Finding H2 Given |\uparrow> and H1

[johnsmi]

Hi,

I'm trying to understand something:
Supposing I have two states:
|$$\uparrow$$> which is an eigen state of H₁
and $$\psi$$ which is the space rep. of an eigen state of H₂

Now, as far as I understand |$$\uparrow$$> $$\otimes$$ $$\psi$$ will be an eigen state of H₁ $$\otimes$$ H₂ (is that true or is it the sum of Hamiltonians?)

But supposing I knew that |$$\uparrow$$> $$\otimes$$ $$\psi$$ was an eigen state of a known H and I knew |$$\uparrow$$> and H₁ how could I find H₂?

Thank you

 

[0xDEADBEEF]

It is the sum of the Hamiltonians and the product of the wavefunctions so H2 = H - H1

 

[johnsmi]

It is the sum of the Hamiltonians and the product of the wavefunctions so H2 = H - H1

Thank you for you reply but it doesn't quite fit with something:

https://www.physicsforums.com/showthread.php?t=416185" is a post I posted with all the details of the question and as you can see it doesn't seem to apply (does it?)

 

[qbert]

Actually, H₂ is exactly H - H₁ provided it is
suitably interpreted. However what you want from your other post
is not H₂ at all. You want an effective Hamiltonian.

I'll try to explain.
Given vector spaces X and Y we can form a NEW vector space
X$\otimes$Y. For any pair of operators (A,B) with
A:X$\rightarrow$X, B:Y$\rightarrow$Y there
exists a unique map A$\otimes$B.

Why is this important?
When you're looking at your hamiltonian as acting on the product
what you start with are the maps A and B above but you need to
be carefull about how they lift. Specifically H₁ lifts
to H₁$\otimes$id₂. Where id₂
is the identity map on the second factor.

Similarly H₂ lifts to id₁$\otimes$H₂.

So that your hamiltonian is actually
H = H₁$\otimes$id₂ + id₁$\otimes$H₂.

Now how can we use this to "factor out" a piece, or find an effective hamiltonian
Suppose we know the eigenbasis of H₂. And further suppose we've
orthonormalized it, then we can look at the matrix elements of H, which stay
in a fixed H₂ subspace.
Let $|m>, |n>$ denote vectors in the first space, and $|\mu>, |\nu>$
denote orthonormal eigenvectors of H₂.

Then
$$<m \otimes \mu| H |n \otimes \nu> = <m \otimes \mu| (H_1 \otimes \text{id}_2) |n \otimes \nu> + <m \otimes \mu| (\text{id}_2 \otimes H_2) |n \otimes \nu>$$
$$= <m|H_1|n><\mu|\nu> + <m|n>\lambda_\nu<\mu|\nu>$$

where $\lambda_\nu$ is the appropriate eigenvalue. Now
if we are only interested in a given state $\nu$ we
can put
$$<m \otimes \nu| H |n \otimes \nu> = <m|H_1|n> + <m|n>\lambda_\nu = <m| H_1 + \lambda_\nu \text{id}_1 |n>$$

This looks then like a hamiltonian only on the 1st factor.

 

[johnsmi]

Wow,
thank you for your detailed post now all I need to do is understand it. I think it will take me some time to properly understand it.

Thank you very much!

 

[johnsmi]

Actually, H₂ is exactly H - H₁ provided it is
suitably interpreted. However what you want from your other post
is not H₂ at all. You want an effective Hamiltonian.

I'll try to explain.
Given vector spaces X and Y we can form a NEW vector space
X$\otimes$Y. For any pair of operators (A,B) with
A:X$\rightarrow$X, B:Y$\rightarrow$Y there
exists a unique map A$\otimes$B.

Why is this important?
When you're looking at your hamiltonian as acting on the product
what you start with are the maps A and B above but you need to
be carefull about how they lift. Specifically H₁ lifts
to H₁$\otimes$id₂. Where id₂
is the identity map on the second factor.

Similarly H₂ lifts to id₁$\otimes$H₂.

So that your hamiltonian is actually
H = H₁$\otimes$id₂ + id₁$\otimes$H₂.

Now how can we use this to "factor out" a piece, or find an effective hamiltonian
Suppose we know the eigenbasis of H₂. And further suppose we've
orthonormalized it, then we can look at the matrix elements of H, which stay
in a fixed H₂ subspace.
Let $|m>, |n>$ denote vectors in the first space, and $|\mu>, |\nu>$
denote orthonormal eigenvectors of H₂.

Then
$$<m \otimes \mu| H |n \otimes \nu> = <m \otimes \mu| (H_1 \otimes \text{id}_2) |n \otimes \nu> + <m \otimes \mu| (\text{id}_2 \otimes H_2) |n \otimes \nu>$$
$$= <m|H_1|n><\mu|\nu> + <m|n>\lambda_\nu<\mu|\nu>$$

where $\lambda_\nu$ is the appropriate eigenvalue. Now
if we are only interested in a given state $\nu$ we
can put
$$<m \otimes \nu| H |n \otimes \nu> = <m|H_1|n> + <m|n>\lambda_\nu = <m| H_1 + \lambda_\nu \text{id}_1 |n>$$

This looks then like a hamiltonian only on the 1st factor.

Your unswer seems to be very usefull since I did find in a related article something like this:
$$<\uparrow (\phi)| \otimes <\phi| \frac{\Pi^2}{2M} |\phi '> \otimes |\uparrow (\phi ')> = <\phi| \frac{Q^2} {2M} |\phi '> \\ where\ Q=\Pi -const.$$

I think I am beginning to understand what you mean evan though there are a few terms you used which are unfamiliar to me.
1) An identity map: I think is an operater that sends a vector to the unit vector in that space?
2) I don't quite understand what you mean by 'lifting'
3) I gues that $\lambda_\nu$ is the eigenvalue we get by applying $H_2|\nu>$ (i.e. $H_2|\nu>=\lambda_\nu|\nu>$)


And finally, at the end you said that:

$|\mu>, |\nu>$ denote orthonormal eigenvectors of H₂.

dosn't that necessarilly mean that: $<\mu|\nu>=0$ ?

 

[0xDEADBEEF]

An identity map sends a vector to itself.

With lifting he means what the operator looks like on the larger space.

 

[johnsmi]

An identity map sends a vector to itself.

With lifting he means what the operator looks like on the larger space.

thank you, the identity map was a stupid mistake of me
But after having a better look in Cohen-Tannoudji I understood that qbert made a slite typo mistake

$$<m \otimes \mu| H |n \otimes \nu> = <m \otimes \mu| (H_1 \otimes \text{id}_2) |n \otimes \nu> + <m \otimes \mu| (\text{id}_2 \otimes H_2) |n \otimes \nu>$$

and it should be

$$<m \otimes \mu| H |n \otimes \nu> = <m \otimes \mu| (H_1 \otimes \text{id}_2) |n \otimes \nu> + <m \otimes \mu| (\text{id}_1 \otimes H_2) |n \otimes \nu>$$

What I still don't understand (even though thanks to the both of you I am understanding much more) is how to get this:
$$<\uparrow (\phi)| \otimes <\phi| \frac{\Pi^2}{2M} |\phi '> \otimes |\uparrow (\phi ')> = <\phi| \frac{Q^2} {2M} |\phi '> \\ where\ Q=\Pi -const.$$

and why isn't $<m|n>=<\mu|\nu>=0$ ?

Thanks

 

[johnsmi]

Can I assume that a spinor (spin state that depends on location) state and an eigenstate of momentum (which also depends on location) are in separate Hilbert spaces or does the momentum operator act on the spinor as well?