# Proving Cauchy-Schartz inequality in Brak-ket notation

• jdstokes
In summary, the conversation discussed proving the Cauchy-Schwarz inequality in Brak-ket notation. The equation (\langle \alpha | + \lambda^\ast\langle \beta |)(|\alpha\rangle+ \lambda|\beta\rangle) = \langle \alpha |\alpha \rangle + |\lambda|^2\langle \beta | \beta \rangle + \lambda \langle \alpha | \beta \rangle + \lambda^\ast \langle \beta | \alpha \rangle \geq 0 was used, and it was shown that \langle \alpha | \alpha \rangle \langle\beta|\beta\rangle \geq \langle \alpha |\beta\rangle \langle \beta | \
jdstokes
[SOLVED] Proving Cauchy-Schartz inequality in Brak-ket notation

$(\langle \alpha | + \lambda^\ast\langle \beta |)(|\alpha\rangle+ \lambda|\beta\rangle) = \langle \alpha |\alpha \rangle + |\lambda|^2\langle \beta | \beta \rangle + \lambda \langle \alpha | \beta \rangle + \lambda^\ast \langle \beta | \alpha \rangle \geq 0$.

Put $\lambda = - \langle \beta| \alpha\rangle/\langle \beta|\beta \rangle$. Then

$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle + |\langle\beta|\alpha\rangle|^2 - 2\langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$.

Not quite what I wanted.

Any help would be appreciated.

jdstokes said:
$(\langle \alpha | + \lambda^\ast\langle \beta |)(|\alpha\rangle+ \lambda|\beta\rangle) = \langle \alpha |\alpha \rangle + |\lambda|^2\langle \beta | \beta \rangle + \lambda \langle \alpha | \beta \rangle + \lambda^\ast \langle \beta | \alpha \rangle \geq 0$.

Put $\lambda = - \langle \beta| \alpha\rangle/\langle \beta|\beta \rangle$. Then

$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle + |\langle\beta|\alpha\rangle|^2 - 2\langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$.
the last line implies

$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle - \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$
so
$\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle \geq \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle$
Isn't this what you wanted to show?

How do you get $\langle \alpha | \alpha \rangle \langle\beta|\beta\rangle - \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0$?

OMG I'm so dumb. No need to answer that. Thanks.

## What is the Cauchy-Schwarz inequality in Brak-ket notation?

The Cauchy-Schwarz inequality in Brak-ket notation is a mathematical expression that shows the relationship between inner products, also known as dot products, of two vectors in a vector space. It states that the absolute value of the inner product of two vectors is always less than or equal to the product of their lengths.

## Why is it important to prove the Cauchy-Schwarz inequality in Brak-ket notation?

Proving the Cauchy-Schwarz inequality in Brak-ket notation is important because it is a fundamental result in mathematics that has many applications in fields such as physics, engineering, and statistics. It also provides a powerful tool for solving many mathematical problems and inequalities.

## What is the process of proving the Cauchy-Schwarz inequality in Brak-ket notation?

The process of proving the Cauchy-Schwarz inequality in Brak-ket notation involves using the properties of inner products, such as linearity and positivity, along with mathematical manipulations and logical reasoning. It is a rigorous process that requires a solid understanding of vector spaces and inner products.

## What are some real-life applications of the Cauchy-Schwarz inequality in Brak-ket notation?

The Cauchy-Schwarz inequality in Brak-ket notation has many real-life applications, including in the study of geometry, physics, and statistics. For example, it is used in the derivation of the Pythagorean theorem, in proving the Heisenberg uncertainty principle in quantum mechanics, and in statistical analysis to determine the correlation between variables.

## Are there any limitations to the Cauchy-Schwarz inequality in Brak-ket notation?

Yes, there are some limitations to the Cauchy-Schwarz inequality in Brak-ket notation. It only holds for inner products in vector spaces that satisfy certain properties, such as linearity and positivity. It also does not hold for all types of vector spaces, such as spaces with non-Euclidean metrics.

• Advanced Physics Homework Help
Replies
8
Views
1K
• Advanced Physics Homework Help
Replies
8
Views
2K
• Advanced Physics Homework Help
Replies
3
Views
901
• Advanced Physics Homework Help
Replies
3
Views
930
• Advanced Physics Homework Help
Replies
7
Views
3K
• Advanced Physics Homework Help
Replies
1
Views
1K
• Quantum Physics
Replies
3
Views
984
• Advanced Physics Homework Help
Replies
4
Views
2K
• Advanced Physics Homework Help
Replies
4
Views
4K
• Quantum Interpretations and Foundations
Replies
2
Views
840