Question 9
The variance σ2X=⟨(X^−⟨X^⟩)2⟩ of an operator, X^, is a measure of how large a range its possible values are spread over (the standard deviation is given by σ=σ2−−√). Suppose that |X⟩ is an eigenstate of some operator X^, what is the variance of X^ in this state? You may assume that |X⟩ is normalized (⟨X|X⟩=1).
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To find the variance of an operator X^ in an eigenstate |X⟩, we need to calculate the expectation value of the squared deviation of the operator from its mean value in that state.
The variance of X^, denoted as σ2X, can be calculated as:
σ2X = ⟨(X^ - ⟨X^⟩)2⟩
Where ⟨X^⟩ is the expectation value of X^ in the state |X⟩.
In the case where |X⟩ is an eigenstate of X^, the eigenvalue equation is:
X^ |X⟩ = X |X⟩
This means that the operator X^ acting on the eigenstate |X⟩ simply gives the eigenvalue X multiplied by the eigenstate:
X^ |X⟩ = X |X⟩
Now, let's find the expectation value of X^ in the state |X⟩. This can be calculated as:
⟨X^⟩ = ⟨X|X^|X⟩
Since |X⟩ is an eigenstate, we can write it as:
⟨X|^2|X⟩ = X⟨X|X⟩
From the normalization condition ⟨X|X⟩ = 1, we can simplify it further:
⟨X^⟩ = X
Substituting this expression for ⟨X^⟩ into the variance formula, we get:
σ2X = ⟨(X^ - X)2⟩
Since X^ acting on |X⟩ gives the eigenvalue X multiplied by |X⟩, we can rewrite the expression as:
σ2X = ⟨(X - X)2⟩
Simplifying it further, we find:
σ2X = ⟨0⟩
The expectation value of 0 is simply 0. Therefore, the variance of X^ in the eigenstate |X⟩ is 0.
In summary, if |X⟩ is an eigenstate of the operator X^, the variance of X^ in this state is 0. This means that the measurement of X^ in this state will always give the same value X, with no spread or variation.
To find the variance of an operator X^ in an eigenstate |X⟩, we can use the following formula:
σ^2X = ⟨(X^ - ⟨X^⟩)^2⟩
Here, ⟨X^⟩ refers to the expectation value of the operator X^ in the given eigenstate.
Since |X⟩ is an eigenstate of the operator X^, we know that X^|X⟩ = x|X⟩, where x is the eigenvalue corresponding to |X⟩.
Now, let's calculate the variance:
⟨(X^ - ⟨X^⟩)^2⟩ = ⟨(X^ - x)^2⟩
Expanding the square, we get:
⟨(X^ - x)^2⟩ = ⟨X^2 - 2X^x + x^2⟩
Since |X⟩ is an eigenstate, X^|X⟩ = x|X⟩, so we can replace X^x with xX^:
⟨(X^ - x)^2⟩ = ⟨X^2 - 2xX^ + x^2⟩
Using the linearity of the expectation value, we can further expand this expression:
⟨X^2 - 2xX^ + x^2⟩ = ⟨X^2⟩ - 2x⟨X^⟩ + x^2⟨1⟩
Since |X⟩ is normalized, ⟨1⟩ = ⟨X|X⟩ = 1.
Therefore, the expression simplifies to:
⟨X^2⟩ - 2x⟨X^⟩ + x^2
This is the variance of the operator X^ in the given eigenstate |X⟩.
Please note that I have made some assumptions and used basic mathematical properties to derive this result.