How to derive the quantum commutation in matrix mechanics?

In summary,The homework statement asks how to derive the quantum commutation relations in matrix form. The attempted solution attempted to derive the commutation from the original equation of Fourier expansion of x(t) and p(t). However, the equation does not seem to work when E is an operator.
  • #1
Adel Makram
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Homework Statement


I would like to know how to derive the quantum commutation relations in matrix form,
$$i \hbar \partial_t x(t)= [x(t),E]$$
$$i \hbar \partial_t P(t)= [P(t),E]$$
Where X(t), P(t) and E are the position, momentum and the energy of the particle, respectively.

2. Homework Equations

The solution of harmonic oscillator as expressed in Fourier expansion;
$$X(t)=\sum X_{mn} e^{i \omega (m-n)t} $$
$$P(t)=\sum P_{mn} e^{i \omega (m-n)t} $$
differentiate X(t) relative to t would give P(t) assuming the mass of the particle=1.
$$\partial_t X(t)=\sum X_{mn} i\omega(m-n) e^{i \omega (m-n)t} $$
$$\partial_t X(t)=\sum X_{mn} i \frac{E_m - E_n}{\hbar} e^{i \omega (m-n)t} $$
But how to arrive to the commutation formula from here?

The Attempt at a Solution


I tried retrospectively from the commutation relation to prove the equivalence with the original equation;
$$i \hbar \partial_t x(t)= [x(t),E]$$
So I started from;
$$=[X(t) E - E X(t)]$$
$$=X_{mn} E_m \delta_{mn} - E_m \delta_{mn} X_{mn}$$
where $$E=E_{mn}=E_m \delta_{mn}$$
But this did not give me;
$$\partial_t X(t)=\sum X_{mn} i \frac{E_m - E_n}{\hbar} e^{i \omega (m-n)t} $$
 
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  • #2
Adel Makram said:

Homework Statement


I would like to know how to derive the quantum commutation relations in matrix form,
$$i \hbar \partial_t x(t)= [x(t),E]$$
$$i \hbar \partial_t P(t)= [P(t),E]$$
Where X(t), P(t) and E are the position, momentum and the energy of the particle, respectively.
Those relations apply for a general operator actually, not only restricted to ##x(t)## and ##p(t)##.
Note that
$$x(t)=e^{iEt/\hbar}xe^{-iEt/\hbar}
$$
Now differentiate this with respect to ##t## (I assume ##E## is an operator).
 
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  • #3
blue_leaf77 said:
Those relations apply for a general operator actually, not only restricted to ##x(t)## and ##p(t)##.
Note that
$$x(t)=e^{iEt/\hbar}xe^{-iEt/\hbar}
$$
Now differentiate this with respect to ##t## (I assume ##E## is an operator).
$$\partial_t x(t)=\partial(e^{iE_mt/\hbar}xe^{-iE_nt/\hbar})$$
$$\partial_t x(t)=\frac{i(E_m-E_n)}{\hbar} (e^{iE_mt/\hbar}xe^{-iE_nt/\hbar})$$
$$i\hbar \partial_t x(t)= (E_n-E_m) (e^{iE_mt/\hbar}xe^{-iE_nt/\hbar})$$
I assumed E is a different energy level. I don`t know how to proceed if E is an operator because for your equation, the exponential function will reduce to 1.
 
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  • #4
I am rather confused with what you actually want to do. I thought you wanted to prove
Adel Makram said:
$$i \hbar \partial_t x(t)= [x(t),E]$$
$$i \hbar \partial_t P(t)= [P(t),E]$$
which are true provided ##E## is the Hamiltonian operator. If ##E## in those equations is a value of certain energy, i.e. a number, then the commutation above should equal zero. This further implies that ##x(t)## is time independent, which contradicts the way it was written. So, what do you want to prove actually?
 
  • #5
I just want to derive those commutation from the original equation of Fourier expansion of x(t) and p(t). In those expansion, ##\omega(m-n)=\frac{E_m-E_n}{\hbar}##
I know how to derive the commutation from your equation,
$$\partial_t x(t)=\partial(e^{iEt/\hbar}xe^{-iEt/\hbar})$$
$$\partial_t x(t)=\frac{i(E)}{\hbar}(e^{iEt/\hbar}xe^{-iEt/\hbar})-(e^{iEt/\hbar}xe^{-iEt/\hbar})\frac{iE}{\hbar}$$
$$\partial_t x(t)=\frac{i(E)}{\hbar}x(t)-x(t)\frac{iE}{\hbar}$$
$$\hbar\partial_t x(t)=i(E x(t)-x(t) E)$$
$$i\hbar\partial_t x(t)=x(t)(E)-(E)x(t)$$
$$i\hbar\partial_t x(t)=[x(t),(E)]$$ as desired.
But I want to do the same from matrix algebra.
 
  • #6
What is ##E##? Is it an operator or a scalar?
 
  • #7
blue_leaf77 said:
What is ##E##? Is it an operator or a scalar?
In your equation, E is an operator, but in ##E_n## and ##E_m## in Fourier expansion are numbers.
 
  • #8
The root of the trouble seems to be the equations:
Adel Makram said:
$$X(t)=\sum X_{mn} e^{i \omega (m-n)t} $$
$$P(t)=\sum P_{mn} e^{i \omega (m-n)t} $$
I honestly never encounter them. ##X_{mn}## is a matrix element of ##X## and then you basically sum them all along the rows and columns. Then somehow it turns out to be equal to the operator ##X(t)##. Where do you get these formula from?
 
  • #9
blue_leaf77 said:
The root of the trouble seems to be the equations:

I honestly never encounter them. ##X_{mn}## is a matrix element of ##X## and then you basically sum them all along the rows and columns. Then somehow it turns out to be equal to the operator ##X(t)##. Where do you get these formula from?
https://en.wikipedia.org/wiki/Matrix_mechanics see the section under Heisenberg reasoning. So, I summat over all m and n by mistake. In the same page, under the matrix basics, the commutation relation is shown without derivation, in matrix form. My question, how can we derive that?
On the other hand, you gave $$x(t)=e^{iE/\hbar t}xe^{-iE/\hbar t}$$
Is this a solution to the harmonic oscillation?
 
  • #10
Adel Makram said:
https://en.wikipedia.org/wiki/Matrix_mechanics see the section under Heisenberg reasoning. So, I summat over all m and n by mistake. In the same page, under the matrix basics, the commutation relation is shown without derivation, in matrix form. My question, how can we derive that?
Now you are referring to the commutation between x and p, which is a different thing from the equations you said you wanted to prove and ask how to derive this.
Adel Makram said:
So, I summat over all m and n by mistake.
So, how do the correct formula look like? I am really curious with these Fouriert transform equation for ##X(t)## and ##P(t)##.
Adel Makram said:
Is this a solution to the harmonic oscillation?
No, that's just how an operator in the Heisenberg picture is defined. But wait, in your original problem statement, harmonic oscillator problem was not mentioned at all. Why are you considering harmonic oscillator?
 
  • #11
blue_leaf77 said:
Now you are referring to the commutation between x and p, which is a different thing from the equations you said you wanted to prove and ask how to derive this.
This is almost the same. the equations I want to prove using matrix algebra are:
$$i\hbar\partial_tx(t)=[x(t),E]$$
and
$$i\hbar\partial_tp(t)=[p(t),E]$$
Now consider a harmonic oscillator
$$H=\frac{P^2}{2}+x$$ assuming f(x)=-1, Also;
$$\partial_tp(t)=f(x)$$
plug this into the second equation,
$$i\hbar=[p(t),E]$$
$$=p(\frac{p^2}{2}+x)-(\frac{p^2}{2}+x)p$$
$$=px-xp=[x,p]$$
So the key to derive the commutation ##[x,p]## is to derive the first two equations which is my question. But it is ok also if you have another way to derive the relation without considering harmonic oscillator. But please let me know who Heisenberg definition of operator is first derived? (in matrix algebra not wave mechanics).
blue_leaf77 said:
So, how do the correct formula look like? I am really curious with these Fouriert transform equation for X(t)X(t)X(t) and P(t)P(t)P(t).
It should look like:
$$x_{mn}(t)=\sum x_{mn}(0)e^{i\omega_{mn}t}$$
I think that was the main essence behind the matrix mechanics, where ##\omega_{mn}## is replacing ##(m-n)\omega## as in classical mechanics. So long as the quantum theory demands introduction of ##\omega_{mn}## as a matrix element, the Fourier coefficient must be also a matrix element ##x_{mn}(0)##
 
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  • #12
Adel Makram said:
Now consider a harmonic oscillator
$$
H=\frac{P^2}{2}+x
$$
No, that's not a Hamiltonian of harmonic oscillator.
Adel Makram said:
assuming f(x)=-1
Just an intermezzo though, you assumption f(x)=-1 is actually true, it's not just an assumption. Using Baker-Hausdorff formula, you can prove that
$$
p(t)=e^{it(p^2/2+ax)}pe^{-it(p^2/2+ax)} = p-at
$$
so that ##\partial_tp(t)=-a##.
Adel Makram said:
It should look like:
$$
x_{mn}(t)=\sum x_{mn}(0)e^{i\omega_{mn}t}
$$
That cannot be right because ##mn## in the left hand side is a fixed indices while ##mn## in the right hand side is the indices for summation. If you are interested in knowing the matrix element of ##x(t)##, it's just ##x_{mn}(t)=x_{mn}(0)e^{i(\omega_m-\omega_n)t}##, no summation is needed.
Adel Makram said:
So the key to derive the commutation [x,p][x,p][x,p] is to derive the first two equations which is my question.
The commutation between x and p is the so-called fundamental commutation relation in quantum mechanics. I believe this is a principle which was not invented through derivation, rather it actually comes from the analogy with the Poisson bracket between x and p in classical mechanics.
From what I comprehend about the current situation, you want to prove
$$
i \hbar \partial_t x(t)= [x(t),E]$$
using matrix algebra, which failed because you started with a wrong expression for ##x_{mn}(t)##. Now that I have told you that ##x_{mn}(t)=x_{mn}(0)e^{i(\omega_m-\omega_n)t}##, you should be able to complete the derivation by showing that
$$
\langle m |[x(t),E]| n\rangle
$$
does becomes ##x_{mn}(0)e^{i(\omega_m-\omega_n)t}##. Or is it not the matrix algebra that you desired?
 
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1. What is the quantum commutation in matrix mechanics?

The quantum commutation in matrix mechanics is a mathematical rule that describes the behavior of quantum mechanical operators. It states that the order in which two operators are applied matters, and their product is not necessarily the same as the product of their individual actions.

2. How is the quantum commutation expressed in matrix mechanics?

The quantum commutation is expressed using the commutator, denoted by [A,B], where A and B are operators. It is defined as [A,B] = AB - BA, and it measures the failure of two operators to commute.

3. What is the significance of the quantum commutation in matrix mechanics?

The quantum commutation is significant because it reveals the non-classical nature of quantum mechanics. It shows that the order of operations is important, and the position and momentum of a particle cannot be simultaneously known with certainty.

4. How can the quantum commutation be derived in matrix mechanics?

The quantum commutation can be derived using the Heisenberg uncertainty principle and the commutation relations of the position and momentum operators. By substituting these operators into the commutator definition, we can obtain the quantum commutation relation.

5. Are there any real-life applications of the quantum commutation in matrix mechanics?

Yes, the quantum commutation is essential in understanding and predicting the behavior of quantum systems, such as atoms and subatomic particles. It is also used in quantum computing and quantum information theory to manipulate and measure quantum states.

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