Action of a non-compound Bra on a compound Ket

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In summary, the student is trying to find the time dependence of a compound system. He has found the eigenvalues and eigenvectors of the Hamiltonian, but does not know how to compute the tensor product of two vectors in different Hilbert spaces. He is asking for help from an expert.
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Homework Statement



Hi. This isn't really a 'problem' as such but it is to do with coursework and seems a little too basic to be posted in the maths section. I need to know the result of an arbitrary bra, <an|, say, representing an eigenvector of a system, acting on a compound ket such as |0>a|1>b.

The reason I'm doing this is, I need to find the time dependence of my compound system. I solved the eigenvalues and eigenvectors of my Hamiltonian which gave me the basis {|a>} upon which I need to expand my initial state vector, which is in compound form, i.e. |n>a|m>b |n>a and |m>b belong to different Hilbert spaces and represent different objects.

Homework Equations


The Attempt at a Solution


Since the eigenvectors I found were derived by considering the interaction part of the Hamiltonian of the total system (my two systems are magnetically coupled), I assume they span the Hilbert spaces of both a and b, so an element from any of my eigenvectors can multiply an element in either an a or b vector. Also I know that |0>a|1>b represents the tensor product of the a and b vectors. So for example

|a>=[1;1], say, then; <a||0>a|1>b= [1 1]*[1;0]a[1 0]b
Is this correct so far? But then I am at a loss, I don't know how to compute the tensor product of two vectors in different Hilbert spaces, and don't know what kind of object this would be once all is said and done. I've looked around the internet but haven't found anything helpful. Any help would be greatly appreciated!

(p.s, sorry about the lack of latex, it was doing weird things when I previewed the post. The notation for vectors is meant to be MATLAB notation, so

[1;1] represents a column vector:
1
1

and [1 0] represents a row vector (1, 0).

Thanks!)
 
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  • #2
What do you mean by |a>=[1;1]? Does it mean [tex]|a\rangle=|1\rangle\otimes|1\rangle[/tex] or something else? If |a> is a state of the total system it must be a sum of tensor products. But is it? This is not completely clear to me from what you wrote.
 
  • #3
It does not make sense to act with a single state ket [tex]\langle a|[/tex] on a tensor product state, unless a itself represents a tensor product state.

In Other words to calculate the action you need both ket and bra to be in the same basis. So either you have to write a in the 2 particle Hilbertspace or you have to convert the compound system.
 
  • #4
Sorry if I was unclear. |a> is not the state of the whole system, it is one of the eigenvectors of the Hamiltonian. Actually, for my purposes, it is an eigenvector of an excitation subspace of the interaction part of the total Hamiltonian. My example |a> was meant to be

[tex]\left(\begin{array}{cc}1\\1\end{array}\right)[/tex]

I just hadn't figured out latex. But say I'm dealing with the first excitation subspace, the eigenvectors I get for the Hamiltonian are

[tex]\left(\begin{array}{cc}1\\1\end{array}\right)[/tex]

and

[tex]\left(\begin{array}{cc}-1\\1\end{array}\right)[/tex].

So my state will look like |[itex]/psi[/itex]>= c1 [itex]\left(\begin{array}{cc}1\\1\end{array}\right)[/itex] + c2 [itex]\left(\begin{array}{cc}-1\\1\end{array}\right)[/itex].
Obviously adding time dependence in this form is easy, which is what I'm aiming to do overall.

I then take an initial state (since I'm in the first excitation subspace I can only have one of my objects excited one level, so I can only pick |1>a|0>b, |0>a|1>b, or some superposition of the two). Say I pick |1>a|0>b. I need to expand this on the basis {|a>} so as to work out the coefficients c1 and c2, so I can then write the time dependence. In order to do this expansion and work out the coefficients I have to compute objects like:

[tex]\left(\begin{array}{cc}1&1\end{array}\right)[/tex] * |0>a|1>b .

The problem is I don't know how to do that! I don't know how to write |0>a|1>b in matrix form... I know it's a tensor product, but since a and b belong to different Hilbert spaces, surely one can't just crudely compute this as if the a and b labels weren't there? And if that's true, how does one compute it?

Hopefully my problem is clearer to you now, thanks for reading.
 
  • #5
betel said:
It does not make sense to act with a single state ket [tex]\langle a|[/tex] on a tensor product state, unless a itself represents a tensor product state.

In Other words to calculate the action you need both ket and bra to be in the same basis. So either you have to write a in the 2 particle Hilbertspace or you have to convert the compound system.

I see. I'll try doing that then. Thanks.
 
  • #6
You should really state the whole problem, otherwise we can just make educated guesses.

Your Hamiltonian gives you two different eigenstates. Ok.
But your initial state is written like it is two particles, each in one of the eigenstates.
 
  • #7
In fact one can define

[tex] \langle a|_1 : |b\rangle_1\otimes|c\rangle_2\mapsto\langle a|b\rangle_1\, |c\rangle_2[/tex]

but we are not sure what is really being looked for.
 
  • #8
I'll explain the problem fully. I was trying to avoid it because it's rather long but I guess I should've just done it straight away.

We consider two magnetically coupled quantum harmonic oscillators. Call them A and B. Our goal is to extract the dynamics of excitation exchange; for example, we might want to prepare the system so that oscillator A is in the first excited state and oscillator B is in the ground state, and see how long it takes for the excitation to be transferred from A to B.

To begin with we write the Hamiltonian in terms of the annihilation and creation operators for A and B: a, a+, b, b+. By using the rotating wave approximation, and ignoring free energy and free interaction terms, we can simplify things greatly.

We end up with the coupling part of the Hamiltonian looking like:

H = a+b + b+a

Now we calculate the matrix elements of this in different subspaces. Take the first excitation subspace, for simplicity. We have,

<0|a<1|b|H|0>a|1>b = 0
<1|a<0|b|H|0>a|1>b = 1
<0|a<1|b|H|1>a|0>b = 1
<1|a<0|b|H|1>a|0>b = 0

The Hamiltonian for the coupling in the first excitation subspace, in matrix form, is then

[tex]\left(\begin{array}{cc}0&1\\1&0\end{array}\right)[/tex]

Now we want to find out about the dynamics. So first we must find the eigenvalues/eigenvectors of this Hamiltonian. This is where my problem lies. I don't need to tell you the eigenvectors of this matrix are

1/[tex]\sqrt{}2[/tex][tex]\left(\begin{array}{cc}1\\1\end{array}\right)[/tex]
and

1/[tex]\sqrt{}2[/tex][tex]\left(\begin{array}{cc}-1\\1\end{array}\right)[/tex]

So to find the dynamics I need to expand an initial state upon the basis given by these eigenvectors, right? Because I need to find the correct complex coefficients associated with the eigenvectors, given an initial state. And since my initial state has to consist of a tensor product state, given the fact I have two coupled oscillators, I end up acting with single bras (the eigenvectos) on tensor product kets (my initial state), because this is how the required coefficients are calculated (at least that's what I thought). So as betel said, I need to either write the eigenvectors in terms of my tensor product states or vice versa. I'm not sure how I'd do this without just running into the single-bra-on-tensor-ket problem again but I shall try and think of a way. Any more help is most welcome!
 
  • #9
Just to make it clearer, say I chose my initial state to be |psi(0)>= |1>a|0>b, and call the eigenvectors

[tex]\left(\begin{array}{cc}1\\1\end{array}\right)[/tex] = |A>and[tex]\left(\begin{array}{cc}-1\\1\end{array}\right)[/tex]= |B>

I would then want to find coefficients such that
|psi(0)> = a1|A> + a2 |B>

To do this I would need to expand |psi(0)> in the eigenvector basis, and to do that, I would need to say

|psi(0)> = <A|psi(0)>|A> + <B|psi(0)>|B>

That's how I end up with my initial problem.

Edit: it's chopping off my vectors for some reason but they're meant to be column vectors with elements 1,1 and -1,1 respectively.
 
  • #10
The Hamiltonian can not be like this. That's for sure. It will be a sum of two tensor products of matrices - that is a 4x4 matrix (or 2x2 block matrix). You need to introduce a basis in your product space. You will have 2x2=4 vectors in this basis.
 
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  • #11
Ok. So now I understand what you want to do. You can either do the full problem as arkaja suggested, i.e. take all 4 basis vectors and working with 4x4 matrices and reduce to the subspace later, or you can do it in your 2D subspace from start.

But you have to be careful with your definitions. If you write your interaction Hamiltonian as 2x2 matrix then the Basisvectors for this matrix will be
[tex] \begin{pmatrix} 1\\0\end{pmatrix} := |0>_a|1>_b[/tex] and [tex] \begin{pmatrix} 0\\1\end{pmatrix} := |1>_a|0>_b[/tex]

So now you can proceed as you wanted before, but you have to keep in mind, that by restricting to the 2D subspace you defined new basisvectors.
 
  • #12
Ah, okay. That makes sense. Thanks a bunch, everyone.
 
  • #13
Hi again, everything worked out nicely with the first excitation subspace, got good results, but now I'm stuck thinking about the second!

The second subspace hamiltonian looks like

[0 1 0]
[1 0 1]
[0 1 0]albeit multiplied by root two, but that's irrelevant.

How can I associate my base states with basis vectors here? It seems nonsensical in this case to just take the columns of the matrix that are associated with each state and say, "this column is the basis vector for that state", because, for one, only two of the columns form a basis, there's a spare one (in fact, two are identical), and I have 3 base states

|0>a|2>b , |1>a|1>b , |2>a|0>b

It seems very wrong to associate the same basis vector with two states, correct?
Moreover, one of the eigenvectors of this matrix is of the form

[tex]\left(\begin{array}{cc}-1\\0\\1\end{array}\right)[/tex]

This can't be written as a linear combination of the column vectors that make up the matrix (or the column space basis vectors).
I'm pretty sure there's a crucial concept I'm unaware of or neglecting, or I'm looking at this whole thing in the wrong way, or something.
How should I proceed?
Thanks in advance!
 
  • #14
How many more excitations you will need? If this is the end, I do not understand why you will not introduce 9 basic vectors |i>a|j>b and work with 9x9 matrices? After all you do not have to do your calculations by hand, you can use some software, and there is a lot of it around.
 
  • #15
Thanks for replying. That seems perfectly reasonable, but basically my supervisor wants it done in this manner so there's not much I can do ;-)
I'll be going up a couple more excitation levels, maybe only one, but obviously we're in computer territory even with a 9x9 matrix, and the project is meant to be purely analytical.
If it's significantly more difficult doing this then working with the whole matrix, then whole matrix and computer it shall have to be, but if it's not crazy to do it working in subspaces that's what I'd like to do.

Edit: Even if I don't solve it this way, the question has been driving me mad for the whole of yesterday and today, because it makes me think there's some important linear algebra or, even worse, physics I'm not understanding!
 
  • #16
Well, you did the same again as before. By calculating a 3x3 matrix you implicitly redefined new basis vectors. You have three different excitations states 11, 02, 20, which give a basis for your three dimensional vector space. If you thought you could read of the basis vectors from the hamiltonian, that's wrong. For the frist excitation space it was just coincidence. Simply forget about 11, 02, 20 and define three basis vectors
[tex]e_1\equiv |0>|2>, e_2\equiv |1>|1>,e_3\equiv |2>|0>[tex] In this basis your Hamiltonian will be of the form described above.
 
  • #17
Okay. This time I really do understand. You're a lifesaver, cheers!
 

1. What is a non-compound Bra?

A non-compound bra is a type of bra that is made from a single piece of fabric and does not have any additional layers or components. It is typically made from stretchy material and has no underwire or padding.

2. What is a compound Ket?

A compound ket is a chemical compound that contains a ketone functional group. It is characterized by a carbon double bond with an oxygen atom, and can be found in a variety of organic molecules such as sugars, fats, and proteins.

3. How does a non-compound bra affect a compound ket?

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5. Are there any benefits to wearing a non-compound bra for someone with a compound ket?

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