Are Probabilities the Same for Operator A in Different Basis?

In summary, the conversation discusses finding the possible eigenvalues and corresponding probabilities of a given operator in a 3D Hilbert-space. The suggested method for finding the coefficients is incorrect and the correct method is provided.
  • #1
Niles
1,866
0

Homework Statement


Hi all.

I have the following state at t=0 in a 3D Hilbert-space (it is in the eigenspace of the 3x3 Hamiltonian):

[tex]
\left| \psi \right\rangle = \frac{1}{{\sqrt 2 }}\left| \psi \right\rangle _1 + \frac{1}{2}\left| \psi \right\rangle _2 + \frac{1}{2}\left| \psi \right\rangle _3.
[/tex]

Now I have an operator representing an observal given by:

[tex]
\hat A = \left( {\begin{array}{*{20}c}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{array}} \right)
[/tex]

I have to find the possible eigenvalues of A and the corresponding probabilities.

The Attempt at a Solution



The possible eigenvalues of A are easy. I am more concerned about the probabilities. I reasoned that they are the same, because the above state at t=0 is independent of the Hilbert space in is written in. So it will look the same if I write it in the eigenspace of A, but the unit-vectors (i.e. the possible states) are now different.

So my attempt: The probabilities are the same, i.e. 1/2, 1/4 and 1/4. Can you confirm this?

Thanks in advance.


Niles.
 
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  • #2
Ok, I think I got it now.

My above suggestion is not correct. I think I have to find the coefficients by writing the state |psi> in the basis of A by the following method:

[tex]
a_{cofficient\,\,1} = \left\langle {{\psi _{a\,\,1} }}
\mathrel{\left | {\vphantom {{\psi _{a\,\,1} } \psi }}
\right. \kern-\nulldelimiterspace}
{\psi } \right\rangle = \sum\limits_i {\left\langle {{\psi _{a\,\,1} }}
\mathrel{\left | {\vphantom {{\psi _{a\,\,1} } {\psi _i }}}
\right. \kern-\nulldelimiterspace}
{{\psi _i }} \right\rangle \left\langle {{\psi _i }}
\mathrel{\left | {\vphantom {{\psi _i } \psi }}
\right. \kern-\nulldelimiterspace}
{\psi } \right\rangle },
[/tex]

where psi_i are the eigenfunctions of the Hamiltonian and psi_{a 1} are the eigenfunctions of A. The above method gives me the first coefficient for the representation of |psi> in the basis of A.

Is this correct?
 

FAQ: Are Probabilities the Same for Operator A in Different Basis?

What is an operator in quantum mechanics?

An operator in quantum mechanics is a mathematical object that represents a physical observable. It acts on a quantum state and produces a new quantum state as a result. Examples of operators include position, momentum, energy, and spin. Operators are essential in calculating probabilities and making predictions about the behavior of quantum systems.

How do probabilities for an operator differ from classical probabilities?

In classical mechanics, probabilities are typically assigned to different possible outcomes of a measurement. In quantum mechanics, probabilities are calculated using operators, which represent the possible outcomes of a measurement. The square of the absolute value of the wavefunction, known as the probability density, represents the probability of finding a particle at a particular position or with a particular momentum.

How are probabilities for an operator calculated?

The probability of obtaining a certain eigenvalue (or measurement result) for an operator is calculated using the Born rule, which states that the probability is equal to the square of the absolute value of the corresponding eigenstate coefficient. This means that the probability of obtaining a particular result is directly related to the amplitude of the state associated with that result.

Can probabilities for an operator be negative?

No, probabilities for an operator cannot be negative. In quantum mechanics, probabilities are always positive numbers, representing the likelihood of obtaining a certain measurement result. This is because the square of the absolute value of the wavefunction, which is used to calculate probabilities, is always positive.

How do probabilities for an operator change over time?

In quantum mechanics, the state of a system evolves over time according to the Schrödinger equation. This means that the probabilities for an operator can change over time, as the system's state changes. The time evolution of probabilities is determined by the Hamiltonian, which represents the total energy of the system.

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