Unitary operators preserve normalization in arbitrary basis

In summary, the conversation discusses proving the independence of an operator's unitarity from basis. The process involves expanding an arbitrary ket at a specific time in an arbitrary basis and using the identity operator to calculate the expansion coefficients. Later, a time-evolved ket is constructed and the relationship between the two sets of coefficients is discussed. The question of how the coefficients change in a basis-switch is brought up.
  • #1
bjnartowt
284
3

Homework Statement



To test my knowledge of Sakurai, I asked myself to: "Prove that an operator being unitary is independent of basis."


The Attempt at a Solution



I want to show the expansion coefficients’ squared magnitudes sum to unity at time “t”, given that they do at time t = t0.
Consider expanding an arbitrary –ket at t0, in the arbitrary basis a’,

[tex]\left| {\alpha ,{t_0}} \right\rangle = {\bf{I}}\left| {\alpha ,{t_0}} \right\rangle = \left( {\sum\nolimits_{a'} {\left| {a'} \right\rangle \left\langle {a'} \right|} } \right)\left| {\alpha ,{t_0}} \right\rangle = \sum\nolimits_{a'} {\left| {a'} \right\rangle \left\langle {a'|\alpha ,{t_0}} \right\rangle } = \sum\nolimits_{a'} {{c_{a'}}({t_0}) \cdot \left| {a'} \right\rangle } [/tex]

Note that the work directly above is a ket that has not yet time-evolved, so the expansion using the identity-operator and the notation of the expansion coefficients[tex]{c_{a'}}({t_0}) = \left\langle {a'|\alpha ,{t_0}} \right\rangle [/tex], is still our “business-as-usual” expansion from Chapter 1.

Now, consider a ket, whatever that ket may be, at some later time “t”. The time “t” is just a label, so we can use the same procedure that we used in [I.8] to construct the time-evolved ket,

[tex]\left| {\alpha ,t} \right\rangle = \sum\nolimits_{a'} {{c_{a'}}(t) \cdot \left| {a'} \right\rangle } [/tex]

So we see that in [I.8] and [I.9], we have two sets of coefficients, and , respectively. I’m not sure how these two sets of coefficients are related to one another...well, I know they sum to unity in both bases. Can't find the expression for how coefficients change in a basis! :-p

What's the expression for the coefficients in an expansion before and after a basis-switch?
 
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  • #2
I am not sure if that is what you want, but:

Write

[tex]|\alpha,t\rangle=U(t-t_0)|\alpha,t_0\rangle[/tex]

Then you can calculate [tex]c_{a''}(t)[/tex] in terms of [tex]c_{a'}(t_0)[/tex]

and matrix elements

[tex]\langle a''|U(t-t_0)|a'\rangle[/tex]
 

1. What is a unitary operator?

A unitary operator is a linear transformation that preserves the inner product of vectors, meaning it maintains both the length and angle between vectors. In other words, it is a type of operator that does not change the magnitude or direction of a vector.

2. How do unitary operators preserve normalization?

Unitary operators preserve normalization by maintaining the length of vectors. This means that if a vector is normalized (has a length of 1), it will remain normalized after it is transformed by a unitary operator. This is because the inner product of the vector with itself will remain 1 after the transformation.

3. What is the significance of unitary operators preserving normalization in arbitrary basis?

The fact that unitary operators preserve normalization in arbitrary basis means that they can be used to transform vectors between different bases without changing their lengths. This is important in quantum mechanics, where vectors represent the state of a system and must be preserved in order to accurately describe the behavior of the system.

4. Can unitary operators also change the phase of a vector?

No, unitary operators do not change the phase of a vector. This is because they only affect the magnitude and direction of a vector, not its overall phase. The phase of a vector is determined by its complex components, which are not affected by unitary transformations.

5. How are unitary operators related to Hermitian operators?

Unitary operators and Hermitian operators are related in that they are both types of linear transformations. Unitary operators preserve the inner product, while Hermitian operators preserve the expectation value. Additionally, they are related through the spectral theorem, which states that every unitary operator can be expressed as a combination of Hermitian operators.

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