Need help for solving an operator equation

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In summary, the conversation discusses the solution of the equation A + [B, X] = 0 in finite and infinite dimensional Hilbert spaces. It is shown that in the finite case, the equation has at least one solution X if Tr(A) = 0. The speaker has proven this using the matrix form of the operators and is now seeking guidance on whether there will always be a solution in infinite dimensions. A suggestion is made to use recursion and block splitting to find a solution and then extend it to infinite dimensions.
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TRXW
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I am working on something and have lead to a problem. I need your help!
Let A and B be self-adjoint operators acting on a finite dimensional Hilbert space. Then, the equation

A + [ B , X ] = 0,
has at least one solution X, iff Tr(A)=0.
([ B , X ] = BX - XB)

I have proved it by taking the matrix form of the operators.
But my question is about countably infinite dimensional Hilbert spaces. There the Trace condition is irrelevant. But will there always be a solution?
I would appreciate your guidance.

Thanks
Leo
 
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(First: it's easy to show that X is anti-Hermitian). Here's a suggestion: use recursion. To begin with, write the solution you already have for finite dimensions to make the recursion explicit. That is, write the NxN matrix in block form, splitting off the first row and column and leaving the other N-1 dimensions in a block. Your unknowns will be similarly split: X00, X0i = - Xi0* and Xij = - Xji*. It helps to suppress the indices and consider these quantities to be a scalar, vector and tensor, respectively.

Then you will have three equations for these three unknowns. For example the first (00) equation will read X·B + XB* = A, and it's easy to show this can always be satisfied. (You can also assume wlog that A is already diagonal.)

If you can do the above, the idea is to recurse N times, each time block-splitting the Aij further by peeling off the first dimension, etc. Then, to prove it for infinite dimensions, all you have to do is forget to stop recursing. :smile:
 

FAQ: Need help for solving an operator equation

What is an operator equation?

An operator equation is a mathematical statement that shows the relationship between two or more mathematical expressions using operators. Operators are symbols that represent mathematical operations, such as addition, subtraction, multiplication, and division.

How do I solve an operator equation?

To solve an operator equation, you need to isolate the variable or unknown term on one side of the equation by using inverse operations. Inverse operations are operations that undo each other, such as addition and subtraction, or multiplication and division. Once the variable is isolated, you can solve for its value by performing the inverse operations on both sides of the equation.

What are some common types of operator equations?

Some common types of operator equations include linear equations, quadratic equations, exponential equations, and logarithmic equations. Linear equations involve variables raised to the first power, while quadratic equations involve variables raised to the second power. Exponential equations involve variables in the exponent, and logarithmic equations involve variables inside a logarithm.

Can you give an example of solving an operator equation?

Sure, let's say we have the equation 2x + 5 = 15. To solve for x, we first need to isolate it on one side of the equation. We can do this by subtracting 5 from both sides, giving us 2x = 10. Next, we divide both sides by 2 to get x = 5. So the solution to the equation is x = 5.

What are some tips for solving operator equations?

Some tips for solving operator equations include starting by simplifying both sides of the equation if possible, using the correct order of operations, checking your work by plugging the solution back into the equation, and being careful with negative signs. It's also important to pay attention to any special rules or properties for the type of equation you are solving, such as the zero product property for solving quadratic equations.

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