Basis Transformation for Wave Function

And for \phi, it acts as above.In summary, the conversation discusses the transformation of a wave function from Cartesian to spherical coordinates, specifically focusing on the effect of the rotation operator on the bra in the spherical wave function. The steps for this transformation are mathematically shown, including the use of the "known result from wave mechanics" in Equation (3.6.9). The conversation also addresses the question of how L_z acts on r, \theta, and \phi, with the conclusion that L_z would be zero for r and \theta, and act as an operator on \phi.
  • #1
Proofrific
12
0

Homework Statement



It's not a homework problem. I'm reading my textbook (Sakurai's Modern QM), and I'm not sure about a step (eq 3.6.6 through 3.6.8). Here it is:

We start with a wave function that's been rotated:
[tex]\langle x' + y' \delta \phi, y' - x' \delta \phi, z' | \alpha \rangle[/tex]

Now, we change the coordinate basis from Cartesian to spherical. That is,
[tex]\langle x', y', z' | \alpha \rangle \to \langle r, \theta, \phi | \alpha \rangle[/tex]

I want to show that the previous wave function transforms to:
[tex]\langle r, \theta, \phi - \delta \phi | \alpha \rangle[/tex]

Homework Equations



See above.

The Attempt at a Solution



Physically, I understand that we're rotating by angle [tex]\delta \phi[/tex] about the z-axis. So, it makes sense that the bra would acquire a [tex]-\delta \phi[/tex] (the ket would acquire a positive). But, I'm not sure how to mathematically show the steps (which is needed for more complicated transforms, such as Sakurai eq. 3.6.10 to 3.6.11).
 
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  • #2
You are essentially acting the rotation operator, [itex]\mathcal{D}[/itex], on the bra in the spherical wavefunction:

[tex]
\langle r,\theta,\phi|\mathcal{D}|\alpha\rangle=\langle r,\theta,\phi\left|1-i\left(\frac{\delta\phi}{\hbar}\right)L_z\right|\alpha\rangle=\langle r,\theta,\phi-\delta\phi|\alpha\rangle
[/tex]

Mathematically, you are multiplying each term in [itex]\langle r,\theta,\phi|[/itex] by [itex]\mathcal{D}[/itex], but since [itex]L_zr=L_z\theta=0[/itex], you are left with

[tex]
\langle r,\theta,\phi|\mathcal{D}=\langle r\mathcal{D},\theta\mathcal{D},\phi\mathcal{D}|\rightarrow\left.\langle r,\theta,\phi\left(1-i\left(\frac{\delta\phi}{\hbar}\right)L_z\right)\right|=\langle r,\theta,\phi-\delta\phi|
[/tex]

Equations (3.6.10) and (3.6.11) use the "known result from wave mechanics" in Equation (3.6.9):

[tex]
\langle\mathbf{x}'|L_z|\alpha\rangle=-i\hbar\frac{\partial}{\partial\phi}\langle\mathbf{x}'|\alpha\rangle
[/tex]

and this is not the same as the second equation above because you do not have the [itex]\propto(1-\delta\phi L_z)[/itex] acting on the bra, you just have [itex]L_x[/itex], [itex]L_y[/itex], or [itex]L_z[/itex]. Converting [itex]L_x[/itex] and [itex]L_y[/itex] from Cartesian into spherical should give you Equations (3.6.10) and (3.6.11) when using the "known result" of Equation (3.6.9)
 
  • #3
jdwood983 said:
Mathematically, you are multiplying each term in [itex]\langle r,\theta,\phi|[/itex] by [itex]\mathcal{D}[/itex], but since [itex]L_zr=L_z\theta=0[/itex], you are left with

Thanks for the help!

How do you know that [itex]L_zr=L_z\theta=0[/itex]? The only thing I remember about [itex]L_z[/itex] is how it acts on an eigenstate: [itex]L_z|l,m\rangle = m \hbar | l,m \rangle[/itex]. How do you know how it acts on [itex]r[/itex], [itex]\theta[/itex], and [itex]\phi[/itex]?
 
  • #4
Proofrific said:
Thanks for the help!

How do you know that [itex]L_zr=L_z\theta=0[/itex]? The only thing I remember about [itex]L_z[/itex] is how it acts on an eigenstate: [itex]L_z|l,m\rangle = m \hbar | l,m \rangle[/itex]. How do you know how it acts on [itex]r[/itex], [itex]\theta[/itex], and [itex]\phi[/itex]?

Well [itex]L_z[/itex] is an operator equivalent to

[tex]
L_z=-i\hbar\frac{\partial}{\partial\phi}
[/tex]

So since [itex]r[/itex] and [itex]\theta[/itex] have no [itex]\phi[/itex] component, [itex]L_z[/itex] acting on them would be zero.
 

Related to Basis Transformation for Wave Function

What is basis transformation for wave function?

Basis transformation for wave function is a mathematical process used to express a wave function in a different set of basis functions. It allows for a different representation of the wave function that may be more suitable for a specific problem or calculation.

Why is basis transformation important?

Basis transformation is important because it allows for a more efficient and accurate representation of the wave function. It can also simplify complex calculations and provide insight into the behavior of the system being studied.

What are the steps involved in basis transformation?

The first step in basis transformation is to choose the new set of basis functions. Next, the wave function is expressed as a linear combination of these new basis functions. The coefficients of the linear combination are determined by solving a set of equations using the original and new basis functions. Finally, the transformed wave function is obtained by substituting the calculated coefficients into the linear combination.

What is the difference between a unitary and non-unitary basis transformation?

A unitary basis transformation preserves the norm of the wave function, meaning that the total probability of the system remains the same. On the other hand, a non-unitary basis transformation does not preserve the norm and can result in a change in the total probability of the system.

Can basis transformation be applied to any type of wave function?

Yes, basis transformation can be applied to any type of wave function as long as the new set of basis functions is chosen appropriately. However, the complexity and accuracy of the transformation may vary depending on the specific wave function and basis functions being used.

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