Suppose we had the following:

$$\langle \psi|\hat A|\psi\rangle$$

but $\psi$ can be broken into factors $\psi_1 \psi_2$ so

$$\langle \psi_1 \psi_2|\hat A|\psi_1 \psi_2\rangle$$

is there some way this can be split up?

Like perhaps?

$$=\langle \psi_1|\hat A|\psi_1\rangle \langle \psi_2|\hat A|\psi_2\rangle$$

I couldn't find any properties like this online or in my textbooks

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    $\begingroup$ That's because there is no such property. Think about $\hat{A}$ being the momentum operator (which is in essence a derivative) and you'll see that it doesn't satisfy this property. $\endgroup$ Feb 20, 2020 at 3:58
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    $\begingroup$ what is the meaning of $|\psi_1\psi_2\rangle$? Is it a tensor product? $\endgroup$ Feb 20, 2020 at 9:07
  • $\begingroup$ @user2723984 The question spawned from a Hydrogen atom problem I didn't understand. The idea was $|\psi \rangle = |\psi_1 \psi_2 \rangle = R_{n,l}(r)Y_{l,m}(\theta ,\phi)$. Please correct me if this approach makes no sense. $\endgroup$ Feb 20, 2020 at 9:39

2 Answers 2


This is a good example of misuse of Dirac's bra and ket notation. Kets are not intended to be a fancy notation for wavefunctions. They should represent states in a more abstract way than using position dependent wavefunctions. According to your clarification comment, you intend $|\psi\rangle$ as a shortened notation for $𝑅_{nl}(r)Y_{lm}(\theta, \phi)$. In a consistent Dirac notation it should be $$ \langle {\bf r}|\psi\rangle = 𝑅_{nl}(r)Y_{lm}(\theta, \phi). $$ Anyway, a part problems of notation, your question can be recast in the simpler formalism of position wavefunctions and operators acting on these functions. Let me indicate as $(f,{\bf A}f)$ the scalar product of a function $f$ by the function ${\bf A}f$ resulting from applying an operator ${\bf A}$ to wavefunction $f$. Then, it becomes the question whether, assuming that the $F$ Is factorized as $f(r,\theta,\phi)=R(r)Y(\theta,\phi)$, there is a simpler structure for $(f,{\bf A}f)$.

The answer is that it depends on ${\bf A}$.

If ${\bf A}= {\bf A_r}+{\bf A_{\theta,\phi}}$, i.e. is the sum of two operators, one acting on functions of $r$ only and not depending on $\theta$ and $\phi$, and the other acting on functions of $\theta,\phi$ and not depending on $r$, it is clear that $$ {\bf A}f = \left( {\bf A_r}+{\bf A_{\theta,\phi}} \right)R(r)Y(\theta,\phi)= \left[ {\bf A_r}R(r) \right]Y(\theta,\phi) + R(r)\left[ {\bf A_{\theta,\phi}} Y(\theta,\phi)\right] $$ and, if functions $R(r)$ and $Y(\theta,\phi)$ are normalized $$ (f,{\bf A}f)=(R,{\bf A_r}R) + (Y,{\bf A_{\theta,\phi}}Y). $$ It would be enough that one of the two operators would contain a function of the other coordinates to prevnt to obtain such result.

Of course, one can recast everything above in the formalism of bra, kets and tensor products between different Hilbert spaces. However I feel that plain examples like this may be the best starting point for a more powerful formalism.


If A is a one particle operator, such as kinetic energy or, effectively, the nuclear potential , then this reduces to

$$\langle \psi_1 \psi_2|\hat A|\psi_1 \psi_2\rangle = \langle \psi_1 |\hat A|\psi_1 \rangle + \langle \psi_2|\hat A| \psi_2\rangle$$

If A is a two particle operator such as the two electron interaction, then no reduction is possible.

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    $\begingroup$ This answer should probably warn about a common abuse of notation: the $\hat{A}$ on both sides of the equation are different operators. We're really talking about a two particle operator $\hat{A}_2 = \hat{A}\otimes \mathbb{1} + \mathbb{1} \otimes \hat{A}$. $\endgroup$
    – Javier
    Feb 20, 2020 at 13:33
  • $\begingroup$ No It is the same operator. It is the sum of two one particle operators. The unit operators are superfluous. If you are a quantum chemist solving a 1000 electron case you don't want to clutter your operators with a thousand unit operators. $\endgroup$
    – my2cts
    Feb 20, 2020 at 13:43
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    $\begingroup$ But clearly OP is not a quantum chemist solving a 1000 electron case; they are a beginner, and these notation tricks should probably be pointed out. $\endgroup$
    – Javier
    Feb 20, 2020 at 13:55
  • $\begingroup$ So don't teach the OP cluttered, useless notation.one day he may be that quantum chemist. $\endgroup$
    – my2cts
    Feb 20, 2020 at 14:11
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    $\begingroup$ @my2cts they would gain the clarity of knowing that operators for different particles act on different Hilbert spaces, and thus they would avoid potential future mistakes by manipulating operators that don't talk to each other; e.g. false commutation relations, resulting to false charges etc. $\endgroup$ Feb 20, 2020 at 23:12

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