solution of schrodinger equation - infinite solutions?

Question

In Griffiths's introductory quantum mechanics book, it states that if $\Psi (x,t)$ is a solution to the Schrodinger's equation, then $A\Psi (x,t)$ must also be a solution, where A is any complex constant.

This seems to suggest that for any particle with defined mass and kinetic/potential energy, there are many possible wave functions, as $A$ can be any complex constant - therefore, it gets impossible to figure out anything from the wave function.

Can anyone show me how I am mistaken?

Thank you very much.

Answer 1

I am confused by the comments. Perhaps I am missing something.

For any linear differential equation with a given solution, a constant times that solution is also a valid solution.

The Schroedinger equation is a linear differential equation for the wavefunction $\Psi$. This means that if $\Psi$ is a solution then so is $A\Psi$ where $A$ is a complex constant. You can fix this constant by further requirements like the normalization condition $\int d^dx |\Psi|^2=1$. Of course as Peter Morgan said, the phase of $A$ remains undetermined by the normalization condition.

I think this is also what Griffiths means. I quote him after he says that the wavefunction should be normalized since the particle has to be somewhere.

Well, a glance at Equation 1.1 [the Schroedinger equation] reveals that if $\Psi(x,t)$ is a solution, so too is $A\Psi(x,t)$, where $A$ is any (complex) constant.

and then he goes on to say that we must pick $A$ such that the wavefunction is normalized.

Answer 2

Suppose you have some operator $\hat a$ for returning some physical property, then the way you find the corresponding physical property is with the integral:

$$\int \Psi^* \hat a \Psi d^3x$$

where $\Psi^*$ is the complex conjugate. Assuming your number $A$ has modulus 1, which is required for normalisation, your integral gets multiplied by $A^*A$ and this is one. That's why multiplying by $A$ makes no difference.

The wavefunction itself cannot be physically observed. To get anything out of it requires some integral like the one above. So it's true, sort of, that the same system can be described by an infinite number of wavefunctions, but your physical predictions will always come out the same.

Answer 3

This is the reason, why the 'correct' definition of expectation values is $$ \langle\hat T\rangle_\psi = \frac{\langle\psi | \hat T | \psi\rangle}{\langle\psi | \psi\rangle}. $$ As $\hat T$ is a linear Operator it is easy to see, that any such $A \in \mathbb{C}$ would cancel out. For normalized states, $\langle\psi | \psi\rangle = 1$, and so one recovers the usual expression $$\langle\hat T\rangle_\psi = \langle\psi | \hat T | \psi\rangle = \int \psi^*(x) \hat T \psi(x) \, dx \,.$$

Answer 4

This is because you're not talking about a normalized wavefunction.

The wavefunction only has a physical meaning when it is interpreted as probability. The value of $\psi\bar\psi=|\psi|^2$ at a point is proportional to the probability of the wavefunction collapsing to any given point.

Since it's just proportional, we can multiply it by an arbitrary constant and not change it's physical meaning.

What we can consider is the normalised wavefunction. In this, we multiply $\psi$ by some complex constant $A$, such that the sum of all the $\psi\bar\psi$ over space is $1$--in this way, $\psi\bar\psi$ becomes the probabilty, instead of just being proportional.

In other words, we find $A$ such that $$\int (A\psi)(\bar A\bar\psi)\mathrm dV=|A|^2\int|\psi|^2\mathrm dV=1$$

Of course, you can see that we still have infinite solutions--since $A$ is a complex number, it can be any number lying on the circle of radius $\frac{1}{\sqrt{\int|\psi|^2\mathrm dV}}$ on the Argand(complex) plane.

But that's no issue--that just means that we can have a phase difference (not sure of this). Its like the phase is a normal wave--changing the phase in a normal sinusoidal wave($Asin(\omega t+\large{\phi)}$, where $\phi$ is the phase constant) doesn't change the wave, it only changes the "where the wave started", or our choice of origin. We get similar things with an integration constant.