Braket Notation: Is <φ|x+y+z|φ> = <φ|x|φ> + <φ|y|φ> + <φ|z|φ>?

[brydustin]

If x,y,z are the position operators.

Is it true that:

<φ|x|φ> + <φ|y|φ> + <φ|z|φ> = <φ | x+y+z| φ> ?

So that if, for example, one wanted to compute <φ|r|φ> (where r =x+y+z), then they would just have to sum the parts.

I know that for scalars, a and b, we have the following:

(a+b)|φ> = a|φ> + b|φ>
But I don't know for sure if this is related at all to the case for operators (especially when they are sandwiched between the bra and the ket.

 

[Chopin]

Yes, it's true that [itex]\langle \psi | A + B | \psi \rangle = \langle \psi | A | \psi \rangle + \langle \psi | B | \psi \rangle[/itex]. But I'm not sure where you're going with the substitution [itex]r = x + y + z[/itex]. If you're trying to convert to spherical coordinates, that's not right--you need to do [itex]r = \sqrt{x^2 + y^2 + z^2}[/itex] instead.

 

[brydustin]

Yes, it's true that [itex]\langle \psi | A + B | \psi \rangle = \langle \psi | A | \psi \rangle + \langle \psi | B | \psi \rangle[/itex]. But I'm not sure where you're going with the substitution [itex]r = x + y + z[/itex]. If you're trying to convert to spherical coordinates, that's not right--you need to do [itex]r = \sqrt{x^2 + y^2 + z^2}[/itex] instead.

I am given the set of overlaps <φ_i|φ_j> and <φ_i|x|φ_j> for all i and j (as well as y and z sets). So that's four sets (the overlaps, and the set with x y and z).

So if I'm given those sets, how do I combine the sets {<φ_i|x|φ_j>, <φ_i|y|φ_j>, <φ_i|z|φ_j>} to get the single set {<φ_i|r|φ_j>}

 

[Chopin]

And the definition of [itex]r[/itex] is [itex]x + y + z[/itex]? Or something else?

 

[brydustin]

Could the following be a solution?


<φ|r|φ> = <φ|x|φ>i-hat + <φ|y|φ>j-hat + <φ|z|φ>k-hat

where i-hat is the unit vector of the x-axis, j-hat is the unit vector of the y-axis, and k-hat is the unit vector of the z-axis.

 

[brydustin]

And the definition of [itex]r[/itex] is [itex]x + y + z[/itex]? Or something else?

I double checked my notes:

r = x*i-hat +y*j-hat + z*k-hat *(by definition)

 

[Chopin]

Ah, I see, they're defining [itex]r[/itex] as a vector-valued operator. In that case, there's actually no new math here, it's just a notation trick. Specifically, it's a way of unifiying three separate operators together into one object that's easy to keep track of. So we have three operators [itex]\hat{x}, \hat{y}, \hat{z}[/itex], and we're defining the new object [itex]\hat{\textbf{r}} = (\hat{x}, \hat{y}, \hat{z})[/itex].

In this case, applying the operator to a state is really applying three separate operators to the state, and using them to make an ordered triple. [itex]\langle \psi | \hat{\textbf{r}} | \psi \rangle = \langle \psi | (\hat{x}, \hat{y}, \hat{z}) | \psi \rangle = (\langle \psi | \hat{x} | \psi \rangle, \langle \psi | \hat{y} | \psi \rangle, \langle \psi | \hat{z} | \psi \rangle)[/itex].

 

[brydustin]

Final question:

What does it mean to do the following:

(<φ|r|φ> - <ψ|r|ψ>)^2

Obviously if <ψ|r|ψ> is a vector, then the difference of two vectors is straightforward. What does it mean to take their square?

 

[Chopin]

Well, think about what it means to take the square of a normal vector. Since [itex]x^2 = x\cdot x[/itex] (where [itex]x[/itex] is a normal number), then the square of a vector must be [itex]\textbf{r}^2 = \textbf{r} \cdot \textbf{r}[/itex]. By the definition of the dot product, [itex]\textbf{r} \cdot \textbf{r} = r_x^2 + r_y^2 + r_z^2[/itex]. The generalization to the quantum case should be pretty straightforward.