How to Find the Expectation Value of Total Energy?

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Homework Help Overview

The discussion revolves around calculating the expectation value of total energy given a specific wavefunction in quantum mechanics. The original poster presents a wavefunction and seeks to understand how to derive the expectation value of total energy, particularly in relation to previously calculated probabilities.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers using the formula for expectation value involving the Hamiltonian operator and questions whether it can be approached through the probabilities derived from the wavefunction.

Discussion Status

Some participants provide insights into the mathematical formulation of expectation values for both discrete and continuous probability distributions. There is a suggestion that the expectation value can be computed using the known probabilities, leading to a proposed expression for total energy. However, the discussion does not reach a consensus on the final approach.

Contextual Notes

The original poster indicates familiarity with calculating probabilities but seeks clarification on how to connect these probabilities to the expectation value of total energy. There is an implication of homework constraints regarding the methods allowed for solving the problem.

Chronos000
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Homework Statement



given a wavefuntion [tex]\Psi[/tex] = (1/sqrt50) (3[tex]\mu[/tex]1 + 5[tex]\mu[/tex]2 - 4[tex]\mu[/tex]3)

what is the expectation value of the total energy?


My thoughts were to calculate <[tex]\Psi[/tex]|[tex]\hat{}H[/tex]|[tex]\Psi[/tex]>

but the previous part to the question asks for the probability of each outcome(which I know how to find). So is there a way to do this using the probabilities?
 
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For a discrete probability distribution [tex]f(x_i)[/tex] [x (and thus f) takes only discrete values], the expectation value of a quantity [tex]x[/tex] is

[tex]<x>= \sum^N_i x_i f(x_i)[/tex],

where x takes on values [tex]x_1, x_2, \ldots , x_N[/tex]

For a continuous probability distribution [tex]g(x)[/tex] [where g and x are continous], the expectation value of x is the limit of the sum, namely the integral

[tex]<x>= \int^{x_{max}}_{x_{min}} x g(x)[/tex].

So if you know the probability distribution (which it seems like you do) the rest is basic maths.
 
so are you saying that the answer is just ET = 9/50 E1 + 25/50 E2 + 16/50 E3 ?
 
Yes.
 

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