Wave Functions of Definite Momentum

[Jimmy87]

Hi,

Apologies if this questions is really easy but it is something quite subtle which is annoying me. In my book of quantum physics it gives a wave function of definite momentum:

ψ = Ae^ipx/ħ

It goes on to say that since there is a momentum 'p' in the exponential then the momentum is known exactly. Therefore the position must be completely unknown. It says the equation shows you this because if you take the absolute square of this wavefunction then you will get a constant i.e. the probability of finding the particle is the same everywhere. This I understand but what is annoying me is the 'x' in the exponential. It doesn't say what this stands for. Isn't x normally position? I thought we didn't know this? You can't enter a definite momentum and position into a wavefunction. What does the 'x' stand for if it is not position?

Thanks.

 

[BvU]

Here $x$ is the argument of the wave function. The probability density as a function of the x coordinate is $\Psi^* \Psi$ which is 1 -- meaning this wave function can not be normalized. Nor can the expectation value for the position, $\int \Psi^* x \Psi / \int \Psi^* \Psi$.

 

[NFuller]

It doesn't say what this stands for. Isn't x normally position?

Yes, $x$ is the position. The wave function is a function of position.

I thought we didn't know this? You can't enter a definite momentum and position into a wavefunction.

You are confusing what the wave function is and what the expectation value of position is. The wave function can be exactly defined as a function of position and momentum. The expectation value of the particle's position can also be exactly known but since the expectation value is a probability, the position itself is not exactly known.

The main issue with the wave function given is that it is not normalizable. This means that that the expectation value and hence the position is completely unknown.

 

[Jimmy87]

Yes, $x$ is the position. The wave function is a function of position.

You are confusing what the wave function is and what the expectation value of position is. The wave function can be exactly defined as a function of position and momentum. The expectation value of the particle's position can also be exactly known but since the expectation value is a probability, the position itself is not exactly known.

The main issue with the wave function given is that it is not normalizable. This means that that the expectation value and hence the position is completely unknown.

Thanks guys. So are you saying that 'x' is the position you are looking at the wavefunction whereas ψψ* is the probability of finding the particle at that 'x' which for a state of definite momentum is completely unknown. Is that right?

 

[Jimmy87]

Sorry, it does go on to say that this is not normalisable so you have to introduce 'L' as your boundary to make it normalisable so you get:

ψ = 1/√Le^ipx/ħ

 

[BvU]

whereas ψψ* is the probability

No. That's the probability density

The probability of finding the particle beween $x$ and $x+dx$ is $\ \ \psi^*\psi\, dx / \int\psi^*\psi$

The probability of fnding the particle somewhere has to be 1 -- in other words $\int \psi^* \psi \, dx = 1$ , a condition that can't be satisfied.

 

[BvU]

Sorry, it does go on to say that this is not normalisable so you have to introduce 'L' as your boundary to make it normalisable so you get:

ψ = 1/√Le^ipx/ħ

Ah, but for that one the momentum expectation value isn't exactly $p$ any more !

 

[Jimmy87]

Ah, but for that one the momentum expectation value isn't exactly $p$ any more !

I think this will explain it better:



This guy is the author of the textbook. At 8mins 40s he talks about the equation ψ = Ae^ipx/ħ and goes on to say at 9mins 10s that you can enter an exact momentum (which he does - 96). He then goes on to say, as you did, that you need to introduce a normalisable term which he gives as 1/√L where 'L' could be the length of the universe. My question is that if you insert an exact momentum of '96' as he does what is the 'x' value you enter representing? It can't be position as you can't know the position if you know the momentum is 96. Is it the position of the wavefunction you are interested in whereas the position I am referring to (which you don't know if you know the momentum is 96) is the integral of ψψ* over some region which must be a constant value if the momentum is known. Is that right?

 

[PeterDonis]

In my book of quantum physics

Which book?

It goes on to say that since there is a momentum 'p' in the exponential then the momentum is known exactly.

That's not quite correct. The reason the momentum is known exactly is that the state you wrote down is an eigenstate of momentum. Just having a single $p$ in the exponential is not the full reason for that.

Does your book explain what eigenstates are?

 

[bhobba]

He then goes on to say, as you did, that you need to introduce a normalisable term which he gives as 1/√L where 'L' could be the length of the universe.

That sounds rather sus to me.

Its not normalizeable - period - and the full answer lies in an advanced area of functional analysis called Rigged Hilbert Spaces,

At your level the way to think of it is as an approximation to some normalizeable function where the domain is very large but finite (its an approximation to what's called a test function which has all sorts of nice properties such as being continuously differentiable and only non-zero on some finite region - technically known as a function of compact support). It could be the length of the universe - but really any really large length will do.

To understand this better, but not yet the full Rigged Hilbert Space answer, you should study distribution theory:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Actually its so useful it's something all physicists and applied mathematicians in general should know.

Thanks
Bill

 

[bhobba]

It says the equation shows you this because if you take the absolute square of this wavefunction then you will get a constant

Take the absolute square - and see what happens - its a constant - yes - but over what region?

As I said the full solution involves some rather advanced math.

Thanks
Bill

 

[bhobba]

Ah, but for that one the momentum expectation value isn't exactly $p$ any more !

Of course - that's why eigenstates of momentum do not actually exist - they are a useful theoretical fiction. But for beginning students just think of it as an approximation to a function like that. Technically as a test function where the region of support is so large it for all practical purposes behaves like the eigenfunction,

Thanks
Bill

 

[BvU]

Ah, but for that one the momentum expectation value isn't exactly $p$ any more !

I used an unfortunate wording here: the expectation value is p allright, but it's no longer an infinitely sharp peak (a delta function). Instead, the p probability distribution has a sharp peak at that p and a width proportional to 1/L. If you are familiar with Fourier transforms: the wave function for x is a product of $e^{ipx\over \hbar}$ and a 'window' of width L. The momentum distribution is then a convolution of a delta function and a $\sin Lp\over Lp$ -- a peak with a width proportional to 1/L.

Shankar Principles of QM (Ch 4, page 138 in 2nd ed, 1994 ) comments on the improper vectors (also in ch 1, p. 67) -- exactly along the lines @bhobba indicates. To me (experimental physicist) he treats enough of the math for a physicist !

 

[bhobba]

To me (experimental physicist) he treats enough of the math for a physicist !

Just so the OP knows this stuff stumped a mathematician as great as Von-Neumann himself. It took other equally as great mathematicians like Grothendieck to even start on the right track to its solution which was completed later by Schwartz and Gelfland (plus probably others as well).

But intuitively its not hard and hinges on a very interesting fact about such pathological functions (called generalized functions). If f is such pathological function, then we can always find a sequence of functions with really nice properties, fn such that fn → f in what's called its weak topology without going into what that is - its not really important to this intuitive view. These fn, if you want to apply Fourier transforms, are known as what's called good functions rather than the functions of compact support I mentioned before - they all are infinitely diffferentiable and fall to zero very fast while not being zero - they are square inrtegrable. Just think of such functions as an fn where n is so large for all practical purposes its the same as the f.

Here is some further reading:
https://web2.ph.utexas.edu/~gleeson/ElectricityMagnetismAppendixB.pdf

I have read Lighthills book but think the one I gave previously is better - but is just a matter of taste.

Thsnks
Bill

 

[PeterDonis]

if you insert an exact momentum of '96' as he does what is the 'x' value you enter representing

Inserting an exact momentum of '96' does not require you to insert an exact position. All inserting an exact momentum does is pick out one particular function out of an infinite number of possible functions of the form $A e^{ipx / \hbar}$: namely, the function $A e^{i 96 x / \hbar}$ (i.e., the function you get when you set $p = 96$ in the general form). But what you have is still a function of $x$; you haven't filled in any particular value for $x$.

Of course that leads to the obvious next question: what do you get if you insert a particular value of $x$, say $x = 45$, into this function? What you get is a complex number called a "probability amplitude" for finding the particle at the particular position $x = 45$; the squared modulus of this complex number gives you the probability of finding the particle at that position. And you can see that, for the particular function in question (indeed, for any function of the same general form), the probability is independent of $x$: it's just $A^2$. That is what it means to say that the position is "completely uncertain" in this state: that there is an equal probability of finding the particle at any position.

What I've said so far actually glosses over a significant complication: that as it stands, the probabilities don't add up to one! You can see this because there are an infinite number of possible positions $x$, and each one has the same probability (whatever $A^2$ is), and so adding up all of the probabilities would give you an infinite answer. This is what all that stuff about the wave functions in question not being "normalizable" refers to. In order to get a function that describes actually realizable probabilities (i.e., so that the probabilities add up to one), you have to put in the normalization factor and so on. But that doesn't change the general points I made above.

 

[DrDu]

Ah, but for that one the momentum expectation value isn't exactly $p$ any more !

It still is if you assume that the wavefunction is periodic $\psi(x+L)=\psi(x)$. Then $\psi$ is still an eigenvalue of p, but the spectrum of p is no longer continuous but discrete. For L of the size of the universe this is not experimentally detectable.

 

[BvU]

Yes. The video continues along that line. One way or another: either you have the particle in a box of size L and get a momentum of width $\propto 1/L\ \$, or you have an single momentum and the particle everywhere.

 

[bhobba]

Yes. The video continues along that line. One way or another: either you have the particle in a box of size L and get a momentum of width $\propto 1/L\ \$, or you have an single momentum and the particle everywhere.

It one way out, but I would prefer simply tell the truth. You simply consider it as an approximation to some good function that has all the nice properties you would like. Beyond that you need first to study distribution theory, then Rigged Hilbert Spaces. Distribution theory is fine for undergrads - in fact the ANU here in Aus includes it as part of their second year curriculum for honors students, but RHS's are advanced and definitely postgraduate territory - still an interesting and surprisingly applicable these days (eg it's also used in White Noise Theory).

Thanks
Bill

 

[Jimmy87]

Inserting an exact momentum of '96' does not require you to insert an exact position. All inserting an exact momentum does is pick out one particular function out of an infinite number of possible functions of the form $A e^{ipx / \hbar}$: namely, the function $A e^{i 96 x / \hbar}$ (i.e., the function you get when you set $p = 96$ in the general form). But what you have is still a function of $x$; you haven't filled in any particular value for $x$.

Of course that leads to the obvious next question: what do you get if you insert a particular value of $x$, say $x = 45$, into this function? What you get is a complex number called a "probability amplitude" for finding the particle at the particular position $x = 45$; the squared modulus of this complex number gives you the probability of finding the particle at that position. And you can see that, for the particular function in question (indeed, for any function of the same general form), the probability is independent of $x$: it's just $A^2$. That is what it means to say that the position is "completely uncertain" in this state: that there is an equal probability of finding the particle at any position.

What I've said so far actually glosses over a significant complication: that as it stands, the probabilities don't add up to one! You can see this because there are an infinite number of possible positions $x$, and each one has the same probability (whatever $A^2$ is), and so adding up all of the probabilities would give you an infinite answer. This is what all that stuff about the wave functions in question not being "normalizable" refers to. In order to get a function that describes actually realizable probabilities (i.e., so that the probabilities add up to one), you have to put in the normalization factor and so on. But that doesn't change the general points I made above.

Perfect, thanks! This is just the answer I was looking for. Also thanks to such great answers from others and all the info you gave me for reading around this. So just to clarify what you said. If you insert an 'x' like '45' and a 'p' of '96' does this mean that at a position of '45' the momentum is '96'? Whereas if you wanted to calculate the probability of finding the particle at x=45 you would have to take the absolute square of the wave function which must be a constant because the absolute square of any complex exponential is always equal to 1? Is that right?

Going back to using p =96 and x =45 - does entering all these values and calculating the wave function as a number give you anything meaningful? For example if you normalised the wave function and therefore got a value for 'A' you would essentially have:

ψ = (some number)e^i(96)(45)/ħ

If you plugged this all into a calculator (like a standard calculation) since you know all the values does this give you anything meaningful? I'm probably guessing not since meaningful data seems to be gotten by doing particular operations on it such as taking the absolute square for the position probability. Also since there is an imaginary number 'i' could you even plug the above expression in and get an answer.

 

[PeterDonis]

just to clarify what you said. If you insert an 'x' like '45' and a 'p' of '96' does this mean that at a position of '45' the momentum is '96'?

No. If we insert $x = 45$ and $p = 96$, we get the complex number $A e^{i 96 * 45 / \hbar}$. This complex number is the value of the wave function for the given values of $x$ and $p$. What does the complex value of the wave function represent?

 

[Jimmy87]

No. If we insert $x = 45$ and $p = 96$, we get the complex number $A e^{i 96 * 45 / \hbar}$. This complex number is the value of the wave function for the given values of $x$ and $p$. What does the complex value of the wave function represent?

Ah right ok. So do you actually plug all this into a calculator and you get a number? If 'A' was '10' can you treat it as a normal equation or does it not work that way. I'm not sure what the complex value of the wave function represents. I know any complex exponential is made up of two trig functions - cosx and isinx.

 

[PeterDonis]

So do you actually plug all this into a calculator and you get a number?

If your calculator can handle complex numbers, sure, you can do that (provided, as you say, that you have a numerical value for A).

I'm not sure what the complex value of the wave function represents.

Then the first thing you should do is go back to the book on quantum physics you are reading and see what it tells you about that. That is one of the most basic ideas of QM, and if you don't understand that, you won't be able to understand anything else.

 

[Jimmy87]

If your calculator can handle complex numbers, sure, you can do that (provided, as you say, that you have a numerical value for A).
Then the first thing you should do is go back to the book on quantum physics you are reading and see what it tells you about that. That is one of the most basic ideas of QM, and if you don't understand that, you won't be able to understand anything else.

Right ok. So if your calculator can handle complex numbers will it literally spit out a number? What units would it have? Or is it dimensionless? I would ask you what this number represents but that is what you asked me to look up - I will do this!

 

[Jimmy87]

If your calculator can handle complex numbers, sure, you can do that (provided, as you say, that you have a numerical value for A).
Then the first thing you should do is go back to the book on quantum physics you are reading and see what it tells you about that. That is one of the most basic ideas of QM, and if you don't understand that, you won't be able to understand anything else.

Ok so I found:

"The wave function is a complex-valued probability amplitude, and the probabilities for the possible results of measurements made on the system can be derived from it. Since the wave function is complex valued, only its relative phase and relative magnitude can be measured—its value does not, in isolation, tell anything about the magnitudes or directions of measurable observables; one has to apply quantum operators, whose eigenvalues correspond to sets of possible results of measurements, to the wave function ψ and calculate the statistical distributions for measurable quantities".

What am I missing as this seems to imply the value of the wave function tells you nothing. However, it does say:

"At a particular instant of time, all values of the wave function Ψ(x, t) are components of a vector."

Is that what you meant? I don't know what that means though :(

Not in my textbook though as its too big to find something so specific :(

 

[bhobba]

What am I missing as this seems to imply the value of the wave function tells you nothing.

To find what it tells you look into the Born Rule:
http://www.math.ru.nl/~landsman/Born.pdf

BTW there is a derivation that is not doubted called Gleason's Theorem:
https://en.wikipedia.org/wiki/Gleason's_theorem

Its very important for the foundations of QM because a careful analysis shows its basis is what's called non-contextuality. This leads to the important Kochen-Specker theorem:
https://en.wikipedia.org/wiki/Kochen–Specker_theorem

You can prove the theorem directly, or as a simple corollary to Gleason.

Thanks
Bill

 

[PeterDonis]

So if your calculator can handle complex numbers will it literally spit out a number?

What do you think? Have you tried it? Note that googling on "complex number calculator" will turn up a number of websites that implement such calculators, if for some reason you don't have one available on your own computer.

What am I missing as this seems to imply the value of the wave function tells you nothing.

Why would you think that? The passage you quoted explicitly tells you "the probabilities for the possible results of measurements made on the system can be derived" from the wave function. Doesn't that make it obvious that the wave function contains useful information?

 

[Jimmy87]

What do you think? Have you tried it? Note that googling on "complex number calculator" will turn up a number of websites that implement such calculators, if for some reason you don't have one available on your own computer.
Why would you think that? The passage you quoted explicitly tells you "the probabilities for the possible results of measurements made on the system can be derived" from the wave function. Doesn't that make it obvious that the wave function contains useful information?

Yes of course it contains useful information because in my earlier posts I said I know you can do various operations to find energy, momentum, position etc but we have been talking about the value of the wavefunction itself for the last few threads. The quote I posted in my last thread says:

"its value does not, in isolation, tell anything about the magnitudes or directions of measurable observables; one has to apply quantum operators"

To me (and I appreciate I am new to QM) this is saying that the value itself (plugging it into a calculator) does not tell you anything meaningful (no measurable observables). If it doesn't tell you about a measurable observable I don't see what use the value itself can be?

 

[PeterDonis]

Yes of course it contains useful information

If it doesn't tell you about a measurable observable I don't see what use the value itself can be?

Now you're contradicting yourself. You obviously recognize that the wave function value contains useful information. That means it must be of some use. Perhaps making use of it requires some other information, like which observable you want to measure; so what?

You appear to be refusing to see the answer to your question that is right in front of you, and that you yourself have already stated.

 

[Jimmy87]

Now you're contradicting yourself. You obviously recognize that the wave function value contains useful information. That means it must be of some use. Perhaps making use of it requires some other information, like which observable you want to measure; so what?

You appear to be refusing to see the answer to your question that is right in front of you, and that you yourself have already stated.

No - I'm talking about the wave function value - the one you said I need to know the answer to - "the complex value of the wavefunction" which I still don't know yet. I totally get that the wavefunction is useful because you can apply energy, momentum, position operators etc on it to get really useful information. However, we have been talking about plugging Ae^ipx/h into a calculator (that can handle complex numbers) and the stuff I quoted in post #24 seems to say that "the wavefunctions value does not, in isolation, tell anything about the magnitudes or directions of measurable observables". Whereas you said "the complex value of the wavefunction is one of the most basic ideas of QM, and if you don't understand that, you won't be able to understand anything else." I still can't find an answer to this. Well, only the one that seems to say it doesn't tell you anything. So I am just confused now.

 

[Jimmy87]

Now you're contradicting yourself. You obviously recognize that the wave function value contains useful information. That means it must be of some use. Perhaps making use of it requires some other information, like which observable you want to measure; so what?

You appear to be refusing to see the answer to your question that is right in front of you, and that you yourself have already stated.

No - I'm talking about the wave function value - the one you said I need to know the answer to - "the complex value of the wavefunction" which I still don't know yet. I totally get that the wavefunction is useful because you can apply energy, momentum, position operators etc on it to get really useful information. However, we have been talking about plugging Ae^ipx/h into a calculator (that can handle complex numbers) and the stuff I quoted in post #24 seems to say that "the wavefunctions value does not, in isolation, tell anything about the magnitudes or directions of measurable observables". Whereas you said "the complex value of the wavefunction is one of the most basic ideas of QM, and if you don't understand that, you won't be able to understand anything else." I still can't find an answer to this. Well, only the one that seems to say it doesn't tell you anything. So I am just confused now.

 

[Jimmy87]

No idea what happened there - sorry!

 

[PeterDonis]

I'm talking about the wave function value - the one you said I need to know the answer to - "the complex value of the wavefunction" which I still don't know yet.

Once again: the answer is right in front of you.

I totally get that the wavefunction is useful because you can apply energy, momentum, position operators etc on it to get really useful information.

And in order to do that, you need to know the (complex) value of the wave function for all possible values of $x$ and $p$. That is the answer to your question.

you said "the complex value of the wavefunction is one of the most basic ideas of QM, and if you don't understand that, you won't be able to understand anything else." I still can't find an answer to this.

Because knowing the complex value of the wave function, as a function of its input variables, is necessary in order to do anything else--because without it, you can't apply any operators to get any predictions for measurements.

 

[Jimmy87]

Once again: the answer is right in front of you.
And in order to do that, you need to know the (complex) value of the wave function for all possible values of $x$ and $p$. That is the answer to your question.
Because knowing the complex value of the wave function, as a function of its input variables, is necessary in order to do anything else--because without it, you can't apply any operators to get any predictions for measurements.

Oh I see now (I think). So when you plot a graph of Ψ against x and Ψ against p, are the values plotted on the graph the complex value of the wave function by doing separate calculations for different values of x and p? Or at least the real part anyway.

 

[PeterDonis]

when you plot a graph of Ψ against x and Ψ against p, are the values plotted on the graph the complex value of the wave function by doing separate calculations for different values of x and p?

Basically, yes, but with some clarifications. If you plot $\Psi$ as a function of $x$, you are plotting the values you get when you evaluate the function $\Psi(x)$ for each different value of $x$. Which function $\Psi(x)$ that is will depend on what value you pick for $p$--or, more generally, what kind of wave function you want to work with (functions of the form $Ae^{ipx}$--leaving out the factor $\hbar$ for simplicity--are not the only possible kinds of wave functions).

Similarly, if you plot $\Psi$ as a function of $p$, you are plotting the values you get when you evaluate the function $\Psi(p)$ for each different value of $p$. Which function $\Psi(p)$ that is will depend on what kind of wave function you want to work with.

The relationship between the functions $\Psi(x)$ and $\Psi(p)$ is that they are Fourier transforms of each other. In the case we are working with, where the function $\Psi(x)$ is $A e^{ipx}$, the corresponding function $\Psi(p)$, obtained by Fourier transforming $\Psi(x)$, is $\delta(p)$, where $\delta$ stands for the Dirac delta function. (Which is not actually a "function", strictly speaking--but for our purposes here we can think of it as one, which is zero for any value of momentum except the single value $p$.)

So if you draw a graph of your $\Psi(x)$, for some particular value of $p$, you will find that it looks like a "corkscrew" winding around the $x$ axis (think of stacking an infinite series of complex planes, one for each value of $x$, and plotting the complex value of $\Psi(x)$ for each value of $x$ in the plane corresponding to that value of $x$). The particular value of $p$ that you pick determines how tightly the corkscrew winds--the higher the value of $p$, the tighter the windings (i.e., the more closely spaced they are).

And if you draw a graph of the $\Psi(p)$ that corresponds to the above $\Psi(x)$, you will find that it is a single "spike" at a particular value of $p$ on the $p$ axis--everywhere else it is zero.

 

[vanhees71]

I don't know, whether you have ever told the OP that a pure quantum state which, up to a phase, is given by a normalizable vector in Hilbert space, $|\psi \rangle$ can be given in position and momentum representation. The eigenstates of position and momentum are in fact not true Hilbert-space vectors but they live in a larger space, the dual of the dense subspace where the position and momentum operators are defined, which are thus only "normalizable to a $\delta$ distribution",
$$\langle p'|p \rangle=\delta(p-p'), \quad \langle x'|x \rangle=\delta(x-x').$$
Further you can show that the generalized momentum eigenstate is given in position representation by
$$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$
I've used "natural units" such that $\hbar=1$, which makes things a bit easier.

Now the most important key to change from one basis, or equivalently from one representation, to another is the completeness relation for generalized position and momentum eigenstates,
$$\int_{\mathbb{R}} \mathrm{d} p |p \rangle \langle p|=\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|=\mathbb{1}.$$
This implies for the position wave function
$$\psi(x) = \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(p),$$
where $\tilde{\psi}(p)$ is the wave function in momentum representation (representing up to a phase factor still the same state as the position wave function!). As you see, you get the position wave function from the momentum wave function by Fourier integration. Of course, you also immediately get the inverse transformation by the same trick of "inserting a unit operator":
$$\tilde{\psi}(p)=\langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x).$$
Here I have used that the generalized position eigenfunction in momentum representation is given by
$$u_x(p) = \langle p|x \rangle = \langle x | p \rangle^*=u_p^*(x)=\frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x).$$