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Suppose I have two operators $J^2$ and$J_z$ where they represent the length of angular momentum and its $z$ component respectively.

Sure, it's legal to write new operators like $J^2-J_z^2$ or $J_x+J_z$ but it's not legal to write $J^2-J_z$ because they don't have the same physical dimension.

However, I want to know what prevent them from becoming operators in ket space mathematically.

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  • $\begingroup$ If you neglect the units then there isn't anything wrong. The sum of two linear operators is a linear operator. Why do you think this is not the case? $\endgroup$ Oct 2 '18 at 17:54
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    $\begingroup$ There is nothing mathematically that would tell you if an operator makes physical sense or not. That is exactly the point where math ends and physics begins. Many consider this a fatal flaw of math, yet its power is to transcend physical reality. $\endgroup$ Oct 2 '18 at 18:57
  • $\begingroup$ @AaronStevens My professor tells us that if you find two operators with different units acting on the same ket then you are making mistakes. I think he is reasonable physically but mathematically this can be correct. This is why I am asking the question. $\endgroup$ Oct 2 '18 at 21:18
  • $\begingroup$ That is what I am saying, and you are correct. $\endgroup$ Oct 2 '18 at 21:38
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    $\begingroup$ @AaronStevens. Yes, that's the point I'm trying to make. But it's not just because of different units that you cannot add $\mathrm{km}$ and $\mathrm{km}^2$. It's because they are of different physical dimension, $\text{length}$ and $\text{length}^2$. $\endgroup$
    – md2perpe
    Oct 3 '18 at 6:16
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If $A,\,B$ are operators you can sum, then $\langle\phi|A+B|\psi\rangle$ exists and is equal to $\langle\phi|A|\psi\rangle+\langle\phi|B|\psi\rangle$; so those terms have the same dimension, and hence so do $A,\,B$.

Suppose $|\phi\rangle$ lives in a Hilbert space $H$ over the field $F$, so $a,\,b\in F$ satisfy $a+b+ab\in F$, making $a,\,b$ dimensionless. While a Hilbert space $H'$ of operators from $H$ to $H$ is closed under summation with coefficients $\in F$, these coefficients are dimensionless, so $J_z$ is not required to live in the same Hilbert space as $J^2$.

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  • $\begingroup$ J^2 and Jz they do live in the same Hilbert space, I don't really get your point. $\endgroup$ Oct 2 '18 at 21:15
  • $\begingroup$ @UniverseMaintainer Technically each dimension for operators obtains a different HS because they map the original kets to different spaces, because expressing the image in the original basis requires dimensionful coefficients. $\endgroup$
    – J.G.
    Oct 2 '18 at 21:21
  • $\begingroup$ @UniverseMaintainer. They act on the same Hilbert space, but they live in different operator spaces. Let $\mathcal{O}(\mathcal{X})$ denote the space of operators with values of physical dimension $\mathcal{X}$, and let $\mathcal{J}$ denote the physical dimension of angular momentum. Then $J_z \in \mathcal{O}(\mathcal{J}),$ but $J^2 \in \mathcal{O}(\mathcal{J}^2).$ $\endgroup$
    – md2perpe
    Oct 2 '18 at 21:31
  • $\begingroup$ @J.G. Do you suggest even in math, operators with different dimensions can not be added? $\endgroup$ Oct 2 '18 at 21:51
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    $\begingroup$ In mathematics physical dimensions normally don't exist. In mathematics we can also construct formal sums of very different linear spaces. So we can write $1\ \rm{m} + 1\ \rm{m}^2$ although it has no deeper meaning than constructing a pair $(1\ \rm{m}, 1\ \rm{m}^2)$ and making this an object in a linear space in a natural way. $\endgroup$
    – md2perpe
    Oct 3 '18 at 12:01
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There’s nothing inherently illegal about something like $\hat A^2+\hat A$. Indeed, functions of an operator $A$ are usually defined by their series expansion, the most common examples being the time evolution $$ U(t)=e^{-it\hat H/\hbar}=1-i\frac{t\hat H}{\hbar} +\frac{1}{2} \left(\frac{i\hat H}{\hbar}\right)^2+\ldots \tag{1} $$ and the rotation about an axis - say $\hat z$: $$ R_z(\alpha)=e^{-i\alpha \hat L_z/\hbar}=1-i\frac{\alpha \hat L_z}{\hbar} +\frac{1}{2}\left(\frac{i\hat L_z}{\hbar}\right)^2 + \ldots \tag{2} $$ Of course, every term in the sums (1) or (2) is dimensionally consistent: in this specific case $t \hat H/\hbar$ and $\alpha\hat L_z/\hbar$ are both dimensionless. They need to be consistent for the same reason that you cannot add something measured in meters with something measured in meters squared. Indeed imagine you work with an eigenbasis of $\hat A$: then there is no physics sense in adding powers of dimensionful eigenvalue, as it would precisely amount to adding quantities having different units.

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Maybe you are worried that there are expressions like $j(j+1)$ in the angular momentum algebra, but really the $J$'s have the same dimensions as $\hbar$ and eigenvalues $\hbar j$ where $j$ is an integer or a half integer.

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Operators can be added if they have the same dimension or if they are dimensionless. So $J^2/h^2 + J_z/h$ makes sense but $J^2 + J_z$ does not.

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