Prove that the norm squared of a superposition of two states is +ve

[Phys12]

TL;DR Summary

Given two wavefunctions $\Psi_1 and \Psi_2$, show that $|\alpha\Psi_1 + \beta\Psi_2|^2 > 0$ such that $\alpha, \beta \in C$ subject to the normalization condition

This is what I have so far: $$ |\alpha\Psi_1 + \beta\Psi_2|^2 = |\alpha|^2|\Psi_1|^2 + |\beta|^2|\Psi_2|^2 + \alpha^*\beta\Psi_1^*\Psi_2 + \alpha\beta^*\Psi_1\Psi_2^* $$

$$=> |\alpha\Psi_1 + \beta\Psi_2|^2 = |\alpha|^2|\Psi_1|^2 + |\beta|^2|\Psi_2|^2 + 2Re(\alpha^*\beta\Psi_1^*\Psi_2) $$

I am having trouble showing that the last term in the above equation is always less than the first two (in terms of magnitude). I think this is what I need to show to make the entire expression positive since the only term that can be negative is the last one. As an aside, I saw this originally in a more specific example where both $\Psi_1$ and $\Psi_2$ were complex exponentials. But I'd imagine this would be true also for any two general wavefunctions, just not sure how to prove that

 

[HomogenousCow]

Are you sure that you’re talking about the complex norms and not the infinite vector space norms?

 

[Phys12]

Are you sure that you’re talking about the complex norms and not the infinite vector space norms?

I think so, at least in the example that the professor did, he took the complex norm. What are infinite vector space norms?

 

[HomogenousCow]

Well $|z|^2 > 0$ is true for any complex number that isn't zero, so I don't know why you're trying to prove this.

 

[Phys12]

Well $|z|^2 > 0$ is true for any complex number that isn't zero, so I don't know why you're trying to prove this.

Well if you look at this video: , from 49:08 till 49:55, he says that you can check that the total quantity is zero (even though the last two terms may not be positive). I'm trying to do that check

 

[Phys12]

You can certainly check that $P>0$ by nothing that $|z|^2 > 0$ for all non-zero complex number $z$.

So just say that the superposed state is a new complex number, z and $|z|^2 > 0$ for all non-zero complex numbers? Is there any way to prove this by expanding the norm-squared value of the equation above as I had started to do?

 

[PeterDonis]

Well $|z|^2 > 0$ is true for any complex number that isn't zero, so I don't know why you're trying to prove this.

The OP is not talking about complex numbers, but about vectors in a Hilbert space. But see below.

just say that the superposed state is a new complex number

Vectors in a Hilbert space are not the same thing as complex numbers.

However, AFAIK part of the definition of a Hilbert space is that the squared norm of any vector that is not the zero vector must be positive, and that a complex linear combination of any two vectors gives another vector in the same Hilbert space. So it does seem like there's something missing from the problem statement if it's supposed to be non-trivial to prove.

 

[PeterDonis]

Are you sure that you’re talking about the complex norms and not the infinite vector space norms?

in the example that the professor did, he took the complex norm

How can you take the complex norm of a vector in a Hilbert space? A vector in a Hilbert space is not the same thing as a complex number.

 

[Phys12]

How can you take the complex norm of a vector in a Hilbert space? A vector in a Hilbert space is not the same thing as a complex number.

Oh, yeah, you're right. So when we talk about taking the norm of a wavefunction, is that taking the norm of a vector in infinite vector space?

 

[Phys12]

However, AFAIK part of the definition of a Hilbert space is that the squared norm of any vector that is not the zero vector must be positive, and that a complex linear combination of any two vectors gives another vector in the same Hilbert space.

I see, that makes sense, thanks!

So it does seem like there's something missing from the problem statement if it's supposed to be non-trivial to prove.

I agree!

 

[PeterDonis]

in the example that the professor did, he took the complex norm

Not of the full Hilbert space vector. All he said was that $|\alpha|$ and $|\beta|$ are complex norms. He did not say that $|\Psi_1|$ and $|\Psi_2|$ are complex norms, or that $|\alpha \Psi_1 + \beta \Psi_2|$ was a complex norm. They aren't.

 

[PeterDonis]

when we talk about taking the norm of a wavefunction, is that taking the norm of a vector in infinite vector space?

It's taking the norm of a vector in a Hilbert space, which is a special kind of vector space. Whether the Hilbert space is finite dimensional or infinite dimensional depends on the specific problem. If we are just talking in general about any possible Hilbert space that could be used in QM, that includes both finite and infinite dimensional Hilbert spaces.

 

[Phys12]

It's taking the norm of a vector in a Hilbert space, which is a special kind of vector space. Whether the Hilbert space is finite dimensional or infinite dimensional depends on the specific problem. If we are just talking in general about any possible Hilbert space that could be used in QM, that includes both finite and infinite dimensional Hilbert spaces.

Right, then it seems like in the lecture, with the amount of information covered so far, we cannot prove that the norm of the superposition of two wavefunctions is always positive, is that right? And because we are not dealing with just complex numbers, the argument that $|z|^2 > 0$ for all $z$'s cannot be used to prove our original claim, correct?

 

[PeterDonis]

it seems like in the lecture, with the amount of information covered so far, we cannot prove that the norm of the superposition of two wavefunctions is always positive, is that right?

Unless the professor has already covered Hilbert space norms, no. But the facts that I gave in a previous post about Hilbert space vectors and norms are sufficient to prove that the norm of any wave function is positive.

because we are not dealing with just complex numbers, the argument that $|z|^2 > 0$ for all $z$'s cannot be used to prove our original claim, correct?

Yes, the properties of complex numbers alone are not enough, you also need the properties of Hilbert space vectors and norms.

 

[PeroK]

Well if you look at this video: , from 49:08 till 49:55, he says that you can check that the total quantity is zero (even though the last two terms may not be positive). I'm trying to do that check

He's got his himself into a real old tangle there! Let's do this the easy way. We have:
$$P(x) = |\alpha \Psi_1(x) + \beta \Psi_2(x)|^2$$
Where $z = \alpha \Psi_1(x) + \beta \Psi_2(x)$ is a complex number. Now $|z|$ and $|z|^2$ are, by definition, non-negative. There is no need to prove anything.

One thing to note, however, is that if we let $z = z_1 + z_2$, then:
$$|z|^2 = |z_1|^2 + |z_2|^2 + z_1z_2^* + z_1^* z_2 = |z_1|^2 + |z_2|^2 + 2Re(z_1z_2^*)$$
And it's not quite so obvious now that this is non-negative. But, the expression is just another way to write the modulus squared, so it must be.

 

[vanhees71]

I'm a bit puzzled what this is about. Of course $|\alpha \Psi_1(x)+\beta \Psi_2(x)|^2=P(x) \geq 0$, because the modulus is a (positive) real number $\geq 0$ and a real number squared is always $\geq 0$ too. So what is to be proven here? It's also not clear what this should have to do with a normalization condition.

Perhaps, you check first again about the precise formulation of the problem we discuss here?

 

[PeterDonis]

We have:
$$P(x) = |\alpha \Psi_1(x) + \beta \Psi_2(x)|^2$$
Where $z = \alpha \Psi_1(x) + \beta \Psi_2(x)$ is a complex number.

It can be considered a complex number once you have picked a particular value of $x$, yes. But considered as a function of $x$, it's a Hilbert space vector.

Also, the lecturer refers to the quantity $|\alpha \Psi_1(x) + \beta \Psi_2(x)|^2$ as a "squared norm" a little earlier. And the functions $\Psi_1(x)$ and $\Psi_2(x)$ themselves are Hilbert space vectors, whose norms are Hilbert space norms.

So I would say the lecturer is not doing a very good job of distinguishing the key concept of a Hilbert space vector from the different (and simpler) concept of a complex number. However...

There is no need to prove anything.

As I noted in an earlier post, this is still true even if we talk about Hilbert space norms instead of squared moduli of complex numbers, since the Hilbert space norm of any nonzero vector is positive. So the lecturer could have made the same point about positivity while still making clear the distinction between Hilbert space vectors and complex numbers.

 

[vanhees71]

I still do not know what you want to prove or what the question is, but the wave functions $\psi(x)=\langle x|\psi \rangle$ build a Hilbert space of square-integrable (integrable in the sense of the Lebesgue integral) functions, called $\mathrm{L}^2(\mathbb{R}$. For any $x$ of course $|\psi(x)|^2$ is just a complex number, and the scalar product of the Hilbert space is given by
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \psi_1^*(x) \psi_2(x).$$
This is a scalar product because, it's a positive definite sesquilinear form, where one assumes that two functions which only differ on a set of Lebesgue-measure 0 are taken as the same function.

The scalar product induces a norm on the Hilbert space,
$$\|\psi \|=\sqrt{\langle \psi|\psi \rangle}.$$
From the positive definiteness of the scalar product you can derive the Cauchy-Schwartz inequality,
[corrected in view of #19]
$$\|\psi_1 \| \|\psi_2 \| \geq |\langle \psi_1|\psi_2 \rangle|.$$
Is it this, what you want to prove?

 

[wrobel]

in the last formula there must be the inverse inequality?

 

[vanhees71]

Yes! Thanks for pointing out the typo. I've corrected it.