# QM Vector Spaces and States

• ryanwilk
In summary, the homework statement is that the vector |p> is given by the function x+2x2 and the operator A = 1/x * d/dx, with x = [0,1]. a) Compute the norm of |p> and find that it is \sqrt{<p|p>} . b) Compute A|p> and find that it belongs to the VS of all real valued, continuous functions on the interval x = [0,1] . c) Find the eigenvalues and eigenvectors of A.

## Homework Statement

The vector |p> is given by the function x+2x2 and the operator A = 1/x * d/dx, with x = [0,1].

a) Compute the norm of |p>

b) Compute A|p>. Does A|p> belong to the VS of all real valued, continuous functions on the interval x = [0,1]?

c) Find the eigenvalues and eigenvectors of A.

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## The Attempt at a Solution

a) For the norm, I have that it should be $$\sqrt{<p|p>}$$ but I don't know how to find the scalar product of x+2x2 with itself

b) A|p> is just 1/x + 4, which isn't continuous at x=0 so no?

c) I have no idea how to turn the operator into a matrix. Once I have the matrix, it should be easy but what is the matrix form of A = 1/x * d/dx? Or do I need to use the eigenvalue equation and say that A|p> = eigenvalue|p>?

Any help would be appreciated.
Thanks!

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I sense somehow that the Hilbert space is $L^2 \left(\left[0,1\right], dx\right)$, so computing the norm of that vector should be easy.

For the point c), you don't need any matrix, just the eigenvalue equation.

bigubau said:
I sense somehow that the Hilbert space is $L^2 \left(\left[0,1\right], dx\right)$, so computing the norm of that vector should be easy.

For the point c), you don't need any matrix, just the eigenvalue equation.

Ah so for the norm, it's just $$\bigg(\int_0^1 (x+2x^2)^2 \mathrm{d}x\bigg)^\frac{1}{2}?$$

The eigenvalue equation will be $$\frac{1}{x}\frac{\mathrm{d}}{\mathrm{d}x} |\psi> = \lambda |\psi>$$ but then how do you solve this, or do you assume that the eigenvector is |p>?

(Actually, |p> clearly isn't an eigenvector...)

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Oh wait, do you just solve it like a 1st order ODE to get $$|\psi> = Aexp\bigg(\frac{x^2}{2}\bigg)$$ which means that the eigenvalue is 1?

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Yes to the first post, kind of yes to to the second post when it comes to the wavefunction (actually $\lambda$ should also be in the exponential), however, the spectrum of the operator (possible values of $\lambda$) is the entire complex plane.

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bigubau said:
Yes to the first post, kind of yes to to the second post when it comes to the wavefunction (actually $\lambda$ should also be in the exponential), however, the spectrum of the operator (possible values of $\lambda$) is the entire complex plane.

Oh there is a $\lambda$, I took it out because it looked a bit strange. So $\lambda$ can be anything?

Yes.

bigubau said:
Yes.

Hmm interesting. Thanks a lot for your help!

## 1. What is a vector space in quantum mechanics?

A vector space in quantum mechanics is a mathematical structure that describes the physical states of a quantum system. It is a collection of vectors that can be added together and multiplied by scalars, and it must satisfy certain axioms such as closure and associativity.

## 2. How are vector spaces used in quantum mechanics?

Vector spaces are used to represent the possible states of a quantum system. In quantum mechanics, these states are described by wavefunctions, which are elements of a vector space. This allows us to mathematically describe and analyze the behavior of quantum systems.

## 3. What are the properties of a quantum state?

A quantum state is represented by a vector in a vector space and has several important properties. It must be normalized, meaning that the sum of the squared magnitudes of its components is equal to 1. It must also be complex, meaning that its components can have both real and imaginary parts. Additionally, it must be able to undergo continuous transformations without changing its physical properties.

## 4. What is the difference between a pure and mixed quantum state?

A pure quantum state is a state that is described by a single wavefunction and has a definite probability of being measured in a certain state. A mixed quantum state, on the other hand, is a combination of multiple pure states and does not have a definite probability of being measured in a particular state. Mixed states are typically used to describe systems with some degree of uncertainty or randomness.

## 5. Can two quantum states be orthogonal?

Yes, two quantum states can be orthogonal, meaning that their inner product is equal to zero. This indicates that the two states are mutually exclusive and cannot coexist in the same system. Orthogonality is an important concept in quantum mechanics and is used to describe the relationship between different states of a system.