Normalizing factor of wave function

[Lagrange fanboy]

TL;DR Summary

Confusion over an equation on page 256 of Quantum Mechanics - The Theoretical Minimum

So on page 256 of Quantum Mechanics - The Theoretical Minimum, it says that the wave function of a momentum eigenvector, with respect to the position eigenbasis is $\psi_p(x)=Ae^{\frac{ipx}{\hbar}}$, and $A$ must be $\frac{1}{\sqrt{2\pi}}$ to keep it a unit vector. However why must $A=\frac{1}{\sqrt {2\pi}}$? $\bar \psi_p=\bar Ae^{-\frac{ipx}{\hbar}}$, so $| \psi_p|^2=A\bar A=|A|^2$. Plugging in $A=\frac{1}{\sqrt{2\pi}}$ that means $\int_{-\infty}^\infty \psi_p\bar\psi_p dx= \int_{-\infty}^\infty \frac{1}{2\pi}dx\not = 1$

 

[PeroK]

TL;DR Summary: Confusion over an equation on page 256 of Quantum Mechanics - The Theoretical Minimum

So on page 256 of Quantum Mechanics - The Theoretical Minimum, it says that the wave function of a momentum eigenvector, with respect to the position eigenbasis is $\psi_p(x)=Ae^{\frac{ipx}{\hbar}}$, and $A$ must be $\frac{1}{\sqrt{2\pi}}$ to keep it a unit vector. However why must $A=\frac{1}{\sqrt {2\pi}}$? $\bar \psi_p=\bar Ae^{-\frac{ipx}{\hbar}}$, so $| \psi_p|^2=A\bar A=|A|^2$. Plugging in $A=\frac{1}{\sqrt{2\pi}}$ that means $\int_{-\infty}^\infty \psi_p\bar\psi_p dx= \int_{-\infty}^\infty \frac{1}{2\pi}dx\not = 1$

The integral you have is divergent, using normal calculus. In QM, however, you can appeal to the following identity:$$\delta(p) = \frac 1 {2\pi}\int_{-\infty}^{+\infty} e^{ipx} \ dx$$

 

[topsquark]

TL;DR Summary: Confusion over an equation on page 256 of Quantum Mechanics - The Theoretical Minimum

So on page 256 of Quantum Mechanics - The Theoretical Minimum, it says that the wave function of a momentum eigenvector, with respect to the position eigenbasis is $\psi_p(x)=Ae^{\frac{ipx}{\hbar}}$, and $A$ must be $\frac{1}{\sqrt{2\pi}}$ to keep it a unit vector. However why must $A=\frac{1}{\sqrt {2\pi}}$? $\bar \psi_p=\bar Ae^{-\frac{ipx}{\hbar}}$, so $| \psi_p|^2=A\bar A=|A|^2$. Plugging in $A=\frac{1}{\sqrt{2\pi}}$ that means $\int_{-\infty}^\infty \psi_p\bar\psi_p dx= \int_{-\infty}^\infty \frac{1}{2\pi}dx\not = 1$

Actually, it turns out that the "wavefunction" $e^{ipx/\hbar}$ can't be renormalized, as you can see from your calculation. The plane waves don't belong to Hilbert space, so we can "normalize" them using any convention we like. The one you mentioned follows the definition of a Fourier transform.

-Dan

 

[Lagrange fanboy]

The integral you have is divergent, using normal calculus. In QM, however, you can appeal to the following identity:$$\delta(p) = \frac 1 {2\pi}\int_{-\infty}^{+\infty} e^{ipx} \ dx$$

But the norm of the wavefunction won't be 1 then, shouldn't this cause a problem down the line when calculating probabilities?

 

[PeterDonis]

the norm of the wavefunction won't be 1 then, shouldn't this cause a problem down the line when calculating probabilities?

The wave function $e^{ipx}$ is not physically realizable, so the fact that it is not normalizable doesn't cause any actual issue. If you want a heuristic probability interpretation of it, it says that the particle is equally likely to be at any $x$ value (i.e., position); some sources will describe this as the position being "completely uncertain". (Conversely, a position eigenstate, $\delta(x)$, can be described as the momentum being "completely uncertain".) But in practice we never need to compute probabilities for such states, only for linear combinations of them (wave packets) that are normalizable.

 

[PeroK]

But the norm of the wavefunction won't be 1 then, shouldn't this cause a problem down the line when calculating probabilities?

These wavefunctions are not physical in themselves, but occur in continuous wave packets. The point of the normalisation constant is to provide an ersatz orthonormality:$$\braket{\psi_p |\psi_q} = \delta(p - q)$$I don't have Susskind's book, so I don't know how and where he explains all this.

 

[Lagrange fanboy]

The wave function $e^{ipx}$ is not physically realizable, so the fact that it is not normalizable doesn't cause any actual issue. If you want a heuristic probability interpretation of it, it says that the particle is equally likely to be at any $x$ value (i.e., position); some sources will describe this as the position being "completely uncertain". (Conversely, a position eigenstate, $\delta(x)$, can be described as the momentum being "completely uncertain".) But in practice we never need to compute probabilities for such states, only for linear combinations of them (wave packets) that are normalizable.

Still, if $Ae^{\frac{ipx}{\hbar}}$ is the wave function for $|p\rangle$ wrt. position basis, then I'd expect $\int_{-\infty}^\infty \langle p|x\rangle \langle x|p\rangle dx = 1$.

 

[PeterDonis]

Still, if $Ae^{\frac{ipx}{\hbar}}$ is the wave function for $|p\rangle$ wrt. position basis, then I'd expect $\int_{-\infty}^\infty \langle p|x\rangle \langle x|p\rangle dx = 1$.

Then your expectation in this case is simply wrong. This expectation is only valid for wave functions that are physically realizable. As I've already said, $e^{ipx}$ is not physically realizable.

 

[PeroK]

Still, if $Ae^{\frac{ipx}{\hbar}}$ is the wave function for $|p\rangle$ wrt. position basis, then I'd expect $\int_{-\infty}^\infty \langle p|x\rangle \langle x|p\rangle dx = 1$.

In QM you often have to expect the unexpected!

 

[topsquark]

Still, if $Ae^{\frac{ipx}{\hbar}}$ is the wave function for $|p\rangle$ wrt. position basis, then I'd expect $\int_{-\infty}^\infty \langle p|x\rangle \langle x|p\rangle dx = 1$.

Quantum Mechanics pretty much only uses wavefunctions (or bras and kets if you are doing Matrix Mechanics), but all of these live in something called "Hilbert space." The function $e^{i p x/\hbar}$ does not actually have a defined norm, so it does not belong to Hilbert space, and thus we don't actually call it a wavefunction. (It often is called a wavefunction, though, even by the instructors.) It is often included along with wavefunctions, though, because it is a useful way to express certain types of behavior.

You just have to keep in mind that the plane wave solutions aren't actually wavefunctions and behave slightly differently.

-Dan

 

[Lagrange fanboy]

Quantum Mechanics pretty much only uses wavefunctions (or bras and kets if you are doing Matrix Mechanics), but all of these live in something called "Hilbert space." The function $e^{i p x/\hbar}$ does not actually have a defined norm, so it does not belong to Hilbert space, and thus we don't actually call it a wavefunction. (It often is called a wavefunction, though, even by the instructors.) It is often included along with wavefunctions, though, because it is a useful way to express certain types of behavior.

You just have to keep in mind that the plane wave solutions aren't actually wavefunctions and behave slightly differently.

-Dan

Wow now I'm really confused. I read on wikipedia that the postulates of quantum mechanics include:

• Quantum states are unit vectors in some Hilbert space
• Observables are Hermitian operators acting on said Hilbert space
• Probability of results are calculated from inner products between the system's current state and the observable's eigenstates

Since $e^{\frac{ipx}{\hbar}}$ isn't in the Hilbert space does that mean the postulates above are inconsistent? What are the correct postulates then?

 

[PeroK]

Wow now I'm really confused. I read on wikipedia that the postulates of quantum mechanics include:

• Quantum states are unit vectors in some Hilbert space
• Observables are Hermitian operators acting on said Hilbert space
• Probability of results are calculated from inner products between the system's current state and the observable's eigenstates

Since $e^{\frac{ipx}{\hbar}}$ isn't in the Hilbert space ...

... it means it isn't a quantum state. In other words, momentum has no eigenstates. It has eigenfunctions outside the Hilbert space, which are mathematically useful, but not themselves quantum states. I'm surprised if Susskind has not said all this.

 

[PeterDonis]

I read on wikipedia

Which is not a good primary source for understanding science, particularly not when a fairly subtle technical issue like this one comes up.

If we were to sharpen up those postulates to take into account the technical issue under discussion, there are at least two ways to do it. One way, the way I was implicitly using in my previous posts, is to limit consideration to states that are physically realizable, which are also normalizable. So the postulates would read like this (note the bolded additions):

• Physically realizable quantum states are unit vectors in some Hilbert space
• Observables are Hermitian operators acting on said Hilbert space which have eigenstates that are physically realizable
• Probability of results of physically realizable measurements are calculated from inner products between the system's current state and the observable's eigenstates

The "states" $e^{ipx}$ are not physically realizable states, and the momentum operator that has those "states" as its eigenstates is not a physically realizable operator (since it would correspond to measuring the momentum with infinite exactness, and physically realizable measurements only have finite accuracy). So the above postulates do not apply to them.

The other method of dealing with the technical issue is to use, instead of just Hilbert space, something called "rigged Hilbert space". This is basically an expansion of Hilbert space to include non-normalizable "states" like $e^{ipx}$ and operators that have those states as eigenstates, with a lot of technical precautions taken to ensure that physically realizable measurements still behave properly. If we take that route, then the postulates would read like this:

• Quantum states are unit vectors in some rigged Hilbert space, which must be normalizable if they correspond to physically realizable states
• Observables are Hermitian operators acting on said rigged Hilbert space
• Probability of results are calculated from inner products between the system's current state and the observable's eigenstates, and then applying corrections as necessary if the states involved are not normalizable

Either approach is valid, but you have to be very careful not to mix the two.

 

[vanhees71]

But the norm of the wavefunction won't be 1 then, shouldn't this cause a problem down the line when calculating probabilities?

That's among the most difficult subtleties in the beginning of learning quantum theory. If you have self-adjoint operators with a continuous spectrum the corresponding "eigenfunctions" almost always are not normalizable to 1 and thus do not belong to the Hilbert space. The math was also a problem in the beginning of quantum theory, and the mathematicians were inspired by Dirac's invention of his distribution to develop modern functional analysis.

The pragmatic physicist's answer is that in this case the "eigenfunctions" are "normalized to a $\delta$ distribution", i.e.,
$$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p^*(x) u_p(x).$$
Now you have
$$\hat{p} u_p(x)=-\mathrm{i} \hbar u_p'(x)=p u_p(x) \; \Rightarrow \; u_p(x)=A(p) \exp(\mathrm{i} p x/\hbar).$$
From the theory of Fourier integrals indeed you get
$$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x A^*(p) A(p') \exp[\mathrm{i} x (p'-p)/\hbar] = 2 \pi |A(p)|^2 \delta[(p-p')/\hbar] = 2 \pi \hbar |A(p)|^2 \delta(p-p').$$
To "normalize the momentum eigenfunction to the $\delta$ distribution" thus implies that you can set
$$A(p)=\frac{1}{\sqrt{2 \pi \hbar}.$$
The normalization factor is only determined up to an arbitrary phase factor, which is unimportant, i.e., you can take the above most simple choice for the normalization factor without loosing anything.

If you want to learn about the "rigged Hilbert space" mentioned in the previous posting, a good starting point is the textbook

L. E. Ballentine, Quantum Mechanics, World Scientific,
Singapore, New Jersey, London, Hong Kong (1998).