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I have a problem understanding three arguments in quantum mechanic:

  1. When we talk about a particle in a central field we have this kind of Hamiltonian: $$H=\frac{p^2}{2m}+V(r)$$ if we use spherical coordinates we can write the Laplacian operator as: $$\nabla^2=\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}(\frac{\partial^2}{\partial \theta^2}+\frac{cos(\theta)}{sin(\theta)}\frac{\partial}{\partial \theta}+\frac{1}{sin^2(\theta)}\frac{\partial^2}{\partial \phi^2})$$ and so $p^2$ in Schordinger's representation as:

$$p^2=-\hbar^2(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r})+\frac{L^2}{r^2}$$ so that the entire Schrodinger's equation becomes:

$$(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}-\frac{L^2}{\hbar^2r^2}-\frac{2mV(r)}{\hbar^2}+\frac{2mE}{\hbar^2})\psi(r,\theta,\phi)=0 \ (1)$$

Now the problem: my professore said that "

since a central Hamiltonian commutes with the operators $L^2$ and $L_z$ (I know why), we write the solution of equation (1) in the factorised form:

$$\psi(r,\theta,\phi)=F(\theta,\phi)R(r)$$

First question: Why, from the fact that a central Hamiltonian commutes with the operators $L^2$ and $L_z$, we come to write the eigenfunction in that "separated" form? I think the answer is related to the fact that if 2 operators commute with each other they have a complete set of simultaneous eigenfunctions, but I can't understand why this lead to the "separate" form of the Hamiltonian eigenfuntion.

  1. The other topic that causes me the same confusion is the following one:

talking about the Hydrogen atom we have $$H=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2}+V(|q_1-q_2|)$$ which becomes, after a change of variables $$H=\frac{P^2}{2M}+\frac{p^2}{2\mu}+V(|q|) \ (2)$$

where $M=m_1+m_2$ and $\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$

Now in my notes I've written that

the equation (2) is a separable variables equation, namely the sum of two terms $H_1$ which only depends from the center of mass variables and $H_2$ which only depends from relative variables, so $H_1$ and $H_2$ commute each other. Since the total Hamiltonian is separable we can search the total eigenfunction in the form $\psi(X,x)=\phi(X)\psi(x)$ (3) where $X$ refers to center of mass variable and $x$ to relative variables.

I can now write equation $(2)$ in Schrodinger's representation:

$$(\frac{-\hbar^2\nabla^2_X}{2M}-\frac{-\hbar^2\nabla^2_x}{2\mu}+V(|X|)\psi(X,x)=E\psi(X,x)) \ (4)$$

then I replace equation $(3)$ into equation $(4)$ obtaining:

$$(-\frac{\hbar^2}{2M}\nabla^2_X\phi(X))\psi(x)+((\frac{-\hbar^2}{2\mu}+V(|x|))\psi(x))\phi(X)=E\phi(X)\psi(x)$$

I can now divide by $\phi(X)\psi(x)$:

$$-\frac{\hbar^2}{2M}\frac{\nabla^2_X\phi(X)}{\phi(X)}-\frac{\hbar^2}{2\mu}\frac{\nabla^2_x\psi(x)}{\psi(x)}+V(|x|)=E$$

which leads to two separated eigenvalues equations:

$$-\frac{\hbar^2}{2M}\nabla^2_X\phi(X)=E_1\phi(X)$$

$$-\frac{\hbar^2}{2\mu}\nabla^2_x\psi(x)+V(|x|)\psi(x)=E_2\psi(x)$$

which I can solve separately.

Here the question is: Why the fact that $H=H_1+H_2$ and $H_1$ commutes with $H_2$ brings me to search the total eigenfunction in the form $\psi(X,x)=\phi(X)\psi(x)$?

  1. In a website (sorry, I can't find it anymore) I've read this when talking about spherical harmonics:

"if we want to determine simultaneous eigenfuntions of $L^2$ and $L_z$ we have to solve this system:

$$L^2|l,m\rangle=\hbar^2l(l+1)|l,m\rangle$$ $$L_z|l,m\rangle=\hbar m|l,m\rangle$$

in Schrodinger's representation:

$$-(\frac{1}{sin(\theta)}\frac{\partial}{\partial{\theta}}(sin(\theta)\frac{\partial}{\partial{\theta}})+\frac{1}{sin^2(\theta)}\frac{\partial^2}{\partial{\phi^2}})\psi_{lm}(r,\theta,\phi)=l(l+1)\psi_{lm}(r,\theta,\phi)$$

$$-i\frac{\partial}{\partial{\phi}}\psi_{lm}(r,\theta,\phi)=m\psi_{lm}(r,\theta,\phi)$$

we can note that operators $L^2$ and $L_z$ don't contain variable $r$ so the $\psi_{lm}(r,\theta,\phi)$ must be of the form $\psi_{lm}(r,\theta,\phi)=f(r)Y_{lm}(\theta,\phi)$.

Third question: Why if operators $L^2$ and $L_z$ don't contain variable $r$ the $\psi_{lm}(r,\theta,\phi)$ must be of the form $\psi_{lm}(r,\theta,\phi)=f(r)Y_{lm}(\theta,\phi)$?

The final question about all this is:

are the three problems I've written related in a certain way? because all three propose an eigenfunction written as a product of eigenfunctions but in the first case this is due to the fact that $H$, $L^2$, $L_z$ commute with each other. In the second case this is due to the fact that $H$ can be written as a sum of Hamiltonians that commute with each other and we speak of separation of variables. In the third case it is due to the fact that the variable $r$ does not appear in the operators $L^2$ and $L_z$. Do all three mean the same thing or are they three different things?

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  • $\begingroup$ I think that is an excellent question for the MathStackExchange $\endgroup$ Jan 17, 2022 at 4:42
  • $\begingroup$ @HolgerFiedler do you think I can post it on MathStackExchange? There's a lot fo physics in it. $\endgroup$
    – Salmon
    Jan 17, 2022 at 12:00
  • $\begingroup$ It seems to me that the hydrogen atom simply serves as an example for a separable problem and what you really want to know is how separability works. Is that observation correct ? $\endgroup$
    – Hans Wurst
    Jan 18, 2022 at 13:31
  • $\begingroup$ @HansWurst Not properly, I would like to know why, in the three problems I have proposed, the wave function is written as a product of different terms but the reason why this is done seems to be different each time. In the first case it is divided into radial part and angular part because the three operators commute while in the third case because the two operators do not contain the variable $r$. Is there anything linking these three cases? $\endgroup$
    – Salmon
    Jan 18, 2022 at 14:32
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    $\begingroup$ There is a very nice discussion of this in Sakurai's "Modern Quantum Mechanics", pages 30-32. For two commuting observables $A$ and $B$, $[A,B]=0$, we can write simultaneous eigenkets $|a',b'\rangle$ such that $A|a',b'\rangle = a'|a',b'\rangle$ and $B|a',b'\rangle=b'|a',b'\rangle$. Sakurai does the proof for a non degenerate spectrum of $A$, but explains how this can be shown to hold also for an $n$-fold degeneracy. There follows a discussion of angular momentum specifically. $\endgroup$ Jan 20, 2022 at 17:54

1 Answer 1

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Note: My approach lacks rigour and I'm rusty. I'm sure there are lots of holes, overly-complicated parts, and perhaps even just outright mistakes. However, I think the gist of the argument motivates why two operators which commute leads one to consider solutions of the form you mention.

Paraphrased: Why does does the fact that $\hat{A}$ and $\hat{B}$ commute in the Hamiltonian $\hat{H}=\hat{A} + \hat{B}$ mean that we can write the solution as factors of the solutions of $\hat{A}$ and $\hat{B}$?

Essentially, the fact that the Hamiltonian's parts, $\hat{A}$ and $\hat{B}$, commute means we can separate the general solution (of the form $\exp(-i\hat{H})$ into two exponential operators.

My approach lacks rigour, but I think it shows the idea:

For the equation $\newcommand{\Ket}[1]{\left|#1\right>}$ $\newcommand{\Bra}[1]{\left<#1\right|}$ $\newcommand{\Braket}[2]{\left<{#1}|{#2}\right>}$ \begin{equation} \hat{H}\Ket{\psi} = i\frac{\partial}{\partial t}\Ket{\psi} \tag{1}\label{SE} \end{equation} the general solution is known to be \begin{align} \Ket{\psi(t)} &= \exp\left(-it\hat{H} \right) \Ket{\psi(0)} \\ &= \exp\left(-it(\hat{A}+\hat{B}) \right) \Ket{\psi(0)} \tag{2} \end{align} You can see this through substitution or look it up in any QM textbook or course notes. The fact that $[A,B]=0$ means we can split the exponential operator, giving \begin{align} \Ket{\psi(t)} &= \exp\left(-it\hat{H} \right) \Ket{\psi(0)} \\ &= \exp\left(-it\hat{A} \right)\exp\left(-it\hat{B} \right) \Ket{\psi(0)} \tag{3} \end{align} Let's try to write this in position representation.

First, let's recognise that we're dealing with more than one spatial coordinate and each operator is acting in mutually exclusive vector spaces, so that $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B$. Alternatively, in position representation, we can see that $\hat{A}$ has an $x$-dependence and $\hat{B}$ has a $y$-dependence (and can be seen to commute by observation).

Also, the time-independent solutions for $\hat{A}$ and $\hat{B}$ separately in position representation are \begin{align} f(x) &= \Braket{x}{f} = \sum_i\lambda_i\Braket{x}{a_i} \\ g(y) &= \Braket{y}{g} = \sum_i\mu_i\Braket{x}{b_i} \tag{4} \end{align} where the eigenfunctions are \begin{align} \hat{A}\Ket{a_i} &= a_i \Ket{a_i} \\ \hat{B}\Ket{b_i} &= b_i \Ket{b_i} \tag{5} \end{align}

Then using $\Braket{x,y}{\psi} = \psi(x,y)$ with $\Bra{x,y} = \Bra{x}\otimes\Bra{y}$, we get \begin{align} \Braket{x,y}{\psi} =& \left[\Bra{x}\exp\left(-it\hat{A} \right) \otimes \Bra{y}\exp\left(-it\hat{B} \right)\right] \Ket{\psi(0)} \tag{6} \end{align}

which we can write using the expansion of the identity as \begin{align} \Braket{x,y}{\psi} =& \left[\sum_n\Bra{x}\exp\left(-it\hat{A} \right) \Ket{a_n}\Bra{a_n} \right. \\ &\otimes \left. \sum_m\Bra{y}\exp\left(-it\hat{B} \right)\Ket{b_m}\Bra{b_m}\right]\psi(0) \tag{7} \end{align} where the operators can now act on their respective eigenstates, giving \begin{align} \Braket{x,y}{\psi(t)} =& \sum_n\Braket{x}{a_n}\exp\left(-ita_n \right)\sum_m\Braket{y}{b_m}\exp\left(-itb_m \right) \\ &\int\int dx^\prime dy^\prime\Braket{a_n}{x^\prime}\Braket{b_m}{y^\prime} \Braket{x^\prime,y^\prime}{\psi(0)} \tag{8}\label{FINALEQ} \end{align} where I've simplified the expression using $$\left(\Bra{x^\prime}\otimes\Bra{y^\prime}\right)\Ket{\psi(0)} = \Braket{x^\prime,y^\prime}{\psi(0)} \tag{9}$$ Note that $\Braket{x}{a_n}$ are the eigenfunctions of $\hat{A}$ and $\Braket{y}{a_m}$ are the eigenfunctions of $\hat{A}$ in position representation.

So, we can see that $\ref{FINALEQ}$ is simply a linear combination of factors of eigenstates of the individual operators $\hat{A}$ and $\hat{B}$: \begin{align} \Braket{x,y}{\psi(t)} =& \sum_{m,n}c_{m,n}f_n(x,t)g_m(y,t) \tag{10} \end{align} This is a general solution.

Ignoring the time-dependence, we can see that just one of those solutions (not constrained by any initial conditions) is $f_n(x,t)g_m(y,t) = \Braket{x}{a}\Braket{y}{b}$ i.e. the multiplication of the eigenstates of the operators separately. Generally, we can consider a linear combination of such and it's still a solution, and in this case, we've worked backwards from that.

I think this is also shows (maybe a bit strong given the lack of rigour... "suggests"?) that the independence of the variables leads us to consider states in different subspaces of $\mathcal{H}$, and hence operators in those different spaces commute, and therefore the general solution in position representation is formed of products of the separate solutions to those subspaces.

It doesn't matter that we're talking about $r$, $\phi$, and $\theta$, or something simpler like $x$ and $y$ - they're just specific representations of the fundamental states. Therefore, we can propose solutions of the form $\psi(x,y) = f(x)g(y)$.

Now, knowing this, whenever we see a Hamiltonian where $H=H_1 + H_2$, and the parts commute, we can jump straight to the conclusion and state $\psi(x,y) = f(x)g(y)$, where $f(x)$ and $g(y)$ are the solutions, and then immediately go about calculating $f(x)$ and $g(y)$.

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  • $\begingroup$ Even though I suggested an answer, I agree with the original comments that you should post it in the mathematics forums in order to a rigorous, comprehensive answer. $\endgroup$
    – Paul
    Jan 19, 2022 at 14:10
  • $\begingroup$ Thank you for your reply but perhaps my question is very badly written, I meant something much simpler. I will try to rewrite it. $\endgroup$
    – Salmon
    Jan 19, 2022 at 15:16
  • $\begingroup$ I thought you were interested in demonstrating why commuting operators in the Hamiltonian lead to separating the solution into functions of different independent variables? $\endgroup$
    – Paul
    Jan 20, 2022 at 19:21
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    $\begingroup$ What I tried to show is that there is a single underlying reason for all of your examples, that being that for commuting operators, which act on different independent variables, eventually leads to a general solution which is a product of the separate solutions. We can create different examples in particular representations, with different combinations of variables, or for particular potentials, but we're always going to get to same conclusion if we satisfy the same criteria. $\endgroup$
    – Paul
    Jan 21, 2022 at 18:54
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    $\begingroup$ I've added a little more to the end of the proposed answer. Basically, where my analysis is valid, we know Hamiltonians of the form $H=H_1+H_2$ have a solution which looks like the product of eigenstates of $H_1$ and $H_2$. Once we know that, we can skip the analysis and jump straight to the conclusion, and then try to work out what those eigenstates actually are. $\endgroup$
    – Paul
    Jan 21, 2022 at 19:02

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