Friday, 18 January 2019

hilbert space - Why does $langle psi_1 psi_2 | H_1|psi_1 psi_2 rangle= langle psi_1 | H_1|psi_1 rangle langle psi_2 | psi_2 rangle$?


In the solution of an exercise, the following reasoning is used:


$1)$ $\langle \psi_1 \psi_2 | H_1|\psi_1 \psi_2 \rangle + \langle \psi_1 \psi_2 | H_2|\psi_1 \psi_2 \rangle = \langle \psi_1 | H_1|\psi_1 \rangle \langle \psi_2 | \psi_2 \rangle + \langle \psi_1 | \psi_1 \rangle \langle \psi_2 | H_2|\psi_2 \rangle$


$2)$ $\langle \psi_1 \psi_2 | H_1|\psi_2 \psi_1 \rangle + \langle \psi_1 \psi_2 | H_2|\psi_2 \psi_1 \rangle = \langle \psi_1 | H_1|\psi_2 \rangle \langle \psi_2 | \psi_1 \rangle + \langle \psi_1 | \psi_2 \rangle \langle \psi_2 | H_2|\psi_1 \rangle$


$H_1$ and $H_2$ refer to the Hamiltonian operator.


May I ask you what property - and if possible the name of this property so that I can look it up on the internet - they used to go from the left hand side to the right hand side for both line $1)$ and $2)$?


I was unable to find anything on the internet (maybe I used the wrong terms ?), even on https://en.wikipedia.org/wiki/Bra%E2%80%93ket_notation#Properties




Answer



What you've written from your text/exercise is an abuse of notation, but it is standard.


The composite state $|\psi \alpha\rangle$ is the tensor product of the states $|\psi\rangle \in \mathcal H_1$ and $|\alpha\rangle \in \mathcal H_2$, which is sometimes written $|\psi\alpha\rangle \equiv |\psi\rangle \otimes |\alpha\rangle$. A very typical example of such a state is a spatial wavefunction attached to a spin-1/2 state, in which case $\mathcal H_1 = L^2(\mathbb R)$ and $\mathcal H_2 = \mathbb C^2$.


The space that such composite states belong to is the tensor product of the Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, usually written $\mathcal H_1 \otimes \mathcal H_2$. The inner product on such a space can be inherited from the inner products on $\mathcal H_1$ and $\mathcal H_2$. Letting $\psi,\phi \in \mathcal H_1$ and $\alpha,\beta\in\mathcal H_2$, we have


$$\langle \psi\alpha|\phi\beta\rangle_{\mathcal H_1 \otimes \mathcal H_2} = \langle\psi|\phi\rangle_\mathcal{H_1} \cdot \langle\alpha|\beta\rangle_{\mathcal H_2}$$


Given two operators $A$ and $B$ which act on $\mathcal H_1$ and $\mathcal H_2$, we can define a new operator which acts on $\mathcal H_1\otimes \mathcal H_2$ like this:


$$\big(A \otimes B\big)\big(|\psi\rangle \otimes |\alpha\rangle\big) = \big(A|\psi\rangle\big)\otimes\big(B|\alpha\rangle\big)$$


Putting everything together (and dropping the subscripts on the inner products,


$$\langle \psi\alpha| \big(A\otimes B\big) |\phi \beta\rangle = \langle \psi|A|\phi\rangle \cdot \langle \alpha|B |\beta\rangle $$





With that out of the way, if you have a composite system made by stitching together the Hilbert spaces $\mathcal H_1$ and $\mathcal H_2$, then the Hamiltonian for the composite system is


$$H = H_1 \otimes \mathbb I_2 + \mathbb I_1 \otimes H_2$$


where $H_{1,2}$ are the Hamiltonian operators acting on $\mathcal H_1$ and $\mathcal H_2$, and $\mathbb I_{1,2}$ are the identity operators on $\mathcal H_1$ and $\mathcal H_2$. Plugging this in to what I wrote above provides you with the answer you're looking for.




More specifically, note that


$$\langle \psi_1\psi_2|H_1\otimes\mathbb I | \psi_2\psi_1 \rangle = \langle \psi_1|H_1|\psi_2\rangle \cdot \langle \psi_2|\mathbb I|\psi_1\rangle = \langle \psi_1|H_1|\psi_2\rangle \cdot \langle \psi_2|\psi_1\rangle$$


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...