I Collapse of wavefunction into a forbidden eigenstate for a free particle

[Kashmir]

For the free particle in QM, the energy and momentum eigenstates are not physically realizable since they are not square integrable. So in that sense a particle cannot have a definite energy or momentum.

What happens during measurement of say momentum or energy ?

So we measure some definite value of momentum or energy which is an eigenvalue of the momentum or Hamiltonian (since the operators commute for a free particle). Then we would in principle collapse the wave function to some eigenstate, but in this case we know that this is not possible (physically realizable).

So what happens during measurement and in what state is the particle after we've measured it's momentum?

Thank you

 

[vanhees71]

It's clear that for a free particle momentum or energy eigenstates cannot be prepared as the corresponding wave functions are not square integrable but belong to the dual of the domain, where position and momentum operators are defined, i.e., they are generalized functions or distributions.

One complete set are of course the momentum eigenstates, i.e., the plane waves,
$$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi \hbar)^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}/\hbar),$$
and they are also the energy eigenstates with the corresponding eigenvalues $E_{\vec{p}}=\vec{p}^2/(2m)$. They are of course not square-integrable but only normalizable to a $\delta$ distribution,
$$\langle \vec{p}|\vec{p}' \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x u_{\vec{p}}^*(\vec{x}) u_{\vec{p}'}(\vec{x})=\delta^{(3)}(\vec{p}-\vec{p}).$$
Of course you can build all square-integrable wave functions in terms of these "eigenstates",
$$\psi(t,\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 p u_{\vec{p}}(\vec{x}) \tilde{\psi}_0(\vec{p}) \exp(-\mathrm{i} E_{\vec{p}} t/\hbar).$$
Here $\tilde{\psi}_0(\vec{p})$ is the wave function in momentum representation at time $t=0$, representing the state in which the particle is prepared at this "initial time", and it must be square integrable.

All these wave functions represent proper physical states and obey the Heisenberg uncertainty relations, $\Delta x_j \Delta p_j \geq \hbar/2$ for $j \in \{1,2,3 \}$, i.e., if you prepare the particle with a pretty well determined momentum in $j$-direction, it's localization in this direction is pretty uncertain and vice versa.

 

[Nugatory]

So what happens during measurement

We don’t know, because quantum mechanics is a theory about measurement results, not measurements.

and in what state is the particle after we've measured it's momentum?

It will be something like a very narrow Gaussian with a peak at the measured value; the more precise our measurement the narrower the Gaussian and the steeper the peak.

 

[Kashmir]

We don’t know, because quantum mechanics is a theory about measurement results, not measurements.It will be something like a very narrow Gaussian with a peak at the measured value; the more precise our measurement the narrower the Gaussian and the steeper the peak.

And what if I measure a single value of momentum what will the wave function in momentum space look like? Or is it not possible to get a single value of momentum?

 

[PeroK]

And what if I measure a single value of momentum what will the wave function in momentum space look like? Or is it not possible to get a single value of momentum?

Yes, it's possible to get a single value. But, what your measurement equipment displays is not the issue. The issue is what your measurement equipment does, in terms of interacting with the particle. This was the basis of the so-called Einstein-Bohr debates, where they discussed how and why the HUP (Uncertainty Principle) was upheld in any experimental setup. This question, strictly speaking, is not the HUP, but is related.

In order to understand measurements of microscopic quantum systems, you must look closely at the precise details of how your apparatus interacts with the particle or system being measured. The question then becomes how you establish that, say, the particle had precisely zero momentum immediately after you measured it?

 

[Kashmir]

Yes, it's possible to get a single value. But, what your measurement equipment displays is not the issue. The issue is what your measurement equipment does, in terms of interacting with the particle. This was the basis of the so-called Einstein-Bohr debates, where they discussed how and why the HUP (Uncertainty Principle) was upheld in any experimental setup. This question, strictly speaking, is not the HUP, but is related.

In order to understand measurements of microscopic quantum systems, you must look closely at the precise details of how your apparatus interacts with the particle or system being measured. The question then becomes how you establish that, say, the particle had precisely zero momentum immediately after you measured it?

if I'm getting just one precise value of momentum, what would be the wavefunction of the state in momentum space?

 

[PeroK]

if I'm getting just one precise value of momentum,

Until you say how you obtained that value of momentum, you can say nothing about the particle. At the very least, you must expect your apparatus to have an experimental error. That in itself (although technically unrelated to QM uncertainty) would undermine your claim to have measured the momentum precisely.

Moreover, from a mathematical point of view, a typical real number requires an infinite amount of information to describe it. A digital display can only display one of a finite number of values.

Before you even get to QM issue (that momentum eigenstates are unphysical), you already have two insurmountable technical problems: one experimental and one mathematical!

 

[Kashmir]

Until you say how you obtained that value of momentum, you can say nothing about the particle. At the very least, you must expect your apparatus to have an experimental error. That in itself (although technically unrelated to QM uncertainty) would undermine your claim to have measured the momentum precisely.

Moreover, from a mathematical point of view, a typical real number requires an infinite amount of information to describe it. A digital display can only display one of a finite number of values.

Before you even get to QM issue (that momentum eigenstates are unphysical), you already have two insurmountable technical problems: one experimental and one mathematical!

I didn't know those problems existed. Thanks for helping me realize them.

Can't we just for the arguments sake assume that we've somehow known the exact value of momentum and try to find the wavefunction. Or is your opinion that it's no use to think that way?

 

[PeroK]

Can't we just for the arguments sake assume that we've somehow known the exact value of momentum and try to find the wavefunction. Or is your opinion that it's no use to think that way?

What QM is telling you is that a point particle is not part of the theory. And that particles are actually wave-packets: with a range of energy/momentum. That's what QM theoretically is telling you.

To imagine a pure point particle that has a definite, precise position in space is going against what QM is saying.

My point is that even before QM tells you this, the idea of experimentally establishing a precise set of real numbers $(x, y, z)$ that describe the position of a particle exactly at some precise time $t$ is not feasible. Even if the theory did not forbid point particles, it's not possible to measure them precisely.

The important thing is that particles are described by wave-packets, which have a range of energy/momentum, and an uncertainty in position that together obey the HUP. That's QM.

 

[Nugatory]

And what if I measure a single value of momentum what will the wave function in momentum space look like? Or is it not possible to get a single value of momentum?

The momentum measurement works pretty much the same way: The wave function written in momentum space ends up a sharply peaked Gaussian. The position space reresentation and momentum space representations are Fourier transformations of one another, so the more peaked the wave function in one representation (that is, the greater the precision with which the observable has been measured) the flatter the other representation is.

 

[Nugatory]

Can't we just for the arguments sake assume that we've somehow known the exact value of momentum and try to find the wavefunction. Or is your opinion that it's no use to think that way?

To say that we know the exact value of the momentum is equivalent to saying that the wave function in momentum space is a delta function. If it were, then the wave function in position space would be a flat line (equal probability of the particle being found anywhere in the universe), a state that is clearly unphysical and also not square-integral.

So no, we can't make that assumption. Neither the position nor the momentum eigenstates are physically realizable.

 

[andresB]

So we measure some definite value of momentum or energy which is an eigenvalue of the momentum or Hamiltonian (since the operators commute for a free particle).

Hold on there. No measurement will ever give you a definite value of position or momentum, there is always an spread of the value.

In principle, In non-relativistic QM, you can make the uncertainty of the position (or momentum, but not both at the same time) of an electron smaller than any given positive number, but that doesn't mean you can actually make it zero.

 

[Kashmir]

Hold on there. No measurement will ever give you a definite value of position or momentum, there is always an spread of the value.

In principle, In non-relativistic QM, you can make the uncertainty of the position (or momentum, but not both at the same time) of an electron smaller than any given positive number, but that doesn't mean you can actually make it zero.

apart from experimental limitations does the theory restrict a precise measurement?

 

[vanhees71]

And what if I measure a single value of momentum what will the wave function in momentum space look like? Or is it not possible to get a single value of momentum?

You get a value with some uncertainty. To describe the particle, if you don't know any other details than that you have measured a momentum with some standard deviation the most objective state you can choose is a mixed state, described by a statistical operator
$$\hat{\rho}=\frac{1}{Z} \exp \left [-\frac{(\hat{\vec{p}}-\vec{p}_0)^2}{2 \Delta p^2} \right]$$
with
$$Z=\mathrm{Tr} \exp \left [-\frac{(\hat{\vec{p}}-\vec{p}_0)^2}{2 \Delta p^2} \right].$$
This association of a statistical description with the given information is following the rule of information theory, according to which one should choose the description of the minimal prejudice in the sense that one should maximize the "missing information measure", which is the von Neuman entropy,
$$S=-\mathrm{Tr} (\hat{\rho} \ln \hat{\rho})$$
under the given constraints, which are here the expectation value and standard deviation of the momentum.

 

[andresB]

apart from experimental limitations does the theory restrict a precise measurement?

Yes. As Nurgatory pointed out, a completely precise knowledge of the eigenvalue of the momentum requires the wavefunction to be non-square integrable, and QM says that you can't have that.

More physically, the more precise you want to measure the position of an electron the more energy you have to put into the measurement. In the limite of complete determination of the position you would require infinite energy. I saw an article here in the forums with the reasoning, but I don't remember the title of it.

 

[vanhees71]

apart from experimental limitations does the theory restrict a precise measurement?

No it doesn't. The uncertainty relation is no restriction in the ability to measure something precisely but to prepare the system in a state, where this observable takes a certain value. In other words, it is impossible to localize a particle precisely at one point. You can only restrict it to some finite, though in principle arbitrarily small, region in space. The same holds for momentum, but if you make one of these observables very determined the other necessarily becomes less determined.

 

[Nugatory]

apart from experimental limitations does the theory restrict a precise measurement?

The theory does not restrict the precision of any single measurement - we can have has many significant digits in our measurement result as our measuring device is good for. In mathematical terms, the Gaussian wave function after measurement can made arbitrarily narrow and steep. However it's still a Gaussian and not the physically unrealizable delta function so if we repeat the measurement there is some probability that we will get a different result.

 

[PeterDonis]

No it doesn't.

Yes, it does. It says that you cannot have a delta function as a result of a measurement; you can only have a Gaussian with some finite width. That is a restriction.

 

[vanhees71]

That's the usual misleading interpretation of the "collapse". A measurement does not necessarily imply the preparation of the measured system in the measured eigenstate. Indeed, you cannot prepare a particle in a state described by a plane wave, because this is not a proper state to begin with. However, this does not imply that you can't measure position or momentum with arbitrary precision.

 

[Nugatory]

No it doesn't.

Yes, it does. It says that you cannot have a delta function as a result of a measurement; you can only have a Gaussian with some finite width. That is a restriction.

You are talking past one another, I think. Vanhees is saying that there is no theoretical limit to the precision of a measurement, which is to say the non-zero width of the Gaussian can be arbitrarily small - but still non-zero.

 

[vanhees71]

And what I also want to say is that a measurement NOT necessarily leads to the preparation of a particle in an eigenstate of the measured observable.

The quantum state describes a preparation procedure. The fact that there are no proper eigenvectors of position and momentum thus implies that you cannot prepare a particle in a state with precise position or precise momentum. This is what's described by the position-momentum uncertainty relation $\Delta x_j \Delta p_j \geq \hbar/2$.

It does not limit in any fundamental way, who precisely you can measure position or momentum of a particle.

 

[Kashmir]

And what I also want to say is that a measurement NOT necessarily leads to the preparation of a particle in an eigenstate of the measured observable, but I think we drift apart into the realm of interpretation...

but my book did say that the collapse is into an eigenstate.

 

[vanhees71]

I know that many books include the collapse postulate, but it is at least very problematic, and the books do not discuss the problems in great detail.

Which textbook are you reading?

 

[Kashmir]

I know that many books include the collapse postulate, but it is at least very problematic, and the books do not discuss the problems in great detail.

Which textbook are you reading?

Mcintyre. Quantum mechanics.

He says "
After a measurement of $A$ that yields the result $a_n$, the quantum system is in a new state that is the normalized projection of the original system ket onto the ket (or kets) corresponding to the result of the measurement:

$\left|\psi^\prime\right> = \frac{P_n\left|\psi\right>}{\sqrt{\left<\psi\right|P_n\left|\psi\right>}}$"

 

[PeterDonis]

my book did say that the collapse is into an eigenstate.

Does your book also say that eigenstates of position and momentum are not normalizable and hence not physically realizable?

If so, then taken literally, your book is saying two contradictory things. Do you think it actually means both of those things literally?

 

[Kashmir]

To say that we know the exact value of the momentum is equivalent to saying that the wave function in momentum space is a delta function.

This implies that a delta function wavefunction in momentum space would give only one value on measurement. So the probability of measuring that particular eigenvalue should be $1$.

If $P$ is the probability density then $P d p=|\varphi(p)|^{2} d p$ will give me the probability of getting a momentum value in a range $dp$ around $p$

If Dirac delta function represents a state that results in only one measured value $p$, we should then have $P d p=|\varphi(p)|^{2} d p =1$ but $\varphi(p)=\delta\left(p-p^{\prime}\right)$ so $\delta^{2}(p-p) d p \neq 1$

So how does a Dirac delta function correspond to a single measurement of $p$ ?

@Nugatory ,@vanhees71 ,@PeroK , can you help me here?

 

[Kashmir]

Does your book also say that eigenstates of position and momentum are not normalizable and hence not physically realizable?

If so, then taken literally, your book is saying two contradictory things. Do you think it actually means both of those things literally?

For me the only way out of the difficulty without violating both of the statements is if we can't know momentum exactly ,even in theory.

If that's true then the wavefunction cannot collapse into an eigenstate because we didn't find a single value of momentum, the collapse will be into a superposition state.

 

[Nugatory]

but my book did say that the collapse is into an eigenstate.

That’s a simplification used by many intro textbooks to avoid some unpleasant mathematical complications associated with continuous spectrum observables. Unfortunately it leads to the contraction that you’re seeing. More advanced texts will straighten out this out with a more mathematically rigorous treatment.

Edit: and I see that @PeroK has already mentioned Sakurai as one that does.

 

[PeroK]

Mcintyre. Quantum mechanics.

He says "
After a measurement of $A$ that yields the result $a_n$, the quantum system is in a new state that is the normalized projection of the original system ket onto the ket (or kets) corresponding to the result of the measurement:

$\left|\psi^\prime\right> = \frac{P_n\left|\psi\right>}{\sqrt{\left<\psi\right|P_n\left|\psi\right>}}$"

From a mathematical point of view, that's explicitly for an operator/measurable with a countable spectrum and, therefore, excludes position and momentum, which have an uncountable spectrum. Sakurai, for example, definitely discusses the two cases separately.

I'd bet McIntyre does the same thing.

 

[PeterDonis]

For me the only way out of the difficulty without violating both of the statements is if we can't know momentum exactly ,even in theory.

If that's true then the wavefunction cannot collapse into an eigenstate because we didn't find a single value of momentum, the collapse will be into a superposition state.

In the sense that a Gaussian in momentum space is a "superposition" of different values of momentum, yes, this is true.

 

[vanhees71]

I'm not very familiar with this book, but I think to remember that he starts with a thorough discussion of the Stern-Gerlach experiment, and that's of course one of the paradigmatic examples, where socalled "von Neumann filter measurements" can, in principle, be performed. Here you measure the spin of a silver atom, which has (as any angular momentum) only discrete eigenvalues, i.e., in this case the eigenstates of the measured observable's representing self-adjoint operator are indeed true states, in which the particle can be prepared, and (in principle) the preparation is very simple: You let the particle go through the inhomogeneous magnetic field, and then you can just filter out particle going in a region, where the particles go with the "unwanted" value of the spin component. E.g., directing the magnetic field in the standard $z$ direction and you want just particles with $s_z=+\hbar/2$ you just use the particles being deflected in the corresponding direction and bump the other particles (which have $s_z=-\hbar/2$) to a wall. That's indeed a "filter measurement" or, more precisely, a preparation through filtering following a measurement.

For a position measurement you cannot argue in this way, as you have observed yourself with your observation detailed in your posting, starting this thread.

One should also be clear that the state of a particle describes a preparation procedure rather than a measurement. So it's rather a question, how to "localize" a particle, i.e., how to put it at some place. One way are traps like the Penning trap, where one uses a static electric quadrupole and a homogeneous static magnetic field to keep particles within the trap, but of course the position of the particle is not exactly fixed but only confined within some finite volume, and the uncertainty relation between position and momentum is always fulfilled.

 

[Kashmir]

This implies that a delta function wavefunction in momentum space would give only one value on measurement. So the probability of measuring that particular eigenvalue should be $1$.

If $P$ is the probability density then $P d p=|\varphi(p)|^{2} d p$ will give me the probability of getting a momentum value in a range $dp$ around $p$

If Dirac delta function represents a state that results in only one measured value $p$, we should then have $P d p=|\varphi(p)|^{2} d p =1$ but $\varphi(p)=\delta\left(p-p^{\prime}\right)$ so $\delta^{2}(p-p) d p \neq 1$

So how does a Dirac delta function correspond to a single measurement of $p$ ?

@Nugatory ,@vanhees71 ,@PeroK , can you help me here?

I've Understood a lot. Just some help here too.

 

[PeterDonis]

how does a Dirac delta function correspond to a single measurement of $p$ ?

Measurements themselves are represented in QM by operators, not wave functions. The momentum operator is $i \hbar \nabla$; its eigestates, which represent possible results of the measurements, are delta functions. Because delta functions are not normalizable, they do not describe physically realizable states, and therefore the momentum operator $i \hbar \nabla$ does not describe a physically realizable measurement. It is a mathematical idealization that is useful for theoretical modeling but cannot be exactly realized in practice.

The actual physically realizable measurement that we describe in practice as "measuring momentum" would therefore not be represented, if we were to insist on being exact, by the operator $i \hbar \nabla$ but by some other operators whose eigenstates represent finite ranges of results. I don't know if anyone has tried to actually write down such an operator explicitly; mathematically it is easier to just use Gaussians in momentum space to represent measurement results and $i \hbar \nabla$ to represent the measurement operator and accept that we are working with a useful approximation.

 

[vanhees71]

This implies that a delta function wavefunction in momentum space would give only one value on measurement. So the probability of measuring that particular eigenvalue should be $1$.

If $P$ is the probability density then $P d p=|\varphi(p)|^{2} d p$ will give me the probability of getting a momentum value in a range $dp$ around $p$

If Dirac delta function represents a state that results in only one measured value $p$, we should then have $P d p=|\varphi(p)|^{2} d p =1$ but $\varphi(p)=\delta\left(p-p^{\prime}\right)$ so $\delta^{2}(p-p) d p \neq 1$

So how does a Dirac delta function correspond to a single measurement of $p$ ?

@Nugatory ,@vanhees71 ,@PeroK , can you help me here?

It does not correspond to a single measurement of $p$ nor does it represent a state of a particle. The (pure) states of a particle can only be represented by a wave function that is square integrable, i.e., which can be normalized to 1:
$$\int_{\mathbb{R}} \mathrm{d} p |\varphi(p)|^2=1.$$
There is no proper square-integrable eigenfunction of the momentum operator $\hat{p}$, but it has only a continuous spectrum $p \in \mathbb{R}$. The corresponding "eigenfunctions" of such "generalized eigenvalues" are "generalized functions" or "distributions". They are not square-integrable.

That's easy to see in the position representation, where $\hat{p}=-\mathrm{i} \hbar \partial_x$. The eigenvalue equation reads
$$-\mathrm{i} \hbar \partial_x u_p(x)=p u_p(x).$$
The solutions are
$$u_p(x)=N \exp(\mathrm{i} p x/\hbar).$$
That's for sure not square integrable, because $|u_p(x)|^2=|N|^2=\text{const}$. Nevertheless all $p \in \mathbb{R}$ are "generalized eigenvalues" of the momentum operator and the eigenfunctions can be "normalized" to a Dirac-$\delta$ distribution,
$$\langle u_p|u_{p'} \rangle=\int_{\mathbb{R}} u_p^*(x) u_{p'}(x)=\delta(p-p').$$
As is known from the theory of Fourier transformations for that we simply have to set the normalization constant to
$$N=\frac{1}{\sqrt{2 \pi \hbar}} \; \Rightarrow \; u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
They build a complete set of orthonormal functions in the sense that you can represent a dense subset of square integrable functions as a Fourier transform,
$$\psi(x)=\langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar) \tilde{\psi}(p)$$
and in the other way
$$\tilde{\psi}(p) = \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} p x/\hbar) \psi(x).$$
It's also clear that the transformation from the "position-space wave function" to the "momentum-space wave function" is a unitary transformation since
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(x) \psi_2(x) = \int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}_1^*(p) \tilde{\psi}_2(p).$$

 

[andresB]

Mcintyre. Quantum mechanics.

He says "
After a measurement of $A$ that yields the result $a_n$, the quantum system is in a new state that is the normalized projection of the original system ket onto the ket (or kets) corresponding to the result of the measurement:

$\left|\psi^\prime\right> = \frac{P_n\left|\psi\right>}{\sqrt{\left<\psi\right|P_n\left|\psi\right>}}$"

Ignoring normalization, notice that the projector you should use in this continuous spectrum case is
$P=\int_{x}^{x+\delta x}\left|x'\right\rangle \left\langle x'\right|dx'$
where $\delta x$ can be as small as you want, but not zero in order to avoid Dirac deltas.

We can write any normalizable state as

$\left|\psi\right\rangle =\int_{-\infty}^{\infty}\psi(x'')\left|x''\right\rangle dx''$

where $\psi(x'')$ is an square integrable function. So, using the prescription you give

$\left|\psi_{post}\right\rangle \sim P\left|\psi\right\rangle =\int_{-\infty}^{\infty}\int_{x}^{x+\delta x}\psi(x'')\left|x'\right\rangle \left\langle x'\right|\left.x''\right\rangle dx'dx''=\int_{-\infty}^{\infty}\int_{x}^{x+\delta x}\psi(x'')\left|x'\right\rangle \delta(x'-x'')dx'dx'' =\int_{x}^{x+\delta x}\psi(x'')\left|x'\right\rangle dx'$So, the final wave function is restricted to an interval $(x,x+\delta x)$. Again, $\delta x$ can be as small as you want but not zero. So the final wavefunction will not be localized at exactly one point.

 

[Kashmir]

It does not correspond to a single measurement of $p$ nor does it represent a state of a particle. The (pure) states of a particle can only be represented by a wave function that is square integrable, i.e., which can be normalized to 1:
$$\int_{\mathbb{R}} \mathrm{d} p |\varphi(p)|^2=1.$$
There is no proper square-integrable eigenfunction of the momentum operator $\hat{p}$, but it has only a continuous spectrum $p \in \mathbb{R}$. The corresponding "eigenfunctions" of such "generalized eigenvalues" are "generalized functions" or "distributions". They are not square-integrable.

That's easy to see in the position representation, where $\hat{p}=-\mathrm{i} \hbar \partial_x$. The eigenvalue equation reads
$$-\mathrm{i} \hbar \partial_x u_p(x)=p u_p(x).$$
The solutions are
$$u_p(x)=N \exp(\mathrm{i} p x/\hbar).$$
That's for sure not square integrable, because $|u_p(x)|^2=|N|^2=\text{const}$. Nevertheless all $p \in \mathbb{R}$ are "generalized eigenvalues" of the momentum operator and the eigenfunctions can be "normalized" to a Dirac-$\delta$ distribution,
$$\langle u_p|u_{p'} \rangle=\int_{\mathbb{R}} u_p^*(x) u_{p'}(x)=\delta(p-p').$$
As is known from the theory of Fourier transformations for that we simply have to set the normalization constant to
$$N=\frac{1}{\sqrt{2 \pi \hbar}} \; \Rightarrow \; u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
They build a complete set of orthonormal functions in the sense that you can represent a dense subset of square integrable functions as a Fourier transform,
$$\psi(x)=\langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x|p \rangle \langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar) \tilde{\psi}(p)$$
and in the other way
$$\tilde{\psi}(p) = \langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi \hbar}} \exp(-\mathrm{i} p x/\hbar) \psi(x).$$
It's also clear that the transformation from the "position-space wave function" to the "momentum-space wave function" is a unitary transformation since
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi_1^*(x) \psi_2(x) = \int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}_1^*(p) \tilde{\psi}_2(p).$$

Thank you so much. I really appreciate that you took time to write it.

I completely understand that delta functions cannot represent any realizable state.

What i was saying is that there is also another problem with relating a delta wavefunction with a single value measurement $p$ since $\delta^{2}(p-p') d p '\neq 1$ at $p'=p$ (i.e even if somehow delta function represented a single value measurement, it fails to give a probability of $1$ as shown above)

 

[PeroK]

Thank you so much. I really appreciate that you took time to write it.

I completely understand that delta functions cannot represent any realizable state.

What i was saying is that there is also another problem with relating a delta wavefunction with a single value measurement $p$ since $\delta^{2}(p-p') d p '\neq 1$ at $p'=p$ (i.e even if somehow delta function represented a single value measurement, it fails to give a probability of $1$ as shown above)

The condition that the generalised eigenfunctions statisfy is Dirac Orthonormality. The eigenfunction $p(p')$ associated with eigenvalue $p$ (in momentum space) is:
$$p(p') = \delta(p - p')$$Now:
$$\langle p_1| p_2 \rangle = \int dp' p_1(p')^*p_2(p') = \int dp'\delta(p_1 - p') \delta(p_2 - p') = \delta(p_1 - p_2)$$In particular, if $p_1 \ne p_2$, then:
$$\langle p_1| p_2 \rangle = 0$$The delta function itself is not square integrable.

 

[atyy]

For the free particle in QM, the energy and momentum eigenstates are not physically realizable since they are not square integrable. So in that sense a particle cannot have a definite energy or momentum.

What happens during measurement of say momentum or energy ?

So we measure some definite value of momentum or energy which is an eigenvalue of the momentum or Hamiltonian (since the operators commute for a free particle). Then we would in principle collapse the wave function to some eigenstate, but in this case we know that this is not possible (physically realizable).

So what happens during measurement and in what state is the particle after we've measured it's momentum?

Thank you

One can ask a similar question about a sharp position measurement. There is some comment on that in section 2.3.2 of https://arxiv.org/abs/0706.3526.The rule that the wave function collapses into an eigenstate of the measurement is not the most general rule for collapse, and cannot apply for position measurements, since the position eigenstate is not an allowed quantum state. There are more general rules for collapse associated with mroe general measurements called POVMs.

 

[throw]

For the free particle in QM, the energy and momentum eigenstates are not physically realizable since they are not square integrable. So in that sense a particle cannot have a definite energy or momentum.

This is just one possible interpretation of continuous spectrum eigenfunctions - it is not the only possible interpretation one can give to continuous spectrum eigenfunctions:

There are two ways out of this difficulty. The first is to give up the concept of absolute probabilities when dealing with wave functions such as (2.13) or (2.16) which are not square integrable. Instead, $|\Psi(\mathbf{r}, t)|^2 d \mathbf{r}$ is then interpreted as the relative probability of finding the particle at time t in a volume element $d\mathbf{r}$ centred about $\mathbf{r}$, so that the ratio $|\Psi(\mathbf{r}_1, t)|^2/|\Psi(\mathbf{r}_2, t)|^2$ gives the probability of finding the particle within a volume element centred around $\mathbf{r} = \mathbf{r}_1$, compared with that of finding it within the same volume element at $\mathbf{r} = \mathbf{r}_2$.

For the particular case of the plane wave $(2.16)$, we see that $|\Psi|^2 = |A|^2$, so that there is an equal chance of finding the particle at any point. The plane wave $(2.16)$ therefore describes the idealised situation of a free particle having a perfectly well-defined momentum, but which is completely delocalised.

This suggests a second way out of the difficulty, which is to give up the requirement that the free particle should have a precisely defined momentum, and to superpose plane waves corresponding to different momenta to form a localised wave packet, which can be normalised to unity.

- Bransden and Joachain, 'Quantum Mechanics'

There are plenty of books which take the alternative approach, for example Landau and Lifshitz 'Quantum Mechanics', and from this perspective many of the statements in this thread are simply indefensible: the origin of the problem really boils down to the difference between discrete and continuous probability distributions, and interpreting the meaning of measuring a single value v.s. theoretical probabilities for finding single values (we don't say single outcomes associated to general continuous probability distributions don't 'exist', only in a mathematical approach to QM is this done for some reason...)

 

[PeroK]

There are plenty of books which take the alternative approach, for example Landau and Lifshitz 'Quantum Mechanics', and from this perspective many of the statements in this thread are simply indefensible:

Are you saying that the OP, who has just started learning QM, should give up McIntyre and study Landau and Lifshitz instead?

 

[vanhees71]

Well, it doesn't hurt to consult other textbooks, if the so far used textbook is obviously not concise enough to help with a very good question as the one asked in #1. Obviously McIntire is a bit too sloppy, if it comes to unbound self-adjoint operators with a continuous spectrum. Instead of Landau and Lifshitz I'd rather suggest to read Messiah. This book in my opinion finds a quite good compromise between mathematical rigidity and the more pragmatic physicists' approach. For a more modern treatment in terms of the "rigged Hilbert space" also Ballentine is very good choice.

 

[andresB]

Checking my copy of Messiah, I find a readable treatment of projector for the continuous spectrum, pages 260 to 270. And a discussion of protective measurements, discrete and continuous cases, is given on pages 297 and 298.

 

[PeroK]

Well, it doesn't hurt to consult other textbooks, ...

It might hurt fundamentally in the sense that the OP may never really learn QM, but just stagger from one textbook to another. It's fine for an expert in the field to compare and contrast McIntyre, L&L, Messiah and Ballentine; but, for a novice that might not work at all.

Personally, I would stick with one textbook for a serious timescale - 4-6 chapters or so. I would focus on digesting the basics. Alternative and subtler analyses can come later.

 

[andresB]

It might hurt fundamentally in the sense that the OP may never really learn QM, but just stagger from one textbook to another. It's fine for an expert in the field to compare and contrast McIntyre, L&L, Messiah and Ballentine; but, for a novice that might not work at all.

Personally, I would stick with one textbook for a serious timescale - 4-6 chapters or so. I would focus on digesting the basics. Alternative and subtler analyses can come later.

While having too many sources to read from can cause some sort of analysis paralysis, the concept of learning from just one book is alien to me.

 

[vanhees71]

Hm, I always consulted other textbooks, if I couldn't answer a question with help of the main textbook I've been studying. For me that's the usual situation when learning a new subject.

 

[PeroK]

While having too many sources to read from can cause some sort of analysis paralysis, the concept of learning from just one book is alien to me.

I'm sceptical of that as an education policy below PhD/post-doctorate level. If you were teaching undergraduate E&M, you really would work from several textbooks concurrently? What about high-school geometry? I can't see it! I suggest it's bravado on your part!

It's different to suggest a choice of textbooks, but not multiple text-books concurrently. Not for the average student. And definitely not for me.

I'm happy to be corrected by those who have taught at university level, but if I was teaching I would stick to one textbook per course. And definitely not present conflicting presentations of the same material at an introductory level.

 

[vanhees71]

That's very strange to me. In our lectures we always got a list of books we can use when studying the material presented in the lectures. As I said above, I never studied a subject with using only a single book. Usually I worked along the content following the notes I took during the lecture and then consulted several textbooks when I couldn't understand something from the notes. In addition, of course, we had problem sessions and could ask the tutors and/or the professor. Last but not least we also worked together in a group of students to solve the problem sets.

 

[Kashmir]

To say that we know the exact value of the momentum is equivalent to saying that the wave function in momentum space is a delta function.

Why do you say that the wave function will be a delta function?

Suppose for arguments sake I knew the exact value of momentum to be $p_0$ .

Let the state corresponding to it be $|\psi\rangle$ . Since $\langle p \mid \psi\rangle$ gives the probability amplitude of measuring momentum around $p$ ,then it follows
$\langle p \mid \psi\rangle=0 \quad ;p \neq p_{0}$ and

$\left\langle p_{0} \mid \psi\right\rangle=1$

which means that the wave function in momentum space is
$\varphi(p)=\left\{\begin{array}{ll}0 & p \neq p_{0} \\ 1 & p=p_{0}\end{array}\right.$which is not a delta function.

 

[PeroK]

$\left\langle p_{0} \mid \psi\right\rangle=1$ , which means that the wave function in momentum space is
$\varphi(p)=\left\{\begin{array}{ll}0 & p \neq p_{0} \\ 1 & p=p_{0}\end{array}\right.$

That's not correct. The inner product of two functions is:
$$\left\langle p_{0} \mid \psi\right\rangle=\int dp' \ p_0(p')^* \psi(p')$$And that integral is zero for your eigenfunction. Note that in terms of square integrable functions, the function you suggest is equivalent to the zero function: $\varphi \equiv 0$

Instead if $p_0(p') = \delta(p_0 - p')$, then:
$$\left\langle p_{0} \mid \psi\right\rangle=\int dp' \ \delta(p_0 - p') \psi(p') = \psi(p_0)$$

 

[vanhees71]

But such a function doesn't exist! A "wave function" describing the situation that the momentum of the particle takes the exact momentum value $p_0$ in the momentum representation representation must be $\langle p|p_0 \rangle=\delta(p-p_0)$, i.e., a $\delta$ distribution. You cannot even take its square! So it's not describing a proper (pure) state of the particle. As already stressed above, you can have arbitrarily well determined values of the momentum but never a preparation in a state where momentum takes an exactly single value.

An example are the Gaussian wave functions:
$$\tilde{\psi}(p)=\frac{1}{\sqrt{\Delta p} (2 \pi)^{1/4}} \exp \left (-\frac{(p-p_0)^2}{4 \Delta p^2} \right ).$$
This describes the situation, where the momentum takes the average value $p_0$ with a standard deviation of $\Delta p$. The probability distribution for measuring the momentum is
$$P(p)=|\tilde{\psi}(p)|^2 = \frac{1}{\sqrt{2 \pi} \Delta p} \exp \left (-\frac{(p-p_0)^2}{2 \Delta p^2} \right ).$$
Indeed, that's the Gaussian distribution with average $\langle p \rangle=p_0$ and $\langle p^2 \rangle-\langle p \rangle^2=\Delta p^2$.

You can make $\Delta p>0$ as small as possible, i.e., making the probability distribution peaking as narrowly around $p_0$ as you like, but never set $\Delta p=0$. Rather the Gaussian goes to the $\delta$-distribution in the limit $\Delta p \rightarrow 0$.