Wave function: vector or scalar?

[lightarrow]

Meant as element of Hilbert space of L^2 functions... etc., the wave function is a vector.

In the abstract description with kets and operators on these, the wave function is the scalar product between a ket |Psi> and the "eigenkets" |x> of the position operator: psi(x) = <x|Psi>.

So: psi is a vector or a scalar?

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lightarrow

 

[ShayanJ]

The wave-function($\psi(x)=\langle x | \psi \rangle$) is a scalar in the sense that it doesn't change under a coordinate transformation $x \rightarrow x'$, i.e. $\psi'(x')=\psi(x)$. In this sense, a vector means a collection of components that transform to linear combinations of themselves under a coordinate transformation. The wave-function is not a vector in this sense.
Here you can read about the notion of a vector space. The wave-function is a member of some vector space, and is said to be a vector in this sense. The vector space that $\psi(x)=\langle x | \psi \rangle$ is a member of, is the space of square-integrable complex valued functions over $\mathbb R$ with the appropriate boundary conditions.

 

[MBPTandDFT]

You are calling two quantities with the same name: the ket |Psi> is the "quantum state", it belongs to the Hilbert space and it is a vector, while the wavefunction Psi(x) is the projection of this state on a basis element (x), and it's a function.

 

[vanhees71]

The object $|\Psi \rangle$ is in the abstract Hilbert space $\mathcal{H}$ (up to equivalence there's basically only one separable Hilbert space). The wave function $\Psi(x)=\langle x|\Psi \rangle$ is the component of $|\Psi \rangle$ with respect to the generalized eigenbasis of the position operator.

In this way there is a one-to-one mapping between the abstract Hilbert space $\mathcal{H}$ and the Hilbert space of square-integrable functions $L^2(\mathbb{R})$.

 

[lightarrow]

The object $|\Psi \rangle$ is in the abstract Hilbert space $\mathcal{H}$ (up to equivalence there's basically only one separable Hilbert space). The wave function $\Psi(x)=\langle x|\Psi \rangle$ is the component of $|\Psi \rangle$ with respect to the generalized eigenbasis of the position operator.

In this way there is a one-to-one mapping between the abstract Hilbert space $\mathcal{H}$ and the Hilbert space of square-integrable functions $L^2(\mathbb{R})$.

Thanks.
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lightarrow

 

[TheInquisitor]

The object $|\Psi \rangle$ is in the abstract Hilbert space $\mathcal{H}$ (up to equivalence there's basically only one separable Hilbert space). The wave function $\Psi(x)=\langle x|\Psi \rangle$ is the component of $|\Psi \rangle$ with respect to the generalized eigenbasis of the position operator.

In this way there is a one-to-one mapping between the abstract Hilbert space $\mathcal{H}$ and the Hilbert space of square-integrable functions $L^2(\mathbb{R})$.

Can you please explain what does this mean?

 

[PeroK]

Can you please explain what does this mean?

How much do you already know about the mathematical formalism underpinning QM?

 

[TheInquisitor]

How much do you already know about the mathematical formalism underpinning QM?

I'm almost 1 month into my course of Quantum Mechanics. So, not much.

 

[PeroK]

I'm almost 1 month into my course of Quantum Mechanics. So, not much.

What parts of the above post did you particularly not understand: Hilbert Space, Dirac notation, abstract inner product, position eigenbasis, wave-function, square-integrable functions?

 

[TheInquisitor]

I understand inner product, eigenbasis and wave function. The rest still confuse me.

 

[PeroK]

I understand inner product, eigenbasis and wave function. The rest still confuse me.

You can look the rest up.

 

[vanhees71]

It's still not clear to me what's not clear about my posting #4. I thought you are familiar with the representation-independent formulation in terms of an abstract Hilbert space and Dirac's bra-ket formalism.