Understanding Exponential of Operator: Expansion and Eigenvalues Explained

In summary, the conversation discusses the expansion of the exponential of an operator and its diagonal elements, as well as the definition of functions of diagonalisable operators. It is noted that non-diagonalisable operators are more complicated to define. Finally, the proof for the expansion is mentioned, involving the use of the spectral theorem for self-adjoint operators.
  • #1
KFC
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Hi there,
I learn from the text that the exponential of an operator could be expanded with a series such that

[itex]e^{\hat{A}} = \sum_{n=0}^{\infty} \frac{\hat{A}^n}{n!}[/itex]
So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]
[itex]e^{\hat{A}}|\psi\rangle[/itex]

will be a matrix with diagonal elements given as [tex]\exp(a_i)[/tex], is that right?

So I am wondering what happen if [itex]\hat{A}[/itex] is now written as a power form, i.e. [itex]\hat{A}^n[/itex], can we conclude that
So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]

[itex]e^{\hat{A}^n}|\psi\rangle[/itex]

gives a matrix with diagonal elements as [itex]\exp(a_i^n)[/itex] ?
 
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  • #2
That's correct - for operators that can be diagonalised.

Its basically how functions of diagonalisable operators are defined - when diagonalised its a function of the eigenvalues.

Non diagonalisable operators are however a bit trickier - there are a few definitions I have seen and even a paper showing they are basically all equivalent - but the point is they are a more difficult proposition.

Thanks
Bill
 
  • #3
If [itex]\hat{A}|\psi\rangle = a_i|\psi\rangle[/itex], then [itex]e^{\hat{A}}|\psi\rangle= \exp(a_i)|\psi\rangle[/itex] and [itex]e^{\hat{A}^n}|\psi\rangle = \exp(a_i^n)|\psi\rangle[/itex]. You should be able to prove this pretty easily using the Taylor series definition you mentioned.
 
  • #4
afaik the proof of

[tex]f(A) = \sum_n f_n A^n[/tex]

where fn are the Taylor coefficients is rather involved mathematically. What you need is the spectral theorem for self-adjoint operators. For us physicists it reads

[tex]f(A) = f(A) \sum_n |a\rangle\langle a| = \sum_n f(a) |a\rangle\langle a| [/tex]

where the states |a> are the eigenstates of A and form a complete set.
 

1. What is an exponential operator?

An exponential operator is a mathematical operator that represents repeated multiplication of a number by itself. It is denoted by the symbol "^" and the number of times it is multiplied is indicated by the exponent.

2. How does the expansion of exponential operator work?

The expansion of an exponential operator is a process of simplifying an expression with multiple exponential terms. It involves using the properties of exponents, such as the product rule and power rule, to combine like terms and simplify the expression.

3. What are eigenvalues in relation to exponential operator?

Eigenvalues are a set of special numbers associated with a matrix that is used in the process of diagonalization. In the context of exponential operator, eigenvalues are the values that satisfy the equation Ax = λx, where A is the matrix and x is the eigenvector.

4. How do eigenvalues relate to exponential operator expansion?

Eigenvalues play a crucial role in the expansion of exponential operator as they determine the diagonal elements of the diagonalized matrix. This allows for the simplification of the exponential expression, making it easier to solve.

5. Why is understanding exponential operator and eigenvalues important?

Understanding exponential operator and eigenvalues is important because they have various applications in mathematical and scientific fields, such as in differential equations, quantum mechanics, and data analysis. They also provide a powerful tool for solving complex mathematical problems and understanding the underlying patterns and relationships in a system.

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