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Consider the well: $$V(x) = \begin{cases} \infty&\text{if }x<0 \\ 0&\text{if }x\in\left(0,L\right) \\ \infty&\text{if }x>L. \end{cases}$$

Solving the time independent Schrödinger equation on the well $$-\frac{\hbar^2}{2m} \frac{∂^2\psi}{∂x^2} =E\psi,$$ will yield one of the solutions $\psi = 0$.

Some books say that this solution is un-normalizable. So we ignore it. Some books say that zero solution has no physical meaning.

I don't understand why the normalizable condition is so important. (Does this condition imply that the particle will not disappear in the worlds?). If $\psi = 0$, what's the meaning of the state? (Does this mean we can't find the particle in the wall? But it's a reasonable solution, why ignore it?)

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  • $\begingroup$ The time-independent equation you mention is an eigenvalue equation, $H\psi = E\psi$, and if an eigenvector of zero were allowed, then any eigenvalue you want is in the spectrum of the Hamiltonian. Physically, that means you could make the potential whatever you want and it would have absolutely no effect on the energy. So, mathematically and physically, allowing a zero eigenvector tends to not make any sense. $\endgroup$
    – penovik
    May 18, 2020 at 19:06

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If ψ=0, what's the meaning of the state

$\psi=0$ isn't (a representation of) a state. It is the position basis representation of the zero ket. A vector space must have a zero vector.

But a state is a ray in Hilbert space so the space of states is not a vector space.

A wavefunction (the position basis representation of a ket) must be normalizable to unity since $\rho=\psi^*\psi$ is interpreted as a probability density. The system you give is the (one) particle in a (1D) box. That is, the system has by stipulation one particle that has 100% probability of being found somewhere within the box.

As a probability density, the solution $\rho = \psi^*\psi = 0$ does not make physical sense.

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  • $\begingroup$ I'm new to quantum physics,$\psi =0$ seems to have some mathematical necessary for this structure. But for now,it seems the interpretation that $\psi = 0$ does not give us extra information seems to be more reasonable for me. $\endgroup$
    – yi li
    May 17, 2020 at 14:38
  • $\begingroup$ @yi_li, under the assumption that you're interested in 'starting off on the right foot' with learning quantum physics, you might reconsider that believing something that is not true is more reasonable than not. Believing that $\psi=0$ is (a representation of) a valid state (even if one that doesn't give us extra information) is to believe something that is false. That's all I have to say about that. $\endgroup$ May 17, 2020 at 15:50
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That's a very legitimate question. Yes, $\psi=0$ is a solution of the above Schrodinger's equation, but most of the text books don't consider it because it is a trivial solution or in other words, $\psi=0$ will not give us extra information about the quantum system. If I formulate this mathematically, we know that the most general wavefunction is written as the linear combination of all the eigenstates. So, $$\psi(x)=\sum_{n=0}^{\infty}c_n\phi_n(x)$$ where $\phi_n's$ are the eigenstate of the system. Adding the solution $\phi=0$ solution to above general $\psi(x)$ will not change the wavefunction, hence there is no extra information being added by adding the trivial solution.

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    $\begingroup$ I disagree only on a detail here. Your function $\psi(x)$ is not a solution of Schrödinger's equation. However, the linear combination $\psi(x,t)=\sum_{n=0}^{\infty}c_n e^{iE_n t/\hbar}\phi_n(x)$ actually would be the most general solution of the time-dependent Schrödinger equation. $\endgroup$ May 17, 2020 at 14:24
  • $\begingroup$ Yes that is correct. But OP has written the time independent Schrodinger's equation and $\psi(x)$ is the general solution to the time independent Schrodinger's equation and obviously not the solution to time dependent one. $\endgroup$
    – sslucifer
    May 17, 2020 at 14:29
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    $\begingroup$ What worries me about this answer is that the requirement that physical solutions be normalizable is not mentioned at all. The solution $\psi = 0$ doesn't correspond to a state in the space of states. $\endgroup$ May 17, 2020 at 14:34
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    $\begingroup$ This $\psi(x)$ does not solve the time-independent Schrödinger equation either. It is easy to disprove for example with $\psi(x)=\phi_1(x)+\phi_2(x)$ (when $E_1\ne E_2$). $\endgroup$ May 17, 2020 at 14:41
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There is no room for a discussion of adding the "solution $\psi(x)=0$" because it is not a solution to the time-independent Schrodinger equation, because it can't be. The time-independent Schrodinger equation is the eigenvalue equation of the Hamiltonian. The eigenvector of an operator is specifically defined to be a non-null vector (the definition of an eigenvector). $\psi(x)=0$ is a null vector, so it can't be an eigenvector.

The fact that it cannot be normalized to one (which is a necessary requirement for a physically realizable state) is only partly relevant. It just means that a physical state cannot be $\psi(x)=0$. This doesn't mean that it can't be an eigenvector of an operator, but being a null vector means that. For example, eigenvectors of the position and the momentum operators are also not physically realizable but they are still relevant because they are eigenvectors.

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  • $\begingroup$ It seems to fit math so well $\endgroup$
    – yi li
    May 17, 2020 at 14:52
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    $\begingroup$ @yili It is a solution of the equation that you have written but it's not a solution of the time independent Schrödinger equation because the time independent Schrödinger equation is an eigenvalue equation and null vectors are, by definition, not eigenvectors. For example, the typical eigenvalue equation for a matrix goes $AX=aX$. Do you say that $X=0$ is an eigenvectors? Of course not! It solves the equation. But that's not the point. The equation as such doesn't capture the full definition of an eigenstate. The full definition is all non-null solutions of such an equation is an eigenvector. $\endgroup$
    – user87745
    May 17, 2020 at 15:02
  • $\begingroup$ That's incorrect. $\psi=0$ is normalizable and its norm is $0$. Moreover, $\psi=0$ IS a solution of the time-dependent Schrodinger equation (with $E=0$), but it is a trivial solution. It doesn't describe any physics because $\vert \psi(x)\vert^2=0$ everywhere. $\endgroup$ May 17, 2020 at 15:19
  • $\begingroup$ @ZeroTheHero By normalizable, I meant normalizable to unity (which is the usual language in physics?). It's not an eigenstate because eigenstates are by definition non-null and thus it's not a solution of an eigenvalue equation. $\endgroup$
    – user87745
    May 17, 2020 at 15:23
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    $\begingroup$ $\psi\equiv0$ is a solution of the TISE. It just isn't a solution of the Schrödinger eigenproblem, which is more than the Schrödinger's equation. The eigenproblem has, in addition to the core equation, also boundary conditions and the constraint of $\psi$ being non-identically-zero. $\endgroup$
    – Ruslan
    May 17, 2020 at 16:10
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It is a trivial solution - solution of no particle (or zero number of particles, where the number of particles is given by the normalization integral.)

Note that the same is true for any homogeneous wave equation - e.g., Maxwell equations also have solution with no field.

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  • $\begingroup$ The way you formulated it seems to imply that there exists a solution with e.g. two particles. But actually, the configuration space of $x\in\mathbb R$ fits exactly one particle, not less, not more. $\endgroup$
    – Ruslan
    May 17, 2020 at 16:15
  • $\begingroup$ In non-interacting case the number of particles is a matter of normalization. E.g., in scattering problems one frequently normalizes the incident flux to the number of particles rather than to one. $\endgroup$
    – Roger V.
    May 17, 2020 at 16:20
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Since the Schrodinger equation is linear, any linear combination of solutions us a solution, so also ψ=0 is one. However, we also demand that the norm if the solution is unity and the zero solution fails this criterion . There are by the way infinitely many solutions of the Schrodinger equation that are not normalised to one.

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