Understanding the Basic Operator Equation for Quantum Mechanics

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In summary, the conversation discusses the expression H(\sum_n w_n |a_n><a_n|) = \sum_n w_n(H|a_n><a_n|-|a_n><a_n|H) and its incorrectness. The correct operators O_L and O_R are introduced and it is shown that they both have equal expectation values in any arbitrary state. The conclusion is that O_R=O_L and both can be used interchangeably. The conversation also mentions that \rho does not satisfy Schrodinger's equation, but instead i\ \hbar \partial_t \rho \neq H \rho must be used. Finally, it is mentioned that the derivation given is valid for time-independent states and would not
  • #1
genericusrnme
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Can someone explain to me how

[itex]H(\sum_n w_n |a_n><a_n|) = \sum_n w_n(H|a_n><a_n|-|a_n><a_n|H)[/itex]

I've done this before and I remember being confused about it before then finding out it was something simple.. I should really start filing my notes away for such an eventuality :frown:
I can't seem to work out what the next step is at all.
 
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  • #2
What you wrote is not correct. If you take an expectation value of your expression in the mth state, you'd find 0 on the RHS. We can find a correct statement if we consider the operators

[tex] O_L = \sum_n w_n (H| a_n \rangle )\langle a_n| ,[/tex]

[tex]O_R = \sum_n w_n | a_n \rangle (\langle a_n|H) .[/tex]

The expectation values in an arbitrary state are equal:

[tex] \langle a_m | O_L | a_m \rangle = \sum_n w_n \langle a_m |H| a_n \rangle \langle a_n| a_m \rangle = w_m \langle a_m |H| a_m \rangle,[/tex]

[tex]\langle a_m | O_R | a_m \rangle = \sum_n w_n \langle a_m | a_n \rangle \langle a_n|H| a_m \rangle = w_m \langle a_m|H| a_m \rangle , [/tex]

where in both cases we used [itex] \langle a_n| a_m \rangle = \delta_{nm}.[/itex]

So in this sense, [itex] O_R=O_L[/itex]: we can let the operator act from the right or left side. This is the same way that expectation values work

[tex] \langle a_n | H | a_m \rangle = \langle a_n | (H | a_m \rangle ) = (\langle a_n | H) | a_m \rangle.[/tex]


We can therefore write

[tex]H \left( \sum_n w_n | a_n \rangle \langle a_n| \right) = \frac{1}{2} \sum_n w_n \Bigl[ ( H | a_n \rangle ) \langle a_n| + | a_n \rangle (\langle a_n|H) \Bigr].[/tex]
 
  • #3
Yeah, I kept coming to that conclusion too yet what I wrote is what is in Sakurai's book..

[itex]i\ \hbar \partial_t \rho = H \rho \neq -[\rho , H][/itex]

Where [itex]\rho = \sum_n w_n |a_n \rangle \langle a_n |[/itex]

I'll attatch an extract
 

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  • #4
genericusrnme said:
Yeah, I kept coming to that conclusion too yet what I wrote is what is in Sakurai's book..

[itex]i\ \hbar \partial_t \rho = H \rho \neq -[\rho , H][/itex]

Where [itex]\rho = \sum_n w_n |a_n \rangle \langle a_n |[/itex]

I'll attatch an extract

OK, the point there is that [itex]\rho[/itex] does not satisfy Schrodinger's equation so [itex]i\ \hbar \partial_t \rho \neq H \rho [/itex]. Instead you have to use

[tex] i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \langle \alpha^{(i)},t_0;t | = \left( i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \right) \langle \alpha^{(i)},t_0;t | + |\alpha^{(i)},t_0;t \rangle \left( i \hbar \frac{\partial}{\partial t} \langle \alpha^{(i)},t_0;t | \right)[/tex]

and

[tex] i \hbar \frac{\partial}{\partial t} \langle \alpha^{(i)},t_0;t | = - \left( i \hbar \frac{\partial}{\partial t} |\alpha^{(i)},t_0;t \rangle \right)^\dagger .[/tex]

The derivation I gave above is valid for time-independent states and wouldn't work here.
 
  • #5
Ah!
I remembered it being something simple :L

Thanks buddy :biggrin:
 

1. What are the basic arithmetic operators and their functions?

The basic arithmetic operators are addition (+), subtraction (-), multiplication (*), division (/), and modulus (%). Addition and subtraction are used to add or subtract numbers, multiplication is used to multiply numbers, division is used to divide numbers, and modulus returns the remainder after division.

2. How are comparison operators used in programming?

Comparison operators are used to compare two values and return a boolean value (true or false) based on the comparison. The most commonly used comparison operators are equal to (==), not equal to (!=), greater than (>), less than (<), greater than or equal to (>=), and less than or equal to (<=).

3. What is the difference between the assignment operator and the equality operator?

The assignment operator (=) is used to assign a value to a variable, while the equality operator (==) is used to compare two values. The assignment operator is used in variable assignment, whereas the equality operator is used in conditional statements and comparisons.

4. How can logical operators be used in programming?

Logical operators are used to combine two or more boolean expressions and return a boolean value based on the result of the combination. The three logical operators are AND (&&), OR (||), and NOT (!). AND returns true if both expressions are true, OR returns true if at least one expression is true, and NOT returns the opposite value of the expression.

5. What is the order of operations for operators in programming?

The order of operations, also known as operator precedence, determines the order in which operators are evaluated in a mathematical expression. In most programming languages, the order of operations is as follows: parentheses, exponentiation, multiplication and division (from left to right), addition and subtraction (from left to right). It is important to use parentheses to specify the order of operations when needed.

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