Normalization Conditions of Wave functions

In summary, the conversation discusses the concept of normalization in quantum mechanics and whether the wave function must approach zero as the variable goes to infinity. The experts conclude that this assumption is not always true and provide counterexamples. They also mention the importance of bounded first derivatives in proving that the wave function approaches zero.
  • #1
fscman
8
1
I am currently reading through Griffiths Quantum Mechanics textbook, and on page 14, Griffiths proves that
[itex]\frac{d}{dt}\int_{-\infty}^{\infty} |\Psi(x,t)|^2 \, dx = \left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex]
and he claims that the right hand side evaluates to zero since [itex]\Psi[/itex] must be normalizable and hence [itex] \lim_{x \to \infty} \Psi(x,t) = 0 [/itex].
My question is whether [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}[/itex] must also equal zero. I ask this because:

Suppose we take the function [itex]\Psi (x,t)=\frac{1}{x} \sin(x^9)[/itex]. Then, we can calculate [itex] \int_{-\infty}^{\infty} \Psi^2 \, dx = 1.14...[/itex] so [itex]\Psi[/itex] IS normalizable. We can find that [itex]\frac{\partial \Psi}{\partial x}=\frac{1}{x^2}\left(9x^9 \cos(x^9)-\sin(x^9)\right)=9x^7 \cos(x^9)-\frac{\sin(x^9)}{x^2}[/itex],
so if we take [itex] \Psi^* \frac{\partial \Psi}{\partial x}=9 x^6 \sin(x^9) \cos (x^9) - \frac{\sin^2(x^9)}{x^3}[/itex], we find that this expression does not approach 0 as x approaches infinity (which renders Griffiths' argument using the fact [itex] \lim_{x \to \infty} \Psi(x,t) = 0 [/itex] to explain why the original integral false; there are functions that tend to zero as x approaches infinity but the derivative can grow arbitrarily large). Of course, in this case, [itex]\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex] happens to vanish since [itex]\Psi^*=\Psi[/itex], but if we choose some general complex function, then I would guess that [itex]\left.\frac{i \hbar}{2m}\left( \Psi^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x} \Psi \right) \right|_{-\infty}^{\infty}[/itex] might not vanish if we do not set [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}[/itex] equal to zero. However, I have never seen this condition in quantum mechanics textbooks. Is there something wrong with my analysis, or is [itex]\lim_{x \to \infty} \frac{\partial \Psi}{\partial x}=0[/itex] an implicit assumption physicists make? Thanks.
 
Last edited:
Physics news on Phys.org
  • #2
I don't have a complete answer, just a few observations.
  • Physicists are really sloppy with these things, so they'll say that ##\psi## must go to 0, even though it may not be true.
  • As you have noted, it isn't true. To prove that ##\psi\to 0##, you will need other assumptions than than square-integrability.
  • We are also assuming that ##\psi## is a solution of the Schrödinger equation. So ##\psi## is differentiable, and therefore continuous.
  • The right-hand side you're working with is 0 if ##\psi\to 0## and ##\partial\psi/\partial x## is bounded.
I suspect that it's possible to prove that if ##\psi## is square-integrable, continuous and has bounded first derivatives, then ##\psi\to 0##. I also suspect that there's something very "unphysical" about wavefunctions with unbounded partial derivatives. I expect them to have infinite energy or something like that.
 
  • #3
You are right!

Physicists sometimes believe that normalization of a wave function requires that it approaches zero at infinity, i.e. lim x→±∞ ψ(x) = 0. This assumption is wrong - there are well-known counter examples. e.g.:

ψ(x) = 1 for x in [n,n+1/n2] for all integers n
ψ(x) = 0 elsewhere

lim x→±∞ ψ(x) does not exist, but of course the infinite sum for the normalization converges which means that ψ(x) is normalizable. You can even construct wave functions with increasing height of the peaks for x→±∞

You may exclude those wave functions based on "physical reasoning" but in pure maths i.e. for the L2 Hilbert space they are perfectly valid.
 
  • #4
Here's a nice example of a continuous square-integrable function that doesn't go to zero as the variable goes to infinity, from this article.

attachment.php?attachmentid=42567&d=1326093702.png
 

Attachments

  • counterexample.png
    counterexample.png
    5.6 KB · Views: 1,460
  • #5
Thanks for your replies! I will look more into the mathematical side of L2 spaces and read the article. I believe the "physical reality" argument for why ψ approaches 0 as x approaches infinity, and I just realized that from the time independent Schrodinger equation, if we require the potential energy to be bounded at infinity, then [itex]\frac{\partial^2 \Phi}{\partial x^2}[/itex] must approach 0 as x approaches infinity if the original function ψ behaves that way. Then, by mathematical arguments involving the mean value theorem, I believe that implies [itex]\lim_{x \to \infty} \frac{\partial \Phi}{\partial x}=0[/itex] also.
 
  • Like
Likes Philethan

Related to Normalization Conditions of Wave functions

1. What is the purpose of normalizing wave functions?

The purpose of normalizing wave functions is to ensure that the total probability of finding a particle in a given region is equal to 1. This is necessary in order to accurately describe the behavior of quantum systems and make meaningful predictions.

2. How is the normalization condition written mathematically?

The normalization condition is written as an integral of the wave function squared over all space, which must equal 1. This is represented by the equation:
∫ |Ψ(x)|² dx = 1

3. Can wave functions be normalized to a different value than 1?

No, wave functions must be normalized to a value of 1 in order to accurately describe the behavior of quantum systems. Normalizing to a different value would alter the probabilities and predictions of the system.

4. What happens if a wave function is not normalized?

If a wave function is not normalized, it means that the total probability of finding a particle in a given region is not equal to 1. This can lead to incorrect predictions and does not accurately represent the behavior of the system.

5. How do you normalize a wave function?

To normalize a wave function, you must first calculate the integral of the wave function squared over all space. Then, you divide the wave function by the square root of this integral. This will ensure that the wave function is normalized to a value of 1.

Similar threads

  • Quantum Physics
Replies
15
Views
356
Replies
7
Views
646
Replies
3
Views
893
Replies
3
Views
516
  • Quantum Physics
Replies
9
Views
984
Replies
24
Views
696
  • Quantum Physics
Replies
20
Views
1K
Replies
29
Views
4K
Replies
17
Views
2K
  • Quantum Physics
Replies
11
Views
1K
Back
Top