Meaning of "Up to a Scale" for Eigenvectors in Quantum Mechanics

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Homework Help Overview

The discussion revolves around the interpretation of the phrase "up to a scale" as it pertains to eigenvectors in quantum mechanics, specifically in the context of their uniqueness and scalar multiples.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the meaning of eigenvectors being unique "up to a scale," with some suggesting it refers to scalar multiples. Others seek clarification on the specific wording used in a quantum mechanics text.

Discussion Status

The conversation includes attempts to clarify the phrase's meaning, with some participants confirming their understanding and others correcting previous statements about the wording. There is an ongoing exploration of the implications of this phrase in the context of eigenvectors.

Contextual Notes

Participants reference a specific theorem from a quantum mechanics textbook, indicating that the phrase is a direct quote from the text. There is a focus on ensuring accurate terminology in the discussion.

Gary Roach
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Homework Statement


What is the meaning of the phrase "up to a scale" as applied to eigenvectors in quantum mechanics.


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The Attempt at a Solution


N/A did an extensive search of the web and my texts. No joy.
 
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I'll guess that this means that two eigenvectors point in the same or opposite direction. IOW, each one is a scalar multiple of the other.
 
I guess I wasn't specific enough in my question. Sorry. The definition I would like is as the phrase applies to the following statement:

ie \Lambda|\omega_i> is an eigenvector of \Omega with eigenvalue \omega_i. Since the vector is unique \underline{up\ to\ a\ scale},

\Lambda|\omega_i > = \lambda_i | \omega_i >
 
Last edited:
Yes, that is exactly what Mark44 was referring to. If v is an eigenvector of linear transformation, A, with eigenvector \lambda, then Av= \lambda v. If u is any "scalar multiple" of v, u= sv for some scalar, s, then, since A is linear, Au= A= (sv)= s(Av)= s(\lambda v)= \lambda(sv)= \lambda u so that u is also an eigenvector with eigenvalue \lambda. That is, the eigenvector is unique "up to a scalar multiple" which is, I presume, what this physics text means by "up to scale". (You might want to recheck the exact wording. "Up to a scale" doesn't seem grammatically correct.)
 
This is from the proof of theorem 13 page 44 of Principles of Quantum Mechanics- 2nd edition by R Shanker. The up to scale is a direct quote.

Thanks all for the help.

Gary R.
 
Gary Roach said:
This is from the proof of theorem 13 page 44 of Principles of Quantum Mechanics- 2nd edition by R Shanker. The up to scale is a direct quote.
But what you said before was "up to a scale" which is not a direct quote.

Thanks all for the help.

Gary R.
 
I rechecked the text. The actual statement is "up to a scale". I goofed in the second message. Sorry

Gary R.
 

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