If |a> is an eigenvector of A, is f(B)|a> an eigenvector of A?

[happyparticle]

Homework Statement:

Show that a vector is an eigenvector of an operator

Relevant Equations:

$A|a\rangle = a|a\rangle$

Hi,
If $|a\rangle$ is an eigenvector of the operator $A$, I know that for any scalar $c \neq 0$ , $c|a\rangle$ is also an eigenvector of $A$

Now, is the ket $F(B)|a\rangle$ an eigenvector of $A$? Where $B$ is an operator and $F(B)$ a function of $B$.

Is there way to show that $F(B)|a\rangle$ is and eigenvector of $A$ and find the eigenvalue?

Thank you!

 

[topsquark]

Homework Statement:: Show that a vector is an eigenvector of an operator
Relevant Equations:: $A|a\rangle = a|a\rangle$

Hi,
If $|a\rangle$ is an eigenvector of the operator $A$, I know that for any scalar $c \neq 0$ , $c|a\rangle$ is also an eigenvector of $A$

Now, is the ket $F(B)|a\rangle$ an eigenvector of $A$? Where $B$ is an operator and $F(B)$ a function of $B$.

Is there way to show that $F(B)|a\rangle$ is and eigenvector of $A$ and find the eigenvalue?

Thank you!

The usual method to discuss something like this is to start with the Taylor expansion of F(B). Since this is a polynomial in B then we are applying B directly to $\mid a \rangle$. So what you are really asking is whether $\mid a \rangle$ is an eigenstate of B.

The only way this can happen is if [A, B] = 0. In this case we can calculate the eigenvalues of B in the $\mid a \rangle$ basis: $B \mid a \rangle = ( \langle a \mid B \rangle ) \mid a \rangle$
(The proof is fairly simple, just use closure twice.)

From there you can construct what happens when we apply F(B) to $\mid a \rangle$.

-Da n

 

[happyparticle]

@topsquark
For instance, If I have $[X,F(P_x)]$ and $X ,P_x$ don't commute so we can't measure a value for X and $P_x$ at the same time so they don't share an eigenvalue ?

I know that [A,F(B)] doesn't commute, but I have to show that $F(B)|a\rangle$ is also an eigenvector of $A$, where $|a\rangle$ is an eigenvector of $A$

So I misunderstood something.

I'm trying to show that $B|x\rangle = iP_x/h|x\rangle$ is an eigenvector of $X$ if $|x\rangle$ is an eigenvector of $X$

 

[PeroK]

Homework Statement:: Show that a vector is an eigenvector of an operator
Relevant Equations:: $A|a\rangle = a|a\rangle$

Hi,
If $|a\rangle$ is an eigenvector of the operator $A$, I know that for any scalar $c \neq 0$ , $c|a\rangle$ is also an eigenvector of $A$

Now, is the ket $F(B)|a\rangle$ an eigenvector of $A$? Where $B$ is an operator and $F(B)$ a function of $B$.

Is there way to show that $F(B)|a\rangle$ is and eigenvector of $A$ and find the eigenvalue?

Thank you!

I suggest you look for a counterexample using the simplest case of 2x2 matrices.

 

[topsquark]

@topsquark
For instance, If I have $[X,F(P_x)]$ and $X ,P_x$ don't commute so we can't measure a value for X and $P_x$ at the same time so they don't share an eigenvalue ?

I know that [A,F(B)] doesn't commute, but I have to show that $F(B)|a\rangle$ is also an eigenvector of $A$, where $|a\rangle$ is an eigenvector of $A$

So I misunderstood something.

I'm trying to show that $B|x\rangle = iP_x/h|x\rangle$ is an eigenvector of $X$ if $|x\rangle$ is an eigenvector of $X$

So the actual problem is
1) Show that if $\mid x\rangle$ is an eigenstate of X then it is also an eigenstate of $P_x$?

Or is it
2) Show that $\mid x \rangle$ is an eigenstate of $F(P_x)$?

In the future, please post the whole question.
1) Clearly not, as
$B \mid x \rangle = \dfrac{d}{dx} \mid x \rangle$
which cannot give us an eigenvalue equation in $\mid x \rangle$ since $\dfrac{d}{dx} \mid x \rangle \neq \text{(constant)} \mid x \rangle$.

2) Because of 1), the only function that F(B) could be is a constant, but if I now have the full question, F(B) is an arbitrary function.

-Dan

 

[PeroK]

Homework Statement:: Show that a vector is an eigenvector of an operator
Relevant Equations:: $A|a\rangle = a|a\rangle$

Hi,
If $|a\rangle$ is an eigenvector of the operator $A$, I know that for any scalar $c \neq 0$ , $c|a\rangle$ is also an eigenvector of $A$

Now, is the ket $F(B)|a\rangle$ an eigenvector of $A$? Where $B$ is an operator and $F(B)$ a function of $B$.

Is there way to show that $F(B)|a\rangle$ is and eigenvector of $A$ and find the eigenvalue?

Thank you!

PS note that you have an arbitrary operator $B$ and an arbitrary function $F$. That means that $F(B)$ is an arbitrary operator. E.g. take any operator $B$ and the identity function as $F$.

Now, as $F(B)$ is arbitrary, then $F(B) \ket a$ can be any vector. I.e. this is an arbitrary vector. So, you're question can be rephrased as:

If $\ket a$ is an eigenvector of $A$, then are all vectors eigenvectors of $A$?

 

[PeroK]

The usual method to discuss something like this is to start with the Taylor expansion of F(B). Since this is a polynomial in B then we are applying B directly to $\mid a \rangle$. So what you are really asking is whether $\mid a \rangle$ is an eigenstate of B.

The only way this can happen is if [A, B] = 0. In this case we can calculate the eigenvalues of B in the $\mid a \rangle$ basis: $B \mid a \rangle = ( \langle a \mid B \rangle ) \mid a \rangle$
(The proof is fairly simple, just use closure twice.)

From there you can construct what happens when we apply F(B) to $\mid a \rangle$.

-Da n

Alternatively, if $B$ commutes with $A$, then:
$$A(B\ket a) = (AB)\ket a = (BA)\ket a = B(A\ket a) = B( \lambda \ket a) = \lambda(B\ket a)$$And we see that $B\ket a$ is also an eigenvector of $A$ with the same eigenvalue, $\lambda$.

Morevover, if $B$ commutes with $A$, then powers of $B$ commute with $A$ and hence any function of $B$ that has a valid power series expansion commutes with $A$

 

[vanhees71]

The usual method to discuss something like this is to start with the Taylor expansion of F(B). Since this is a polynomial in B then we are applying B directly to $\mid a \rangle$. So what you are really asking is whether $\mid a \rangle$ is an eigenstate of B.

The only way this can happen is if [A, B] = 0. In this case we can calculate the eigenvalues of B in the $\mid a \rangle$ basis: $B \mid a \rangle = ( \langle a \mid B \rangle ) \mid a \rangle$
(The proof is fairly simple, just use closure twice.)

From there you can construct what happens when we apply F(B) to $\mid a \rangle$.

-Da n

That's not true. An important counter-example are the annihilation and creation operators of the harmonic oscillator $\hat{a}$ and $\hat{a}^{\dagger}$ and the "phonon-number operator" $\hat{N}=\hat{a}^{\dagger} \hat{a}$. Since $[\hat{a},\hat{a}^{\dagger}$ it's easy to show that if $|n \rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $n$, then $\hat{a}|n \rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $(n-1)$ (or the null vector) and $\hat{a}^{\dagger}|n \rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $(n+1)$. Since $\hat{N}$ is a positive definite self-adjoint operator, it's easy to show that there's an eigenvector of $\hat{N}$, $|\Omega \rangle$ such that $\hat{a} |\Omega \rangle=0$, i.e., it's the eigenvector of $\hat{N}$ with eigenvalue $0$, and thus the eigenvalues of $\hat{N}$ must be $n \in \{0,1,2,3,\ldots \}=\mathbb{N}_0$.

 

[happyparticle]

My problem is that I can't show that my operators commutes.
Since $X$ and $P_x$ don't commute, I think that any function of $P_x$ doesn't commute with $X$.
Thus, $[X,e^{ip_x}]$ doesn't commute.

 

[topsquark]

That's not true. An important counter-example are the annihilation and creation operators of the harmonic oscillator $\hat{a}$ and $\hat{a}^{\dagger}$ and the "phonon-number operator" $\hat{N}=\hat{a}^{\dagger} \hat{a}$. Since $[\hat{a},\hat{a}^{\dagger}$ it's easy to show that if $|n \rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $n$, then $\hat{a}|n \rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $(n-1)$ (or the null vector) and $\hat{a}^{\dagger}|n \rangle$ is an eigenvector of $\hat{N}$ with eigenvalue $(n+1)$. Since $\hat{N}$ is a positive definite self-adjoint operator, it's easy to show that there's an eigenvector of $\hat{N}$, $|\Omega \rangle$ such that $\hat{a} |\Omega \rangle=0$, i.e., it's the eigenvector of $\hat{N}$ with eigenvalue $0$, and thus the eigenvalues of $\hat{N}$ must be $n \in \{0,1,2,3,\ldots \}=\mathbb{N}_0$.

Huh! Okay, I see what I did now. I learned this out of Sakuri, but what I didn't realize is that here I was arguing the converse of what he was saying. I went back and reviewed it and, for some reason, I had it in my head that compatible observables was a biconditional with common eigenstates. Clearly that isn't the case.

Thanks for the correction!

-Dan

 

[topsquark]

My problem is that I can't show that my operators commutes.
Since $X$ and $P_x$ don't commute, I think that any function of $P_x$ doesn't commute with $X$.
Thus, $[X,e^{ip_x}]$ doesn't commute.

Right. As I mentioned above we expand $e^{i p_x}$ as a Taylor series.
Explicitly:
$e^{i p_x} = 1 + i p_x + \dfrac{i^2}{2} p_x^2 + \dfrac{i^3}{3!} p_x^3 + \dots$

So
$[x, e^{i p_x} ] = [x , 1 + i p_x + \dfrac{i^2}{2} p_x^2 + \dfrac{i^3}{3!} p_x^3 + \dots ]$

$= [x,x] + i [x, p_x] + \dfrac{i^2}{2!} [x, p_x^2 ] + \dfrac{i^3}{3!} [ x, p_x^3 ] + \dots$

and
$[x, p_x^n] = i \hbar p_x^{n-1}$

So
$[x, e^{i p_x} ] = i i \hbar + \dfrac{i^2}{2!} 2 i \hbar p_x + \dfrac{i^3}{3!} 3 i \hbar p_x^2 + \dots$

$= - \hbar - \hbar p_x + \dfrac{1}{2} p_x^2 + \dots$

Eventually
$[x, e^{i p_x} ] = - \hbar e^{i p_x} \neq 0$

(Which we could have gotten to in one step by $[x, g(p_x)] = i \hbar g^{ \prime } (p_x)$.)

-Dan

 

[PeroK]

My problem is that I can't show that my operators commutes.
Since $X$ and $P_x$ don't commute, I think that any function of $P_x$ doesn't commute with $X$.

That definitely can't be true. A function can transform $P_x$ into any other operator. A trivial example would be $(P_x)^0 = I$.

Thus, $[X,e^{ip_x}]$ doesn't commute.

This doesn't make sense. You should say that the commutator is non-zero: thus $[X,e^{ip_x}] \ne 0$.

 

[vanhees71]

Huh! Okay, I see what I did now. I learned this out of Sakuri, but what I didn't realize is that here I was arguing the converse of what he was saying. I went back and reviewed it and, for some reason, I had it in my head that compatible observables was a biconditional with common eigenstates. Clearly that isn't the case.

Thanks for the correction!

-Dan

No, now you misunderstood what I said. Of course, if you want to have a complete set of common eigenvectors of two self-adjoint operators these operators must commute. That's easy to prove: Assume that there is a common complete set of orthonormalized eigenvectors,
$$\hat{A}|a,b \rangle=a|a,b \rangle, \quad \hat{B}|a,b \rangle=b |a,b \rangle.$$
Then any vector can be written as
$$|\psi \rangle=\sum_{a,b} \psi_{ab} |a,b \rangle.$$
Here I assumed that all the eigenvalues are discrete. If you have continuous spectra or if part of the spectrum of the one or the other operator is continuous, you just have to add the usual integrals, but this doesn't change too much on the argument (at the level of "robust mathematics" used by physicists in these matters ;-)). Now we have
$$\hat{A} \hat{B} |\psi \rangle=\sum_{a,b} \psi_{ab} b \hat{A}|a,b \rangle = \sum_{a,b} \psi_{ab} a b|a,b \rangle$$
and
$$\hat{B} \hat{A} |\psi \rangle=\sum_{a,b} \psi_{ab} a \hat{B}|a,b \rangle=\sum_{a,b} \psi_{ab} a b |a,b \rangle,$$
but this shows that for all vectors
$$\hat{A} \hat{B} |\psi \rangle=\hat{B} \hat{A} \psi \rangle \; \Rightarrow \; \hat{A} \hat{B}=\hat{B} \hat{A},$$
i.e., that the operators commute.

 

[happyparticle]

That definitely can't be true. A function can transform $P_x$ into any other operator. A trivial example would be $(P_x)^0 = I$.

This doesn't make sense. You should say that the commutator is non-zero: thus $[X,e^{ip_x}] \ne 0$.

I'm not sure to understand. I thought that if 2 operators A and B doesn't commute then $[A,B] \neq 0$

 

[PeroK]

I'm not sure to understand. I thought that if 2 operators A and B doesn't commute then $[A,B] \neq 0$

You've repeatedly used the terminology "$[A,B]$ doesn't commute". Which is nonsensical:

I know that [A,F(B)] doesn't commute,

Thus, $[X,e^{ip_x}]$ doesn't commute.

That's what I was pointing out.

 

[happyparticle]

I'm still confuse. Is it the term I used that is wrong or should I say $A$ and $B$ don't commute instead of $[A,B]$ doesn't commute.

 

[PeroK]

I'm still confuse. Is it the term I used that is wrong or should I say $A$ and $B$ don't commute instead of $[A,B]$ doesn't commute.

Yes.

 

[happyparticle]

Alright, thank you