Is Quantum Probability Calculation of |e-ix|^2 Correct?

[quietrain]

Probability = |<x|x>|²

but if |x> = e^-ix

then prob = e^+ixe^-ix = 1

but that is not |e^-ix|² right?

|e^-ix|² becomes e^-2ix?

thanks!

edit: i was watching a lecture on quantum mechanics and the lecturer wrote

P_x = |a e^-iEt|² = a*a

 

[Fredrik]

The absolute value of a complex number z is $|z|=\sqrt{z^*z}$, so for all real numbers x, $$|e^{-ix}|^2=e^{-ix}e^{ix}=e^{-ix+ix}=e^0=1.$$

 

[vanhees71]

I also hope that the lecturer mentioned that the plane wave (momentum eigenstate) is not representing a state in quantum theory. Only square integrable functions represent states!

 

[quietrain]

The absolute value of a complex number z is $|z|=\sqrt{z^*z}$, so for all real numbers x, $$|e^{-ix}|^2=e^{-ix}e^{ix}=e^{-ix+ix}=e^0=1.$$

wow thank you i never knew that the absolute of a complex number is that.

i thought its just changing it to +

 

[quietrain]

I also hope that the lecturer mentioned that the plane wave (momentum eigenstate) is not representing a state in quantum theory. Only square integrable functions represent states!

1) are you talking about -ih(bar)d/dx? the momentum operator?

2) also i thought a state means like if light passes through a polarizer, then if its polarized in the horizontal position, that means its a state |x> ? , if vertical means |y> ?

3) anyway, the lecturer wrote something like this
P_x = |a e-iEt|2 = a*a
where P_x is probability

4) btw, what do you mean by square integrable functions? you mean this?
http://en.wikipedia.org/wiki/Square-integrable_function

so what's the catch? is it got to do with normalizing?

 

[Fredrik]

1) are you talking about -ih(bar)d/dx? the momentum operator?
...
4) btw, what do you mean by square integrable functions? you mean this?
http://en.wikipedia.org/wiki/Square-integrable_function

so what's the catch? is it got to do with normalizing?

He's talking about the eigenfunctions of that operator. They are the functions $u_p$ defined by $u_p(x)=e^{-ipx/\hbar}$ for all x. The catch is that they are not square integrable. If you plug one of the $u_p$ functions into the formula for the norm of a square integrable function, the result is $\infty$.

If these functions should be called "states" or not is to some extent a matter of taste.

Are you still having difficulties with (3)? The absolute value of a complex number z=x+iy where x and y are real is $|z|=\sqrt{z^*z}=\sqrt{x^2+y^2}$. What does that tell you about |wz| where w is another complex number?

By the way, the function <,> defined on that Wikipedia page is a semi-inner product, not an inner product. The article got that detail wrong.

 

[quietrain]

He's talking about the eigenfunctions of that operator. They are the functions $u_p$ defined by $u_p(x)=e^{-ipx/\hbar}$ for all x. The catch is that they are not square integrable. If you plug one of the $u_p$ functions into the formula for the norm of a square integrable function, the result is $\infty$.

If these functions should be called "states" or not is to some extent a matter of taste.

Are you still having difficulties with (3)? The absolute value of a complex number z=x+iy where x and y are real is $|z|=\sqrt{z^*z}=\sqrt{x^2+y^2}$. What does that tell you about |wz| where w is another complex number?

By the way, the function <,> defined on that Wikipedia page is a semi-inner product, not an inner product. The article got that detail wrong.

oh, so is there anything i should do to make an eigenfunction square integrable? or there's nothing i can do about it?

i understand (3),
but with regards to the |wz|, do you mean i have to squareroot [(wz)*(wz)] ? which means i complex conjugate the whole of wz this time?

with regards to semi-inner product,
a site said this
"Hence, a semi-inner product on a vector space is just like an inner product, but for which <v|v> can be zero (even if v=0 )."

i thought as long as v=v it is 1, if its not = to each other, its 0?
so in a semi-inner product, if <0|0>, then it is 0 and not 1? this is the catch?

 

[Fredrik]

i understand (3),
but with regards to the |wz|, do you mean i have to squareroot [(wz)*(wz)] ? which means i complex conjugate the whole of wz this time?

Yes, that's what the definition of the absolute value says. You really need to do the following (easy) exercise. Write w=a+ib, z=c+id, where a,b,c,d are real. Define the absolute value by $|z|^2=\sqrt{c^2+d^2}$ and the complex conjugate by z*=c-id. Show that (wz)*=w*z* and that |wz|=|w||z|. (This is the kind of stuff you should have done within the first hour after you learned about complex numbers).

"Hence, a semi-inner product on a vector space is just like an inner product, but for which <v|v> can be zero (even if v=0 )."

i thought as long as v=v it is 1, if its not = to each other, its 0?
so in a semi-inner product, if <0|0>, then it is 0 and not 1? this is the catch?

Both inner products and semi-inner products are linear in one of the variables and antilinear in the other. So if either of x or y is =0, then <x,y>=0. (For example, if y=0, we have <x,y>=<x,0y>=0<x,y>=0). The difference is that inner products satisfy the condition if <v,v>=0, then v=0. The function <,> defined in that article is a semi-inner product, because for example, if f is the function defined by $$f(x)=\begin{cases}0 &\text{if}\ x\neq 0\\ 1 &\text{if}\ x=0,\end{cases}$$ we have <f,f>=0.

u≠v obviously doesn't imply <u,v>=0. That would actually mean that <v,v>=(1/2)<v,2v>=0 for all v, so that we actually have <u,v>=0 for all u,v.

You are very careless with your mathematical statements. I assume that you understand that it's not possible for v to not be =v.

 

[jambaugh]

probability = |<x|x>|²

but if |x> = e^-ix

then prob = e^+ixe^-ix = 1

$1^2 = 1$

 

[quietrain]

Yes, that's what the definition of the absolute value says. You really need to do the following (easy) exercise. Write w=a+ib, z=c+id, where a,b,c,d are real. Define the absolute value by $|z|^2=\sqrt{c^2+d^2}$ and the complex conjugate by z*=c-id. Show that (wz)*=w*z* and that |wz|=|w||z|. (This is the kind of stuff you should have done within the first hour after you learned about complex numbers).

[(a+ib)(c+id)]*
=ac-bd-iad-ibc
but how do i go from here to (a-ib)(c-id)? long division?
or just expand (a-ib)(c-id) = ac-bd-iad-ibc ?

so for |wz| = sqrt[(wz)*(wz)] = sqrt(w*z*wz) = sqrt(w*wz*z)= sqrt(w*w)sqrt(z*z)=|w||z| ?

Both inner products and semi-inner products are linear in one of the variables and antilinear in the other. So if either of x or y is =0, then <x,y>=0. (For example, if y=0, we have <x,y>=<x,0y>=0<x,y>=0). The difference is that inner products satisfy the condition if <v,v>=0, then v=0. The function <,> defined in that article is a semi-inner product, because for example, if f is the function defined by $$f(x)=\begin{cases}0 &\text{if}\ x\neq 0\\ 1 &\text{if}\ x=0,\end{cases}$$ we have <f,f>=0.

oh so for normal inner product <v|v> = 1
but if <v|v> = 0, then it is a semi-inner product and v can only be = 0? so the inner product of something with itself can only be 0 if it is 0 itself? <0|0> = 0, nothing else?thats the catch to semi-inner products?

u≠v obviously doesn't imply <u,v>=0. That would actually mean that <v,v>=(1/2)<v,2v>=0 for all v, so that we actually have <u,v>=0 for all u,v.

You are very careless with your mathematical statements. I assume that you understand that it's not possible for v to not be =v.

i thought <u|v> = 0 if u≠v ? or is that the kronecker delta?
oh if u is orthogonal to v then <u|v> is 0 right?

also, i don't really understand this part
<v,v>=(1/2)<v,2v>=0 for all v, so that we actually have <u,v>=0 for all u,v.

why will having half of the inner product of v,2v be equal to 0 which equal <v|v> ? is this showing orthogonality?

 

[quietrain]

$1^2 = 1$

haha i see

 

[jtbell]

i thought <u|v> = 0 if u≠v ?[/url]

Not necessarily.

oh if u is orthogonal to v then <u|v> is 0 right?

Right. (and vice versa: if <u|v> = 0 then u and v are orthogonal.)

 

[ZealScience]

|e-ix|2 is not equivalent to (e^-ix)², right?

 

[jtbell]

|e-ix|2 is not equivalent to (e^-ix)², right?

Right.

$$|e^{ix}|^2 = (e^{ix})(e^{ix})^* = (e^{ix})(e^{-ix}) = e^{ix-ix} = e^0 = 1$$

$$(e^{ix})^2 = (e^{ix})(e^{ix}) = e^{ix+ix} = e^{2ix}$$

 

[ZealScience]

Right.

$$|e^{ix}|^2 = (e^{ix})(e^{ix})^* = (e^{ix})(e^{-ix}) = e^{ix-ix} = e^0 = 1$$

$$(e^{ix})^2 = (e^{ix})(e^{ix}) = e^{ix+ix} = e^{2ix}$$

Then OP is not quite true...

 

[Fredrik]

[(a+ib)(c+id)]*
=ac-bd-iad-ibc
[...]
(a-ib)(c-id) = ac-bd-iad-ibc

This is the easiest way to do it.

so for |wz| = sqrt[(wz)*(wz)] = sqrt(w*z*wz) = sqrt(w*wz*z)= sqrt(w*w)sqrt(z*z)=|w||z| ?

This is correct, but to justify the fourth equality, you need to prove either that z*z is real for all complex z, or that sqrt(wz)=(sqrt w)(sqrt z) for all complex w,z. The first option is the easiest: z*z=(c-id)(c+id)=c^2+d^2.

oh so for normal inner product <v|v> = 1

For all non-zero v, yes.

but if <v|v> = 0, then it is a semi-inner product and v can only be = 0?

No. If it's an inner product, this equality implies v=0. If it's a semi-inner product, it doesn't.

so the inner product of something with itself can only be 0 if it is 0 itself? <0|0> = 0, nothing else?

Yes.

i thought <u|v> = 0 if u≠v ? or is that the kronecker delta?

Only if you have already specified that u and v are members of the same orthonormal subset of a Hilbert space. It's certainly not true for arbitrary u,v.

oh if u is orthogonal to v then <u|v> is 0 right?

Yes. The definition of "orthogonal" says that u and v are said to be orthogonal if <u,v>=0.

also, i don't really understand this part
<v,v>=(1/2)<v,2v>=0 for all v, so that we actually have <u,v>=0 for all u,v.

why will having half of the inner product of v,2v be equal to 0 which equal <v|v> ? is this showing orthogonality?

I was showing you a consequence of your claim that for all u,v, we have <u,v>=0. Both inner products and semi-inner product are linear in the second variable, so for any complex c, we have <v,v>=(1/c)<v,cv>. Choose c≠1. Now your claim implies that the second factor on the right is 0. So we find that for all v, we have <v,v>=0. Since you had already assumed that <u,v>=0 whenever u≠v, the conclusion is that for all u,v, we have <u,v>=0. This would be a completely useless semi-inner product. If it's an inner product, it even implies that the only vector in the inner product space is the zero vector.

 

[vanhees71]

It cannot be stressed often enough. That's why I put it here again and in bold letters:

A wave function represents a state in quantum mechanics if and ONLY if it is square integrable and thus normalizable to 1!

Momentum "eigenstates", i.e., plane waves are NOT square integrable and thus do not represent a state. They belong to another space, namely the dual space of the dense subspace in Hilbert space, where the momentum operator is defined, i.e., it is to be considered as a generalized function (or distribution).

The same is true for any "eigenstate" for a "eigenvalue" of an operator in the continuous part of its spectrum.

E.g., for the generalized eigenvector of the position operator to the "eigenvalue", $x_0$ one has

$$u_{x_0}(x)=\langle x|x_0 \rangle = \delta(x-x_0),$$

where I consider a one-dimensional problem for simplicity's sake.

The generalized eigenvector of the momentum operator in the position representation is

$$u_{p_0}(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p).$$

I've set $\hbar=1$ for simplicity.

Generalized eigenstates are normalized to a $\delta$ distribution. Indeed, one has

$$\langle p'|p \rangle=\int \mathrm{d} x \langle p'|x \rangle \langle x|p \rangle = \int \mathrm{d} x \frac{1}{2 \pi} \exp[\mathrm{i} x(p-p')]=\delta(p-p').$$

For "real states", represented ONLY by square integrable wave functions in the position representation, for any two observables the uncertainty relation holds. For position and momentum it reads

$$\delta x \delta p \geq 1/2,$$

and this shows that there doesn't exist any real state, for which either position or momentum are precisely determined. You can make the uncertainty in one of the observables arbitrarily small at the expense of the accuracy the other is known, but you can never make the uncertainty of either variable really 0! Thus, there cannot exist true eigenstates for position or momentum, because for those the uncertainty would vanish!

 

[Fredrik]

vanhees71, did you perhaps post in the wrong thread by accident? Your reply would make more sense as a reply to post 25 in this thread.

 

[vanhees71]

Maybe, but also in this thread I've seen some strange ideas like

$$\langle x| x \rangle =1 \quad (\mathrm{WRONG}).$$

Generalized position eigenstates are distributions and not normalizable to 1 and it doesn't make any sense to take this scalar product. The correct expression is

$$\langle x' | x\rangle =\delta(x'-x),$$

but that I tried to explain in my previous posting.

 

[quietrain]

thx fredrik i understand now i hope

with regards to vanhees,

It cannot be stressed often enough. That's why I put it here again and in bold letters:

A wave function represents a state in quantum mechanics if and ONLY if it is square integrable and thus normalizable to 1!

erm does it mean that if i can normalize something, it means it is square integrable? like adding a 1/sqrt2 in front of the expression?

Momentum "eigenstates", i.e., plane waves are NOT square integrable and thus do not represent a state. They belong to another space, namely the dual space of the dense subspace in Hilbert space, where the momentum operator is defined, i.e., it is to be considered as a generalized function (or distribution).

The same is true for any "eigenstate" for a "eigenvalue" of an operator in the continuous part of its spectrum.

do you mean that the momentum operator is not a state? i watched a quantum lecture on youtube, it says operators are observables? states are wavefunctions of the system? something like that?

E.g., for the generalized eigenvector of the position operator to the "eigenvalue", $x_0$ one has

$$u_{x_0}(x)=\langle x|x_0 \rangle = \delta(x-x_0),$$

where I consider a one-dimensional problem for simplicity's sake.

The generalized eigenvector of the momentum operator in the position representation is

$$u_{p_0}(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p).$$

I've set $\hbar=1$ for simplicity.

ah i have seen the lecturer talk about this, is this about the eigen vectors of position operator as delta functions while eigen vectors of momentum being cos and sine waves?

Generalized eigenstates are normalized to a $\delta$ distribution. Indeed, one has

$$\langle p'|p \rangle=\int \mathrm{d} x \langle p'|x \rangle \langle x|p \rangle = \int \mathrm{d} x \frac{1}{2 \pi} \exp[\mathrm{i} x(p-p')]=\delta(p-p').$$

i got lost here, are you trying to say that states are normalized to a delta function and not to 1? what is p prime, p', x' etc?

also it reminds me of what fredrik said earlier

If you plug one of the up functions into the formula for the norm of a square integrable function, the result is ∞.

is this the formula for the norm of a square integrable function?

Thus, there cannot exist true eigenstates for position or momentum, because for those the uncertainty would vanish!

? so the momentum eigenstate e^-ipx/h is 'false' ??

 

[quietrain]

Maybe, but also in this thread I've seen some strange ideas like

$$\langle x| x \rangle =1 \quad (\mathrm{WRONG}).$$

Generalized position eigenstates are distributions and not normalizable to 1 and it doesn't make any sense to take this scalar product. The correct expression is

$$\langle x' | x\rangle =\delta(x'-x),$$

but that I tried to explain in my previous posting.

oh shucks, i remember the lecturer saying something about eigenstates of position being delta functions , but why do you take the inner product of x' and x to show this? what is x' too?

also, position eigenstates are not normalizable to 1?i remembered the lecturer saying something about to normalize wavefunctions i just need to take the squareroot of the sum of square of the individual elements of the matrix and make it 1?

so i can't do this to position eigenstates? then which ones can i normalize to 1?

 

[vanhees71]

I think, it's important to get the math of quantum theory very clear from the very beginning. So here's the structure of "quantum kinematics" in a nutshell (and not always too rigorous):

(1) Pure states of a quantum system are represented by (rays of) vectors in a Hilbert space, which is a complex vector space with a scalar product that is complete.

(2) Observables are represented by self-adjoint operators that are defined on a dense subspace in Hilbert space. This range of definition is usually not the whole Hilbert space.

(3) Possible outcomes of measurements of an observable are the spectral values of the corresponding operators. In physics usually one doesn't treat this very rigorously and just speacs about spectral values as eigenvalues and defines also generalized eigenstates, which generally are not Hilbert-space vectors but belong to the dual space of the definition range of the operators. This dual space define distributions (generalized functions) rather than true states.

(4) The probability (density) to measure a certain eigenvalue $\lambda$ of an observable, given the state of the system to be $|\psi \rangle$ is given by the modulus squared of the wave function

$$\psi(\lambda)=\langle \lambda|\psi \rangle,$$

where $|\lambda \rangle$ is the (generalized) eigenvector of the operator representing the observable with spectral value, [\itex]\lambda[/itex].

To give a good probability distribution, these wave functions must be normalizable, i.e., they must fulfill

$$\int \mathrm{d} \lambda |\psi(\lambda)|^2=1,$$

where I assumed for simplicity that the observable has only a continuous spectrum. Otherwise one would have also to sum over the discrete eigenvalues.

Let's take as an example the momentum operator and work in the position representation, i.e., we use the space of square integrable functions of position as the Hilbert space of states. We also look at the motion of a particle in only one space direction. Then the momentum operator is given by (with $\hbar=1$)

$$\hat{p}=-\mathrm{i} \frac{\partial}{\partial x}.$$

Now let's look for (generalized) eigenvectors of this operator. We must have

$$\hat{p}u_p(x)=-\mathrm{i} \partial_x u_p(x)=p u_p(x)$$

This differential equation has the solution

$$u_p(x)=N_{p} \exp(\mathrm{i} p x), \quad N_{p}=\text{const}.$$

This for no value of p is a square integrable function, but to have it at least bounded for all real values of x, the generalized eigenvalues must be real, i.e., $p \in \mathbb{R}$. So the possible values of momentum are all real numbers, i.e., you only have a continuous spectrum. Correspondingly the generalized eigenvectors cannot be normlized to 1, but you can normalized them in the generalized sense to a $\delta$ distribution:

$$\langle p' |p\rangle=\int_{\mathbb{R}} \mathrm{d} x u_{p'}^*(x) u_p(x)=N_{p'}^* N_p \int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i}x(p-p')]=2 \pi |N_{p}|^2 \delta(p'-p) \; \quad \Rightarrow N_{p}=\frac{1}{\sqrt{2 \pi}}.$$

Of course $N_{p}$ is only determined up to a phase factor, but this doesn't matter anyway, so I define the normalization factor to be positive real. Thus, we have

$$u_p(x)=\frac{1}{\sqrt{2 \pi}} \exp[\mathrm{i} x p].$$

Given a true state $|\psi \rangle$, the wave function $\psi(x)=\langle x | \psi \rangle$ must be square integrable, and its modulus squared gives the probability distribution to find the particle at position x. To get the probability distribution for the momentum, you find

$$\tilde{\psi}(p)=\langle p|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p| x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p^*(x) \psi(x)=\int_{\mathbb{R}} \frac{\mathrm{d} x}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x).$$

Of course, if you have given the state in terms of the momentum-wave function, you get in the same way the inverse of this transformation, which is just a Fourier transformation, by

$$\psi(x)=\int_{\mathbb{R}} \frac{\mathrm{d} p}{\sqrt{2 \pi}} \exp(+\mathrm{i} p x) \tilde{\psi}(p).$$

I hope that with this, mathematically not rigorous, clarifications the whole issue of continuous eigenvalues and generalized eigenvectors becomes a bit more clear.

 

[quietrain]

yes it certainly helps

but just a question

why is <p|x> = u_p(x)*

how am i suppose to interpret this?

the inner product of p and x is equal to the eigenvector of the operator p? the (x) means its in the position basis?