What is the complex conjugate of the momentum operator?

  • #1
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Hello, i am kind of confused about something.
What is the complex conjugate of the momentum operator? I don't mean the Hermitian adjoint, because i know that the Hermitian adjoint of the momentum operator is the momentum operator.
Thanks!
 

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  • #2
blue_leaf77
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Complex conjugate is usually applied to scalars, as far as I know. But sometimes it gets loosely used for operators and vectors as well to actually mean the adjoint.
 
  • #3
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Complex conjugate is usually applied to scalars, as far as I know. But sometimes it gets loosely used for operators and vectors as well to actually mean the adjoint.
Thanks for the reply!
I have seen it being used without meaning the adjoint.
I i think i have figured it out if you can verify it for me. Because the adjoint of the first derivative is minus the derivative, the adjoint of the momentum operator is itself(because there is an i). So, the complex conjugate of the momentum operator of it is just minus the operator. Is this right?
 
  • #4
blue_leaf77
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I i think i have figured it out if you can verify it for me. Because the adjoint of the first derivative is minus the derivative, the adjoint of the momentum operator is itself(because there is an i). So, the complex conjugate of the momentum operator of it is just minus the operator. Is this right?
Do you mean when taking the complex conjugate of ##-i\hbar \partial_x##, you will get minus of that expression?
Remember that, such an operation is not how an adjoint is defined. The adjoint of an operator ##A## is defined through the relation ##\langle \phi |A \psi\rangle = \langle A^*\phi | \psi\rangle## for all ##|\psi\rangle## and ##|\phi\rangle##.
 
  • #5
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Do you mean when taking the complex conjugate of ##-i\hbar \partial_x##, you will get minus of that expression?
Remember that, such an operation is not how an adjoint is defined. The adjoint of an operator ##A## is defined through the relation ##\langle \phi |A \psi\rangle = \langle A^*\phi | \psi\rangle## for all ##|\psi\rangle## and ##|\phi\rangle##.
Wouldn't that be A dagger rather than A*? I mean, i know that the adjoint of p is p, but i think its complex conjugate(in matrix representation just complex conjugating the components and not taking the inverse which would make it its adjoint) is -p.
 
  • #6
blue_leaf77
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Wouldn't that be A dagger rather than A*?
Yes, I meant a dagger.
i think its complex conjugate(in matrix representation just complex conjugating the components and not taking the inverse which would make it its adjoint) is -p
The matrix element of p is ##p_{ij} = -i\hbar \int u_j^*(x)\frac{\partial}{\partial x} u_i(x) \, dx##. The complex conjugate of this is ##\bar{p}_{ij} = i\hbar \int u_j(x)\frac{\partial}{\partial x} u_i^*(x) \, dx \neq -p_{ij}##.
What does an inverse operation have to do in the current discussion?
 
  • #7
vanhees71
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Thanks for the reply!
I have seen it being used without meaning the adjoint.
I i think i have figured it out if you can verify it for me. Because the adjoint of the first derivative is minus the derivative, the adjoint of the momentum operator is itself(because there is an i). So, the complex conjugate of the momentum operator of it is just minus the operator. Is this right?
Where have you seen that? I'd be very careful with any text claiming there is something as the complex conjugate of an operator!
 
  • #8
vanhees71
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Yes, I meant a dagger.

The matrix element of p is ##p_{ij} = -i\hbar \int u_j^*(x)\frac{\partial}{\partial x} u_i(x) \, dx##. The complex conjugate of this is ##\bar{p}_{ij} = i\hbar \int u_j(x)\frac{\partial}{\partial x} u_i^*(x) \, dx \neq -p_{ij}##.
What does an inverse operation have to do in the current discussion?
Of course, if you take the matrix elements of an operator as a matrix (with usually infinitely many rows and columns), then the matrix elements of the adjoint operator consists of the adjoint matrix, i.e., with the complex conjugated matrix elements and rows and columns interchanged (transposed matrix with the conjugate complex matrix elements). Let ##\hat{A}## be the operator. Its matrix elements with respect to an orthonormal basis are
$$A_{jk}=\langle j|\hat{A} k \rangle,$$
and the matrix elements of the adjoint operator are
$$(A^{\dagger})_{jk} = \langle j|\hat{A}^{\dagger} k \rangle = \langle \hat{A} j|k \rangle=(\langle k|\hat{A} j \rangle)^*=A_{kj}^*.$$
 
  • #9
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Yes, I meant a dagger.

The matrix element of p is ##p_{ij} = -i\hbar \int u_j^*(x)\frac{\partial}{\partial x} u_i(x) \, dx##. The complex conjugate of this is ##\bar{p}_{ij} = i\hbar \int u_j(x)\frac{\partial}{\partial x} u_i^*(x) \, dx \neq -p_{ij}##.
What does an inverse operation have to do in the current discussion?
Sorry, i meant transpose
 
  • #10
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Where have you seen that? I'd be very careful with any text claiming there is something as the complex conjugate of an operator!
I have seen a proof using the complex conjugate of an inner product <u|T|u> (u a vector and T an operator). Then in the proof it wrote the inner product as an integral, and then it wrote it equal to the whole integral conjugated with all the parts in it complex conjugated(so as if it is not conjugated) in order to do some calculations easier.
It did not use the adjoint because it just conjugated every part inside the integral.
 
  • #11
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Of course, if you take the matrix elements of an operator as a matrix (with usually infinitely many rows and columns), then the matrix elements of the adjoint operator consists of the adjoint matrix, i.e., with the complex conjugated matrix elements and rows and columns interchanged (transposed matrix with the conjugate complex matrix elements). Let ##\hat{A}## be the operator. Its matrix elements with respect to an orthonormal basis are
$$A_{jk}=\langle j|\hat{A} k \rangle,$$
and the matrix elements of the adjoint operator are
$$(A^{\dagger})_{jk} = \langle j|\hat{A}^{\dagger} k \rangle = \langle \hat{A} j|k \rangle=(\langle k|\hat{A} j \rangle)^*=A_{kj}^*.$$
If p=-ihd/dx (i will write hbar as h), then if you just complex conjugate it, won't you take p*=+ihd/dx? Isn't only the i that is imaginary?
 
  • #12
stevendaryl
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Hello, i am kind of confused about something.
What is the complex conjugate of the momentum operator? I don't mean the Hermitian adjoint, because i know that the Hermitian adjoint of the momentum operator is the momentum operator.
Thanks!

As blue leaf77 says, people don't typically talk about the complex conjugate of anything other than scalars. That's because whether an operator is imaginary or not depends on how you represent it. In introductions to quantum mechanics, wave functions are usually written as functions of position, [itex]\psi(x)[/itex]. As an operator on such functions, [itex]p[/itex] is represented as [itex]-i \hbar \frac{\partial}{\partial x}[/itex], and it seems to be imaginary, because of the presence of the "i". But you can just as well work with momentum representations. Instead of dealing with [itex]\psi(x)[/itex], you can deal with its Fourier transform, [itex]\tilde{\psi}(k)[/itex]. In terms of [itex]\tilde{\psi}[/itex], the momentum operator is trivial:

[itex]p \tilde{\psi}(k) = \hbar k \tilde{\psi}(k)[/itex]

But the position operator is more complicated:

[itex]x \tilde{\psi}(k) = i \frac{\partial}{\partial k} \tilde{\psi}(k)[/itex]

So in terms of [itex]\tilde{\psi}(k)[/itex], it seems that [itex]p[/itex] is real, and [itex]x[/itex] is imaginary.

So "complex conjugate" is ambiguous unless you say whether you're using a position representation or momentum representation. On the other hand, "adjoint" is independent of what representation you are using. The adjoint of x is x, and the adjoint of p is p.
 
  • #13
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Yes, I meant a dagger.

The matrix element of p is ##p_{ij} = -i\hbar \int u_j^*(x)\frac{\partial}{\partial x} u_i(x) \, dx##. The complex conjugate of this is ##\bar{p}_{ij} = i\hbar \int u_j(x)\frac{\partial}{\partial x} u_i^*(x) \, dx \neq -p_{ij}##.
What does an inverse operation have to do in the current discussion?
I think that the matrix corresponding to the momentum operator is all zero except from two diagonals. The one above and the one below the "middle" diagonal(i hope you understand! :) ). The one of the two is all 1's and the other is all -1's. So whle you are correct stating that Pij=(-Pji)*
As blue leaf77 says, people don't typically talk about the complex conjugate of anything other than scalars. That's because whether an operator is imaginary or not depends on how you represent it. In introductions to quantum mechanics, wave functions are usually written as functions of position, [itex]\psi(x)[/itex]. As an operator on such functions, [itex]p[/itex] is represented as [itex]-i \hbar \frac{\partial}{\partial x}[/itex], and it seems to be imaginary, because of the presence of the "i". But you can just as well work with momentum representations. Instead of dealing with [itex]\psi(x)[/itex], you can deal with its Fourier transform, [itex]\tilde{\psi}(k)[/itex]. In terms of [itex]\tilde{\psi}[/itex], the momentum operator is trivial:

[itex]p \tilde{\psi}(k) = \hbar k \tilde{\psi}(k)[/itex]

But the position operator is more complicated:

[itex]x \tilde{\psi}(k) = i \frac{\partial}{\partial k} \tilde{\psi}(k)[/itex]

So in terms of [itex]\tilde{\psi}(k)[/itex], it seems that [itex]p[/itex] is real, and [itex]x[/itex] is imaginary.

So "complex conjugate" is ambiguous unless you say whether you're using a position representation or momentum representation. On the other hand, "adjoint" is independent of what representation you are using. The adjoint of x is x, and the adjoint of p is p.
Oh, yes, you are right. Thanks.
So, i should clarify that i am working on position representation.
 
  • #14
vanhees71
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I have seen a proof using the complex conjugate of an inner product <u|T|u> (u a vector and T an operator). Then in the proof it wrote the inner product as an integral, and then it wrote it equal to the whole integral conjugated with all the parts in it complex conjugated(so as if it is not conjugated) in order to do some calculations easier.
It did not use the adjoint because it just conjugated every part inside the integral.
That's of course fine, because only the complex conjugate to complex numbers were taken. You can take complex conjugates of matrix elements of operators, but not of the operators themselves! It just doesn't make sense, because it's simply not defined!
 
  • #15
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That's of course fine, because only the complex conjugate to complex numbers were taken. You can take complex conjugates of matrix elements of operators, but not of the operators themselves! It just doesn't make sense, because it's simply not defined!
So, indeed p*=(-ihd/dx)*=+ihd/dx=-p ? (h stands for hbar)
 
  • #16
vanhees71
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I don't understand your notation. That's not a complex number but a differential operator, and thus the complex conjugate symbol doesn't make any sense!
 
  • #17
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I don't understand your notation. That's not a complex number but a differential operator, and thus the complex conjugate symbol doesn't make any sense!
Yeah, sorry. I should think about it in terms of inner products and what they give as a result. So, in the previous comment where i mentioned conjugating an inner product( as integral), i should first let the operator act on a ket( like P|Ψ>=-ih d(|Ψ>)/dx) and then complex conjugate( like [-ih d(|Ψ>)/dx]*=ih d(|Ψ>*)/dx). Right? I think i got what you meant, so if we just let it act on a ket then it is ok.
 
  • #18
blue_leaf77
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I think that the matrix corresponding to the momentum operator is all zero except from two diagonals. The one above and the one below the "middle" diagonal(i hope you understand! :) ). The one of the two is all 1's and the other is all -1's. So whle you are correct stating that Pij=(-Pji)*
Yes I understand what you mean but such kind of matrix form cannot represent that of momentum operator because the matrix you describe is not Hermitian.
What is the complex conjugate of the momentum operator?
Unfortunately it's meaningless to answer that question because the complex conjugate of an operator is not defined. The question would have made sense had you said adjoint instead of complex conjugate. Complex conjugate is for scalar, e.g. you can say the complex conjugate of the matrix element of some operator because a matrix element is a scalar.
p*=(-ihd/dx)*=+ihd/dx=-p
Ok, in the sense that you are just changing the sign of i, that equation may be justified. But it can also be dangerous because a misinterpretation of the mathematical operation like that can lead to concluding ##p^*=-p##. It just doesn't make sense. If you read it, it's as if the complex conjugate of an observable quantity called momentum equals to its minus, but momentum is real so it's weird.
 
  • #19
blue_leaf77
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if we just let it act on a ket then it is ok.
Yes, that one makes more sense. Just that, rather than a ket, you should say wavefunction ##\langle x|\psi\rangle = \psi(x)## because it's a scalar.
 
  • #20
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Yes I understand what you mean but such kind of matrix form cannot represent that of momentum operator because the matrix you describe is not Hermitian.

Unfortunately it's meaningless to answer that question because the complex conjugate of an operator is not defined. The question would have made sense had you said adjoint instead of complex conjugate. Complex conjugate is for scalar, e.g. you can say the complex conjugate of the matrix element of some operator because a matrix element is a scalar.

Ok, in the sense that you are just changing the sign of i, that equation may be justified. But it can also be dangerous because a misinterpretation of the mathematical operation like that can lead to concluding ##p^*=-p##. It just doesn't make sense. If you read it, it's as if the complex conjugate of an observable quantity called momentum equals to its minus, but momentum is real so it's weird.
Yes, i think i can understand. Please check my previous comment.
 
  • #21
vanhees71
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Yeah, sorry. I should think about it in terms of inner products and what they give as a result. So, in the previous comment where i mentioned conjugating an inner product( as integral), i should first let the operator act on a ket( like P|Ψ>=-ih d(|Ψ>)/dx) and then complex conjugate( like [-ih d(|Ψ>)/dx]*=ih d(|Ψ>*)/dx). Right? I think i got what you meant, so if we just let it act on a ket then it is ok.
This doesn't make sense either. ##|\Psi \rangle## doesn't depend on ##x##. You can apply the differential operators only to wave functions not Dirac kets. The relation is
$$\Psi(x)=\langle x|\Psi \rangle, \quad -\mathrm{i} \hbar \frac{\partial}{\partial x} \Psi(x)=\langle x|\hat{p} \Psi \rangle.$$
Of course
$$[-\mathrm{i} \hbar \partial_x \Psi(x)]^*=+\mathrm{i} \hbar \partial_x \Psi^*(x).$$
 
  • #22
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This doesn't make sense either. ##|\Psi \rangle## doesn't depend on ##x##. You can apply the differential operators only to wave functions not Dirac kets. The relation is
$$\Psi(x)=\langle x|\Psi \rangle, \quad -\mathrm{i} \hbar \frac{\partial}{\partial x} \Psi(x)=\langle x|\hat{p} \Psi \rangle.$$
Of course
$$[-\mathrm{i} \hbar \partial_x \Psi(x)]^*=+\mathrm{i} \hbar \partial_x \Psi^*(x).$$
Oh, ok. I was acting on a ket. So, the whole point is that bras and kets and operators are abstract on their own, so the only thing that we can act on is on a braket, a number and not one vector that is not defined on a basis. Right?
 
  • #23
vanhees71
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No, you can do in fact a lot in the representation-free formulation by Dirac. That's his great invention (besides the relativistic wave equation for spin-1/2 particles named after him). You don't need a basis to formulate quantum theory and to do a lot of manipulations to simplify your work when finally you have to choose a basis to solve some given problem.

Of course, to solve a problem completely in the abstract formalism is only possible for very simple cases, i.e., systems with a large symmetry. The most important examples are: energy eigenvalues and eigenvectors of the harmonic oscillator and the hydrogen atom, eigenvalues and eigenvectors of position, momentum, and angular-momentum operators.
 
  • #24
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No, you can do in fact a lot in the representation-free formulation by Dirac. That's his great invention (besides the relativistic wave equation for spin-1/2 particles named after him). You don't need a basis to formulate quantum theory and to do a lot of manipulations to simplify your work when finally you have to choose a basis to solve some given problem.

Of course, to solve a problem completely in the abstract formalism is only possible for very simple cases, i.e., systems with a large symmetry. The most important examples are: energy eigenvalues and eigenvectors of the harmonic oscillator and the hydrogen atom, eigenvalues and eigenvectors of position, momentum, and angular-momentum operators.
So, why it is not allowed in some cases while in other cases it is ok? Is there a rule or something?
 
  • #25
vanhees71
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What do you mean by "it is not allowed"? It's only sometimes too difficult to solve a problem in the abstract formalism, while it is easier to use wave functions and differential equation instead of the abstract algebra of the abstract formalism.
 

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