I Can Operators Have Multiple Partial Derivatives in Quantum Mechanics?

[Happiness]

Suppose Q=2x+t and x=t², then ∂Q/∂t=1.
But Q can also be written as Q=x+t²+t, then ∂Q/∂t=2t+1.
We now have 2 different answers. But I think there can only be one correct answer.

In reference to the equation in the image, no matter we write Q=2x+t or Q=x+t²+t, <Q> should be the same, so the LHS should be the same. But when we have 2 different answers for ∂Q/∂t, the RHS would not be the same. So we have a contradiction.

Q is an observable of a quantum system. LHS stands for left hand side of the equation.

 

[weirdoguy]

then ∂Q/∂t=1

No, because x is also a function of t so it's derivative is not 0.

 

[Happiness]

No, because x is also a function of t so it's derivative is not 0.

∂Q/∂t refers to the partial derivative, not the total derivative. The book goes on to talk about operators that explicitly depend on time (see attached images).

In doing the partial derivative ∂Q/∂t, we do not care about how x depends on t, but only how Q explicitly depends on t.

But no matter how you write Q, <Q> and hence the LHS should be the same, while the RHS are not the same, hence the contradiction.

 

[DrClaude]

In doing the partial derivative ∂Q/∂t, we do not care about how x depends on t, but only how Q explicitly depends on t.

A partial derivative is a derivative with respect to a variable where other variables are held constant. If $x=t^2$, then $x$ isn't held constant.

Note that this has nothing to do with quantum mechanics. It is calculus. For a given $f(x,y)$, $\partial f / \partial y$ can have only one solution.

 

[Happiness]

A partial derivative is a derivative with respect to a variable where other variables are held constant. If $x=t^2$, then $x$ isn't held constant.

Note that this has nothing to do with quantum mechanics. It is calculus. For a given $f(x,y)$, $\partial f / \partial y$ can have only one solution.

Let's consider the Q in footnote 22 (see image).
Q=$\frac{1}{2}$mω²x², where ω is a function of t.
Then is ∂Q/∂t = mω$\frac{dω}{dt}$x²?

Even though for simple harmonic motion, x is also a function of t (ie. x=x₀$\sin$(ωt), where x₀ is the amplitude), we do not care how x depends on t when calculating ∂Q/∂t. Is this correct?

I think, in quantum mechanics, it would be more accurate to say <x>=x₀$\sin$(ωt). But I don't know whether this point helps in this discussion. After all, it doesn't seem to affect the value of <Q> in my original post, where Q=2x+t, so <Q>=2<x>+t. Is this correct?

(In my original post, instead of x=t², it would be more accurate to say <x>=t². But this doesn't seem to change the value of <Q>. So it doesn't seem to resolve the contradiction.)

 

[DrClaude]

Even though for simple harmonic motion, x is also a function of t (ie. x=x0sin(ωt), where x0 is the amplitude), we do not care how x depends on t when calculating ∂Q/∂t. Is this correct?

That's not correct. $x$ is here the coordinate (in position representation; more generically, it would have been better to use here $\hat{x}$, the position operator). It has no time dependence. For a quantum mechanics oscillator, its position is in its state, that is, in its wave function. What you would have is
$$
\braket{x} = \braket{\psi(t) | \hat{x} | \psi(t)} = x_0 \sin(\omega t).
$$

 

[weirdoguy]

In doing the partial derivative ∂Q/∂t, we do not care about how x depends on t

Of course we do, otherwise we run into problems, as you can see yourself.

 

[vanhees71]

First one has to clarify things, because the OP makes no sense to begin with. One should note that in QM $t$ is not an observable and thus not an operator but, of course, the position $x$ is, and it's represented by a self-adjoint operator $\hat{x}$.

Now in the Schrödinger picture of time evolution (pure) states are represented by time-dependent state vectors $|\psi(t) \rangle$, which fulfill the Schrödinger equation,
$$\mathrm{i} \hbar \partial_t |\psi(t) \rangle=\hat{H} |\psi(t) \rangle, \qquad(1)$$
where $\hat{H}$ is the Hamilton operator of the system. It's also a self-adjoint operator and represents the total energy of the system. Taking the adjoint of this equation, you get
$$-\mathrm{i} \hbar \partial_t \langle \psi(t) \rangle =\langle \psi(t) |\hat{H}^{\dagger}=\langle \psi(t) | \hat{H}. \qquad (2)$$

Note that in the Schrödinger picture $\hat{x}$ is not dependent on time. So the question is how to define the velocity $v$, which in classical physics is defined as the time derivative of $x$ along the trajectory of the particle. In QM we like $\hat{v}$ to be a self-adjoint operator for which
$$\langle v \rangle=\frac{\mathrm{d}}{\mathrm{d} t} \langle x \rangle.$$
Now the expectation value for the system in the state represented by $|\psi(t) \rangle$ is given by
$$\langle x \rangle=\langle \psi(t)|\hat{x}|\psi(t) \rangle,$$
and we can calculate the time derivative, using the Eqs. (1) and (2):
$$\frac{\mathrm{d}}{\mathrm{d} t} = (\partial_t \langle \psi(t)|\hat{x}|\psi(t) \rangle + \langle \psi(t) |\hat{x}|(\partial_t \psi(t) \rangle) = \frac{\mathrm{i}}{\hbar} (\langle \psi(t)| \hat{H} \hat{x}|\psi(t) \rangle - \langle \psi(t)|\hat{x} \hat{H}|\psi(t) \rangle=\langle \psi(t)| \mathrm{i}/\hbar [\hat{H},\hat{x}]|\psi(t).$$
Since this should hold true for all $|\psi(t) \rangle$ we conclude that
$$\hat{v}=\frac{\mathrm{i}}{\hbar} [\hat{H},\hat{x}] = \frac{1}{\mathrm{i} \hbar} [\hat{x},\hat{H}],$$
is the operator that represents the velocity. One thus defines a "covariant time dervative" for operators that represent an observable by
$$\mathring{\hat{A}}(\hat{x},\hat{p})=\frac{1}{\mathrm{i} \hbar} [\hat{A}(\hat{x},\hat{p}),\hat{H}].$$
If you have an operator that also explicitly depends on time, then you have of course also the corresponding partial time derivative, i.e.,
$$\mathring{\hat{Q}}(\hat{x},\hat{p},t)
=\frac{1}{\mathrm{i} \hbar} [\hat{Q}(\hat{x},\hat{p}),\hat{H}] + \partial_t \hat{Q}(\hat{x},\hat{p},t).$$