Normalising the wavefunction - two answers

[laser]

Homework Statement

Notes only takes positive answer

Relevant Equations

See description

Hi, when normalising the wavefunction

I get two answers. Is this correct? My notes only has 1/sqrt(L) = A.

 

[Orodruin]

Normalization of wavefunction is always up to an overall phase factor $e^{i\phi}$. You can always multiply by such a factor without affecting normalization so not only are there two possibilities, there us an infinite number of possibilities.

Edit: To be more explicit: The condition $|A|^2 L = 1$ is solved by any $A$ on the form
$$
A = \frac{e^{i\phi}}{\sqrt L}
$$
with $\phi$ an arbitrary real constant.

 

[laser]

$$
A = \frac{e^{i\phi}}{\sqrt L}
$$
with $\phi$ an arbitrary real constant.

If $\phi$ is fixed, am I correct in saying that there are only two solutions in the form $\frac{Be^{i\phi}}{\sqrt L}$, which is B = 1 and B = -1?

 

[Orodruin]

If $\phi$ is fixed, am I correct in saying that there are only two solutions in the form $\frac{Be^{i\phi}}{\sqrt L}$, which is B = 1 and B = -1?

No. There are an infinite number of solutions, which correspond to shifting $\phi$. Once you have the form $e^{i\phi}$ that is the general solution already. There is no point in introducing yet another multiplicative constant. Multiplying by $-1$ specifically just shifts the value of $\phi$ by $\pi$.

 

[laser]

No. There are an infinite number of solutions, which correspond to shifting $\phi$. Once you have the form $e^{i\phi}$ that is the general solution already. There is no point in introducing yet another multiplicative constant. Multiplying by $-1$ specifically just shifts the value of $\phi$ by $\pi$.

I don't follow. By "shifting $\phi$", you are changing the value of $\phi$.

 

[renormalize]

I don't follow. By "shifting $\phi$", you are changing the value of $\phi$.

Yes, that's because $\phi$ is an arbitrary parameter that is not physically measurable. It can be set to $\phi =0 \Rightarrow A=1$ or $\phi =\pi \Rightarrow A=-1$ or to any other value whatsoever without changing the physical predictions that stem from the wavefunction. On the other hand, the phase difference between two quantum states is a physical observable.

 

[laser]

Yes, that's because $\phi$ is an arbitrary parameter that is not physically measurable. It can be set to $\phi =0 \Rightarrow A=1$ or $\phi =\pi \Rightarrow A=-1$ or to any other value whatsoever without changing the physical predictions that stem from the wavefunction. On the other hand, the phase difference between two quantum states is a physical observable.

I understand, but for a wavefunction of definite momentum, $\phi=\frac{px}{\hbar}$.

Edit: Sorry if I am not making any sense, it seems like I have a misconception here which I don't see yet.

To help clarify, you guys are saying that if I have a -1/sqrt(L), it is equivalent to using the 1/sqrt(L) form given that I shift the angle $\phi$ by pi. I am fine with that.

But I claim that -1/sqrt(L) is acceptable without shifting the angle by pi.

 

[renormalize]

I understand, but for a wavefunction of definite momentum, $\phi=\frac{px}{\hbar}$.

No, that's not right. In the notation of your original post:$$\psi_{p}\left(x\right)=Ae^{ipx/\hbar}=\left|A\right|e^{i\phi}e^{ipx/\hbar}=\left|A\right|e^{i\left(px/\hbar+\phi\right)}$$So $\phi$ is the (arbitrary) constant phase of the complex normalization-constant $A$, whereas $px/\hbar$ is the spatially-dependent phase of the complex wavefunction. Don't confuse your phases!

 

[laser]

No, that's not right. In the notation of your original post:$$\psi_{p}\left(x\right)=Ae^{ipx/\hbar}=\left|A\right|e^{i\phi}e^{ipx/\hbar}=\left|A\right|e^{i\left(px/\hbar+\phi\right)}$$So $\phi$ is the (arbitrary) constant phase of the complex normalization-constant $A$, whereas $px/\hbar$ is the spatially-dependent phase of the complex wavefunction. Don't confuse your phases!

Thank you, that makes sense

 

[Orodruin]

But I claim that -1/sqrt(L) is acceptable without shifting the angle by pi.

This is wrong. Using -1 instead of 1 is changing the phase by $\pi$. You could also use $i$, $-i$ or any other phase shift (multiply by a complex number of norm 1).

Edit: The general solution to $|B| = 1$ is $B = e^{i\phi}$ with $\phi \in \mathbb R$ (or in $[0,2\pi)$ if you want to avoid duplicates). So the general solution is not $A = \pm 1/\sqrt L$, it is $A= e^{i\phi}/\sqrt L$ with $\phi$ an arbitrary constant real number.

 

[laser]

This is wrong. Using -1 instead of 1 is changing the phase by $\pi$. You could also use $i$, $-i$ or any other phase shift (multiply by a complex number of norm 1).

Edit: The general solution to $|B| = 1$ is $B = e^{i\phi}$ with $\phi \in \mathbb R$ (or in $[0,2\pi)$ if you want to avoid duplicates). So the general solution is not $A = \pm 1/\sqrt L$, it is $A= e^{i\phi}/\sqrt L$ with $\phi$ an arbitrary constant real number.

Yes sorry I misread your original comment, and thought that the angle was px/hbar, my bad! It makes sense now :) - I'm just used to seeing |A| only in real numbers, in which case then A is only +-A.

 

[laser]

This is wrong. Using -1 instead of 1 is changing the phase by $\pi$. You could also use $i$, $-i$ or any other phase shift (multiply by a complex number of norm 1).

Edit: The general solution to $|B| = 1$ is $B = e^{i\phi}$ with $\phi \in \mathbb R$ (or in $[0,2\pi)$ if you want to avoid duplicates). So the general solution is not $A = \pm 1/\sqrt L$, it is $A= e^{i\phi}/\sqrt L$ with $\phi$ an arbitrary constant real number.

Is it standard in physics to omit this e^(iangle) term? As in the original post, it was just 1/sqrt(L).

 

[Orodruin]

Is it standard in physics to omit this e^(iangle) term? As in the original post, it was just 1/sqrt(L).

If you just want a normalised wave function, you can pick any value for the angle. Choosing $\phi = 0$ is then perfectly permissible.