. Normally I am used to integrals written in this way (with $dx$ on the right side):
$$int f(x)dx$$

In this manner of notation, I can easily see the integrand as it is sandwiched by the integral sign and the $dx$.
However, I often see integrals written in this way (with $dx$ beside the integral sign):
$$int dx f(x)$$
Is this notation not ambiguous? This is especially confusing for me when used in products, as I cannot identify what is the integrand sometimes. For example, I don't understand which is true in the following (when evaluating time evolution operator): $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$



The last line is especially confusing for me as I'm not sure if the integrand changes. Could I please get clarification for these different notations? Is there a reason for such notation? (If I'm not wrong, it is to group the integrals and the integrands in separate places for convenience? I'm not sure if it sacrifices clarity for this though.)



EDIT: There is also an issue of when operators are involved:
$$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$
How do you know which operator is in the integrand? And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?










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  • $begingroup$
    Related: physics.stackexchange.com/q/200378/2451, math.stackexchange.com/q/387572/11127
    $endgroup$
    – Qmechanic
    2 hours ago


















2












$begingroup$


I've been learning about Quantum Dynamics, time evolution operators, etc. I am confused about the notation used in integrals. Normally I am used to integrals written in this way (with $dx$ on the right side):
$$int f(x)dx$$

In this manner of notation, I can easily see the integrand as it is sandwiched by the integral sign and the $dx$.
However, I often see integrals written in this way (with $dx$ beside the integral sign):
$$int dx f(x)$$
Is this notation not ambiguous? This is especially confusing for me when used in products, as I cannot identify what is the integrand sometimes. For example, I don't understand which is true in the following (when evaluating time evolution operator): $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$



The last line is especially confusing for me as I'm not sure if the integrand changes. Could I please get clarification for these different notations? Is there a reason for such notation? (If I'm not wrong, it is to group the integrals and the integrands in separate places for convenience? I'm not sure if it sacrifices clarity for this though.)



EDIT: There is also an issue of when operators are involved:
$$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$
How do you know which operator is in the integrand? And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?










share|cite|improve this question











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  • $begingroup$
    Related: physics.stackexchange.com/q/200378/2451, math.stackexchange.com/q/387572/11127
    $endgroup$
    – Qmechanic
    2 hours ago
















2












2








2





$begingroup$


I've been learning about Quantum Dynamics, time evolution operators, etc. I am confused about the notation used in integrals. Normally I am used to integrals written in this way (with $dx$ on the right side):
$$int f(x)dx$$

In this manner of notation, I can easily see the integrand as it is sandwiched by the integral sign and the $dx$.
However, I often see integrals written in this way (with $dx$ beside the integral sign):
$$int dx f(x)$$
Is this notation not ambiguous? This is especially confusing for me when used in products, as I cannot identify what is the integrand sometimes. For example, I don't understand which is true in the following (when evaluating time evolution operator): $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$



The last line is especially confusing for me as I'm not sure if the integrand changes. Could I please get clarification for these different notations? Is there a reason for such notation? (If I'm not wrong, it is to group the integrals and the integrands in separate places for convenience? I'm not sure if it sacrifices clarity for this though.)



EDIT: There is also an issue of when operators are involved:
$$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$
How do you know which operator is in the integrand? And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?










share|cite|improve this question











$endgroup$




I've been learning about Quantum Dynamics, time evolution operators, etc. I am confused about the notation used in integrals. Normally I am used to integrals written in this way (with $dx$ on the right side):
$$int f(x)dx$$

In this manner of notation, I can easily see the integrand as it is sandwiched by the integral sign and the $dx$.
However, I often see integrals written in this way (with $dx$ beside the integral sign):
$$int dx f(x)$$
Is this notation not ambiguous? This is especially confusing for me when used in products, as I cannot identify what is the integrand sometimes. For example, I don't understand which is true in the following (when evaluating time evolution operator): $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$



The last line is especially confusing for me as I'm not sure if the integrand changes. Could I please get clarification for these different notations? Is there a reason for such notation? (If I'm not wrong, it is to group the integrals and the integrands in separate places for convenience? I'm not sure if it sacrifices clarity for this though.)



EDIT: There is also an issue of when operators are involved:
$$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$
How do you know which operator is in the integrand? And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?







quantum-mechanics notation integration






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edited 2 hours ago







Hexiang Chang

















asked 2 hours ago









Hexiang ChangHexiang Chang

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  • $begingroup$
    Related: physics.stackexchange.com/q/200378/2451, math.stackexchange.com/q/387572/11127
    $endgroup$
    – Qmechanic
    2 hours ago




















  • $begingroup$
    Related: physics.stackexchange.com/q/200378/2451, math.stackexchange.com/q/387572/11127
    $endgroup$
    – Qmechanic
    2 hours ago


















$begingroup$
Related: physics.stackexchange.com/q/200378/2451, math.stackexchange.com/q/387572/11127
$endgroup$
– Qmechanic
2 hours ago






$begingroup$
Related: physics.stackexchange.com/q/200378/2451, math.stackexchange.com/q/387572/11127
$endgroup$
– Qmechanic
2 hours ago












2 Answers
2






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4












$begingroup$

I started seeing $$int dx f(x)$$ in my freshman year of undergraduate.
It's pretty common and the more you learn about integration the more it makes sense.



Now, regarding this part:




$$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$




All of the equals signs there are correct.
Integrals factor like this:
$$
int dx int dy , f(x) , g(y) =
left( int dx , f(x) right) left( int dy ,g(y) right) , ,
$$

which is all you did there.
In fact, these are all the same:
begin{align}
int int dx , dy , f(x) g(y)
&= int dx int dy , f(x) g(y) \
&= int dx , dy , f(x) g(y) \
&= left( int dx , f(x) right) left( int dy , g(y) right) \
&= left( int dx , f(x) right) left( int dx , g(x) right) \
end{align}

Note, however, that you cannot factor something like this:
$$
int_0^t f(t') left( int_0^{t'} dt'' f(t'') right) dt'
$$

because the limit of the second integral depends on the first integral's integration variable.
You can, however, write it as
$$
int_0^t dt' f(t') int_0^{t'} dt'' f(t'') , .
$$



Operators




There is also an issue of when operators are involved:
$$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$




There's really no difference.
The key is to remember that the $dx$ really doesn't mean anything other than to remind you which variable(s) in the integrand is being integrated.
By convention we tend to write the $dx$ either at the front or at the end.
I've never seen it written in the middle like that.
I think everyone would know what you mean, but putting the $dx$ is the middle of the integrands runs the risk that a reader won't notice them.




How do you know which operator is in the integrand?




Ok that's a good question!
It really comes down to the fact that notation has to be clear.
If you use the symbol $x$ to denote both an integration variable and a not-integrated variable, that's just asking for trouble.
It also shouldn't ever happen because integration variables are consumed by the integral, so they can't be referred to anywhere else in an equation.
For example, this makes no sense:
$$ g(x) = int_0^1 sin(x) dx$$
because there's no "free" $x$ on the right hand side.




And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?




Well, you certainly would not write
$$ int dx hat F(x) hat G(x) neq left( int dx hat F(x) right) hat G(x) , .$$
That just makes no sense.






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  • $begingroup$
    Thank you for a great answer! Could you please address my edit as well?
    $endgroup$
    – Hexiang Chang
    2 hours ago










  • $begingroup$
    I'm so sorry, I forgot that I needed to ask that as well
    $endgroup$
    – Hexiang Chang
    1 hour ago



















1












$begingroup$

The notation is not ambiguous; it's purely convention. The correspondence is



$$
left(
int_{t_1}^{t_2} H(t) dt right) left( int_{t'_1}^{t'_2}H(t',t)dt' right)
iff int_{t_1}^{t_2}int_{t'_1}^{t'_2}H(t)H(t',t)dt' dt.
$$



That is instead of evaluating "inside out" we evaluate the integrals from right to left.





If you square an integral as



$$ left(int ^t_{t_0} dt' H(t')right)^2 $$



you should know that in general these two integrals don't talk to one another, except in very special cases. That is, they are completely separate entities.






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    2 Answers
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    2 Answers
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    active

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    4












    $begingroup$

    I started seeing $$int dx f(x)$$ in my freshman year of undergraduate.
    It's pretty common and the more you learn about integration the more it makes sense.



    Now, regarding this part:




    $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$




    All of the equals signs there are correct.
    Integrals factor like this:
    $$
    int dx int dy , f(x) , g(y) =
    left( int dx , f(x) right) left( int dy ,g(y) right) , ,
    $$

    which is all you did there.
    In fact, these are all the same:
    begin{align}
    int int dx , dy , f(x) g(y)
    &= int dx int dy , f(x) g(y) \
    &= int dx , dy , f(x) g(y) \
    &= left( int dx , f(x) right) left( int dy , g(y) right) \
    &= left( int dx , f(x) right) left( int dx , g(x) right) \
    end{align}

    Note, however, that you cannot factor something like this:
    $$
    int_0^t f(t') left( int_0^{t'} dt'' f(t'') right) dt'
    $$

    because the limit of the second integral depends on the first integral's integration variable.
    You can, however, write it as
    $$
    int_0^t dt' f(t') int_0^{t'} dt'' f(t'') , .
    $$



    Operators




    There is also an issue of when operators are involved:
    $$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$




    There's really no difference.
    The key is to remember that the $dx$ really doesn't mean anything other than to remind you which variable(s) in the integrand is being integrated.
    By convention we tend to write the $dx$ either at the front or at the end.
    I've never seen it written in the middle like that.
    I think everyone would know what you mean, but putting the $dx$ is the middle of the integrands runs the risk that a reader won't notice them.




    How do you know which operator is in the integrand?




    Ok that's a good question!
    It really comes down to the fact that notation has to be clear.
    If you use the symbol $x$ to denote both an integration variable and a not-integrated variable, that's just asking for trouble.
    It also shouldn't ever happen because integration variables are consumed by the integral, so they can't be referred to anywhere else in an equation.
    For example, this makes no sense:
    $$ g(x) = int_0^1 sin(x) dx$$
    because there's no "free" $x$ on the right hand side.




    And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?




    Well, you certainly would not write
    $$ int dx hat F(x) hat G(x) neq left( int dx hat F(x) right) hat G(x) , .$$
    That just makes no sense.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for a great answer! Could you please address my edit as well?
      $endgroup$
      – Hexiang Chang
      2 hours ago










    • $begingroup$
      I'm so sorry, I forgot that I needed to ask that as well
      $endgroup$
      – Hexiang Chang
      1 hour ago
















    4












    $begingroup$

    I started seeing $$int dx f(x)$$ in my freshman year of undergraduate.
    It's pretty common and the more you learn about integration the more it makes sense.



    Now, regarding this part:




    $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$




    All of the equals signs there are correct.
    Integrals factor like this:
    $$
    int dx int dy , f(x) , g(y) =
    left( int dx , f(x) right) left( int dy ,g(y) right) , ,
    $$

    which is all you did there.
    In fact, these are all the same:
    begin{align}
    int int dx , dy , f(x) g(y)
    &= int dx int dy , f(x) g(y) \
    &= int dx , dy , f(x) g(y) \
    &= left( int dx , f(x) right) left( int dy , g(y) right) \
    &= left( int dx , f(x) right) left( int dx , g(x) right) \
    end{align}

    Note, however, that you cannot factor something like this:
    $$
    int_0^t f(t') left( int_0^{t'} dt'' f(t'') right) dt'
    $$

    because the limit of the second integral depends on the first integral's integration variable.
    You can, however, write it as
    $$
    int_0^t dt' f(t') int_0^{t'} dt'' f(t'') , .
    $$



    Operators




    There is also an issue of when operators are involved:
    $$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$




    There's really no difference.
    The key is to remember that the $dx$ really doesn't mean anything other than to remind you which variable(s) in the integrand is being integrated.
    By convention we tend to write the $dx$ either at the front or at the end.
    I've never seen it written in the middle like that.
    I think everyone would know what you mean, but putting the $dx$ is the middle of the integrands runs the risk that a reader won't notice them.




    How do you know which operator is in the integrand?




    Ok that's a good question!
    It really comes down to the fact that notation has to be clear.
    If you use the symbol $x$ to denote both an integration variable and a not-integrated variable, that's just asking for trouble.
    It also shouldn't ever happen because integration variables are consumed by the integral, so they can't be referred to anywhere else in an equation.
    For example, this makes no sense:
    $$ g(x) = int_0^1 sin(x) dx$$
    because there's no "free" $x$ on the right hand side.




    And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?




    Well, you certainly would not write
    $$ int dx hat F(x) hat G(x) neq left( int dx hat F(x) right) hat G(x) , .$$
    That just makes no sense.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for a great answer! Could you please address my edit as well?
      $endgroup$
      – Hexiang Chang
      2 hours ago










    • $begingroup$
      I'm so sorry, I forgot that I needed to ask that as well
      $endgroup$
      – Hexiang Chang
      1 hour ago














    4












    4








    4





    $begingroup$

    I started seeing $$int dx f(x)$$ in my freshman year of undergraduate.
    It's pretty common and the more you learn about integration the more it makes sense.



    Now, regarding this part:




    $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$




    All of the equals signs there are correct.
    Integrals factor like this:
    $$
    int dx int dy , f(x) , g(y) =
    left( int dx , f(x) right) left( int dy ,g(y) right) , ,
    $$

    which is all you did there.
    In fact, these are all the same:
    begin{align}
    int int dx , dy , f(x) g(y)
    &= int dx int dy , f(x) g(y) \
    &= int dx , dy , f(x) g(y) \
    &= left( int dx , f(x) right) left( int dy , g(y) right) \
    &= left( int dx , f(x) right) left( int dx , g(x) right) \
    end{align}

    Note, however, that you cannot factor something like this:
    $$
    int_0^t f(t') left( int_0^{t'} dt'' f(t'') right) dt'
    $$

    because the limit of the second integral depends on the first integral's integration variable.
    You can, however, write it as
    $$
    int_0^t dt' f(t') int_0^{t'} dt'' f(t'') , .
    $$



    Operators




    There is also an issue of when operators are involved:
    $$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$




    There's really no difference.
    The key is to remember that the $dx$ really doesn't mean anything other than to remind you which variable(s) in the integrand is being integrated.
    By convention we tend to write the $dx$ either at the front or at the end.
    I've never seen it written in the middle like that.
    I think everyone would know what you mean, but putting the $dx$ is the middle of the integrands runs the risk that a reader won't notice them.




    How do you know which operator is in the integrand?




    Ok that's a good question!
    It really comes down to the fact that notation has to be clear.
    If you use the symbol $x$ to denote both an integration variable and a not-integrated variable, that's just asking for trouble.
    It also shouldn't ever happen because integration variables are consumed by the integral, so they can't be referred to anywhere else in an equation.
    For example, this makes no sense:
    $$ g(x) = int_0^1 sin(x) dx$$
    because there's no "free" $x$ on the right hand side.




    And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?




    Well, you certainly would not write
    $$ int dx hat F(x) hat G(x) neq left( int dx hat F(x) right) hat G(x) , .$$
    That just makes no sense.






    share|cite|improve this answer











    $endgroup$



    I started seeing $$int dx f(x)$$ in my freshman year of undergraduate.
    It's pretty common and the more you learn about integration the more it makes sense.



    Now, regarding this part:




    $$begin{align}left(int ^t_{t_0} dt' H(t')right)^2stackrel{?}{=}int ^t_{t_0} H(t') dt'int ^t_{t_0} H(t'')dt''\stackrel{?}{=}int ^t_{t_0} dt' H(t')int ^t_{t_0} dt'' H(t'')\stackrel{?}{=}int ^t_{t_0} dt'int ^t_{t_0} dt'' H(t') H(t'') end{align}$$




    All of the equals signs there are correct.
    Integrals factor like this:
    $$
    int dx int dy , f(x) , g(y) =
    left( int dx , f(x) right) left( int dy ,g(y) right) , ,
    $$

    which is all you did there.
    In fact, these are all the same:
    begin{align}
    int int dx , dy , f(x) g(y)
    &= int dx int dy , f(x) g(y) \
    &= int dx , dy , f(x) g(y) \
    &= left( int dx , f(x) right) left( int dy , g(y) right) \
    &= left( int dx , f(x) right) left( int dx , g(x) right) \
    end{align}

    Note, however, that you cannot factor something like this:
    $$
    int_0^t f(t') left( int_0^{t'} dt'' f(t'') right) dt'
    $$

    because the limit of the second integral depends on the first integral's integration variable.
    You can, however, write it as
    $$
    int_0^t dt' f(t') int_0^{t'} dt'' f(t'') , .
    $$



    Operators




    There is also an issue of when operators are involved:
    $$begin{align}int dx hat{F}(x) hat{G}(x)stackrel{?}{=}int hat{F}(x)dxhat{G}(x)\stackrel{?}{=}int hat{F}(x)hat{G}(x)dxend{align}$$




    There's really no difference.
    The key is to remember that the $dx$ really doesn't mean anything other than to remind you which variable(s) in the integrand is being integrated.
    By convention we tend to write the $dx$ either at the front or at the end.
    I've never seen it written in the middle like that.
    I think everyone would know what you mean, but putting the $dx$ is the middle of the integrands runs the risk that a reader won't notice them.




    How do you know which operator is in the integrand?




    Ok that's a good question!
    It really comes down to the fact that notation has to be clear.
    If you use the symbol $x$ to denote both an integration variable and a not-integrated variable, that's just asking for trouble.
    It also shouldn't ever happen because integration variables are consumed by the integral, so they can't be referred to anywhere else in an equation.
    For example, this makes no sense:
    $$ g(x) = int_0^1 sin(x) dx$$
    because there's no "free" $x$ on the right hand side.




    And assuming the general case where $hat{F}$ and $hat{G}$ do not commute, you cannot write the integral with $hat{G}(x)$ on the left of the integral. How is this not ambiguous?




    Well, you certainly would not write
    $$ int dx hat F(x) hat G(x) neq left( int dx hat F(x) right) hat G(x) , .$$
    That just makes no sense.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 2 hours ago









    DanielSankDanielSank

    17.6k45178




    17.6k45178












    • $begingroup$
      Thank you for a great answer! Could you please address my edit as well?
      $endgroup$
      – Hexiang Chang
      2 hours ago










    • $begingroup$
      I'm so sorry, I forgot that I needed to ask that as well
      $endgroup$
      – Hexiang Chang
      1 hour ago


















    • $begingroup$
      Thank you for a great answer! Could you please address my edit as well?
      $endgroup$
      – Hexiang Chang
      2 hours ago










    • $begingroup$
      I'm so sorry, I forgot that I needed to ask that as well
      $endgroup$
      – Hexiang Chang
      1 hour ago
















    $begingroup$
    Thank you for a great answer! Could you please address my edit as well?
    $endgroup$
    – Hexiang Chang
    2 hours ago




    $begingroup$
    Thank you for a great answer! Could you please address my edit as well?
    $endgroup$
    – Hexiang Chang
    2 hours ago












    $begingroup$
    I'm so sorry, I forgot that I needed to ask that as well
    $endgroup$
    – Hexiang Chang
    1 hour ago




    $begingroup$
    I'm so sorry, I forgot that I needed to ask that as well
    $endgroup$
    – Hexiang Chang
    1 hour ago











    1












    $begingroup$

    The notation is not ambiguous; it's purely convention. The correspondence is



    $$
    left(
    int_{t_1}^{t_2} H(t) dt right) left( int_{t'_1}^{t'_2}H(t',t)dt' right)
    iff int_{t_1}^{t_2}int_{t'_1}^{t'_2}H(t)H(t',t)dt' dt.
    $$



    That is instead of evaluating "inside out" we evaluate the integrals from right to left.





    If you square an integral as



    $$ left(int ^t_{t_0} dt' H(t')right)^2 $$



    you should know that in general these two integrals don't talk to one another, except in very special cases. That is, they are completely separate entities.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The notation is not ambiguous; it's purely convention. The correspondence is



      $$
      left(
      int_{t_1}^{t_2} H(t) dt right) left( int_{t'_1}^{t'_2}H(t',t)dt' right)
      iff int_{t_1}^{t_2}int_{t'_1}^{t'_2}H(t)H(t',t)dt' dt.
      $$



      That is instead of evaluating "inside out" we evaluate the integrals from right to left.





      If you square an integral as



      $$ left(int ^t_{t_0} dt' H(t')right)^2 $$



      you should know that in general these two integrals don't talk to one another, except in very special cases. That is, they are completely separate entities.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The notation is not ambiguous; it's purely convention. The correspondence is



        $$
        left(
        int_{t_1}^{t_2} H(t) dt right) left( int_{t'_1}^{t'_2}H(t',t)dt' right)
        iff int_{t_1}^{t_2}int_{t'_1}^{t'_2}H(t)H(t',t)dt' dt.
        $$



        That is instead of evaluating "inside out" we evaluate the integrals from right to left.





        If you square an integral as



        $$ left(int ^t_{t_0} dt' H(t')right)^2 $$



        you should know that in general these two integrals don't talk to one another, except in very special cases. That is, they are completely separate entities.






        share|cite|improve this answer











        $endgroup$



        The notation is not ambiguous; it's purely convention. The correspondence is



        $$
        left(
        int_{t_1}^{t_2} H(t) dt right) left( int_{t'_1}^{t'_2}H(t',t)dt' right)
        iff int_{t_1}^{t_2}int_{t'_1}^{t'_2}H(t)H(t',t)dt' dt.
        $$



        That is instead of evaluating "inside out" we evaluate the integrals from right to left.





        If you square an integral as



        $$ left(int ^t_{t_0} dt' H(t')right)^2 $$



        you should know that in general these two integrals don't talk to one another, except in very special cases. That is, they are completely separate entities.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 hours ago

























        answered 2 hours ago









        InertialObserverInertialObserver

        3,2181027




        3,2181027






























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