0
$\begingroup$

I am trying to get my head around the difference between wavefunction and statevector. I looked at the previous answers in this site and don't full understand. Can you some explain the following.

Example: Consider the particle in a infinite potentional:

Here the wavefunction in terms of energy is $ψ_{n}=\sqrt{\frac{2}{L}}\sin(\frac{nπ}{L}x)$.

A state vector is terms of energy eigenfunction is $|ψ⟩=\sum_nc_i|i⟩$

  • Here $|i⟩$ is the energy eigenfunction given by $ψ_{n}=\sqrt{\frac{2}{L}}\sin(\frac{nπ}{L}x)$
  • $c_i$ is some constant

A state vector written in terms of wave function: $|\psi\rangle = \int d^3r\;\psi(\mathbf{r})|\mathbf{r}\rangle$ (taken form State Vector vs wave function)

  • Consider the case where the particle is not in superpostion but in one state {n}.
  • Here |r⟩=|i⟩= $ψ_{n}=\sqrt{\frac{2}{L}}\sin(\frac{nπ}{L}x)$
  • Thus: $|ψ⟩=\int \psi_{n} \psi_{n} d^3r$ (???)
$\endgroup$
4
  • 2
    $\begingroup$ Where did you get $ |\psi \rangle = \int \psi_n \psi_n dx$ ? $\endgroup$
    – Tensor
    Commented Jul 14, 2023 at 12:45
  • $\begingroup$ @Tensor I added more information. What would the correct equation for |ψ⟩ be $\endgroup$ Commented Jul 14, 2023 at 12:53
  • 1
    $\begingroup$ As a simpler example - a vector can be represented abstractly as $\vec{v}$ or as a set of 3 numbers $(v_x,v_y,v_z)$. The latter representation describes the components of $\vec{v}$ in the $(\vec{x},\vec{y},\vec{z})$ basis. If this makes sense to you, then using the same language -- "A state can be represented abstractly as $| \psi \rangle$ or as $\psi(x)$. The latter representation describes the components of $|\psi \rangle$ in the $|x\rangle$ basis." $\endgroup$
    – Prahar
    Commented Jul 14, 2023 at 13:02
  • $\begingroup$ I'm not an expert on this. But as far as I know, state vectors are equivalent to wavefunctions, but written in different notation. Most of the confusion usually come from the notations. I think this link give a pretty decent introduction. For quick test, consider ground state solution to Schrodinger equation, and how will you include spin into the solution. $\endgroup$
    – Tensor
    Commented Jul 14, 2023 at 13:03

1 Answer 1

2
$\begingroup$

In some sense this looks like you kind of have the right idea... I might add the caveat "if that's how you want to think about it". One thing that's definitely wrong is you're messing up index notation in a few places, and I think it's likely that if you fix that things will make a lot more sense to you.

Make sure any index that shows up in your equation is either defined by the summation variable or appears on the left hand side of the equation!! $$\sum_n c_i |i\rangle\text{ }\color{red}\times$$ dosnt make sense... $i$ is not defined here. What you mean is $$\sum_i c_i |i\rangle\text{ }\color{green}\checkmark$$ Then later you have $$|\psi\rangle=\int \psi \psi_n dx\text{ }\color{red}\times$$ It should really set you off that the left side is a state vector and the right side is just a number. What you mean is $$|\psi\rangle=\sum_i\int \psi \psi_i dx |\psi_i\rangle\text{ }\color{green}\checkmark$$ So that we get $$ c_i=\int \psi \psi_i \text{ }\color{green}\checkmark $$ And finally you make another kind of error - defining $n$ in two different ways. $n$ can't be both the summation variable and a variable on the left hand side of the equation. So if you want to define $n$ as the wave function in question, then the summation variable needs to be different, say $i$. $$ |\psi_n\rangle=\int\psi_n\psi_n d^3r\text{ }\color{red}\times $$ $$|\psi_n\rangle=\sum_i\int \psi_n \psi_i dx |\psi_i\rangle=\sum_i\delta_{ni} |\psi_i\rangle=|\psi_n\rangle\text{ }\color{green}\checkmark$$ Where I have used the orthonormality of $\psi_i$ to complete the integral and then the sum.

$\endgroup$
3
  • $\begingroup$ So in the example i have given are $|\psi_{n}⟩$ and $\psi_n$ the same. With them being $ψ_{n}=\sqrt{\frac{2}{L}}\sin(\frac{nπ}{L}x)$. Also where did you get $$|\psi\rangle=\sum_i\int \psi \psi_i dx |\psi_i\rangle\text{ }\color{green}\checkmark$$from, the equation i found in the link was different. $\endgroup$ Commented Jul 14, 2023 at 13:07
  • $\begingroup$ @PhysicsQuestion In the linked question, they focus particularly on converting a state vector into a wavefunction by applying all the same equations I have provided in the position basis. So instead of having discrete different vectors (in an arbitrary basis) $|\psi_n\rangle$ and discrete wavefunctions $\psi_n$, they have a continuous basis of position-space eigenfunctions $|\psi_x\rangle$, where $\psi_x(x')=\delta(x'-x)$. So maybe my equations look different, but they're actually the same equations, and the people in the link have inserted a particular set of "basis" wavefunctions. $\endgroup$
    – AXensen
    Commented Jul 14, 2023 at 13:17
  • $\begingroup$ @PhysicsQuestion I'd say the two are the "same thing" in some sense... they represent the same physical state for the particle. But in another sense they are different - one is a special notation that is a faster way of writing the wavefunction in the position-basis. When you write $\psi_n=$ I expect you to write a function of $x$ on the other side. Might be nice to write it as $\psi_n(x)$. An explicit equation that relates the two is $\psi_n(x)=\langle x|\psi_n\rangle=\int \psi_n(x')\psi_{x} dx'$ $=\int \psi_n(x')\delta(x-x') dx'=\psi_n(x)$ $\endgroup$
    – AXensen
    Commented Jul 14, 2023 at 13:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.