Time Reversal by complex conjugating Schrodinger Equation

Question

I am reading Sakurai's chapter on time-reversal symmetry. He argues that we can show that if $\psi(x,t)$ is a solution to the Schrodinger Equation (using $\hbar=1$):

$$ i\partial_t \psi(x,t)=\left(-\frac{\nabla^2}{2m}+V(x)\right) \psi(x,t) \tag{1}$$

then $\psi^*(x,-t)$ is also a solution. In the book, he says we can show this by complex conjugating the equation above. This leads to:

$$ -i\partial_t \psi^*(x,t)=\left(-\frac{\nabla^2}{2m}+V(x)\right) \psi^*(x,t) \tag{2}$$

Now, in order to verify that $\psi^*(x,-t)$ is a solution, I expect to arrive at

$$ i\partial_t \psi^*(x,-t)=\left(-\frac{\nabla^2}{2m}+V(x)\right) \psi^*(x,-t) \tag{3}$$

But all I can do is use $-i\partial_t\psi^*(x,-t)=i\partial_t\psi^*(x,t)$ to get

$$ i\partial_t \psi^*(x,t)=\left(-\frac{\nabla^2}{2m}+V(x)\right) \psi^*(x,-t) \tag{4}$$

How can I show $\psi^*(x,-t)$ satisfy the Schrodinger Equation?

Answer 1

First thing to note is that under time reversal $t\to-t$, the time derivative also changes sign $\partial_t \to \partial_{-t}=-\partial_t$. So the LHS of our Schrödinger equation goes as:

$$i\partial_t\psi(x,t)\to i\partial_{-t}\psi(x,-t)=-i\partial_t\psi(x,-t)$$

So in your eq. $(3)$ you should have an additional minus sign that then solves your problem.

Answer 2

The trouble here is tied to the bad notation surrounding partial derivatives. At times like these I find it useful to use very clear - but very cumbersome - notation to clarify the concepts, and then use sloppier shorthand once the concept is well-understood.

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Consider the function $f:\mathbb R\times \mathbb R \rightarrow \mathbb R$ which eats two numbers and spits out the square of the first number plus the cube of the second number. What is $\partial f/\partial w$?

If you are confused, it is for good reason. What on earth is $w$? My question makes no sense at all. On the other hand, I could ask for the derivative of $f$ with respect to its first entry; the result, which I will call $\partial_1 f$, is the function which eats two numbers and spits out two times the first number. Similarly, the derivative with respect to the second entry is the function $\partial_2 f$ which eats two numbers and spits out three times the square of the second number.

Now obviously, expressing all of this using English words is incredibly tedious, and it is much easier to simply write $$f(x,y)=x^2+y^3 \qquad \big(\partial_1f\big)(x,y) = 2x \qquad \big(\partial_2f\big)(x,y) = 3y^2$$ However, it is crucial to note that the $x$'s and $y$'s which appear above are dummy variables. In the absence of any additional information, the symbol $\partial f/\partial x$ simply makes no sense. What does make sense is the following:

$$\frac{\partial}{\partial x} f(x,y) = \big(\partial_1 f\big)(x,y) $$

To get the general idea, here are a few other expressions which make sense:

$$\frac{\partial}{\partial x}f(x^2, y) = \big(\partial_1 f\big)(x^2,y) \cdot 2x \qquad \frac{\partial}{\partial x} f(y,x) = \big(\partial_2 f\big)(y,x) \qquad (1)$$ Observe that $\partial/\partial x$ is a derivative with respect to a variable, while $\partial_{1}$ is a derivative with respect to an entry.

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With this bit of notation in hand, let's address your question. Let $\partial_1,\partial_2,\partial_3$ refer to the derivatives with respect to the position entries and $\partial_4$ refer to the derivative with respect to the time entry. The Schrodinger equation (without reference to dummy variables) becomes

$$\left[i\partial_4 + \frac{1}{2m}(\partial_1^2+\partial_2^2+\partial_3^2)-V\right]\psi = 0 \qquad (2)$$ Complex conjugation gives $$\left[-i\partial_4 + \frac{1}{2m}(\partial_1^2+\partial_2^2+\partial_3^2)-V\right]\psi^* = 0 \qquad (3)$$

Now define $\rho(\mathbf x,t) = \psi^*(\mathbf x,-t)$. For the spatial variables we have e.g. $$\big(\partial_1\rho\big)(\mathbf x,t) =\frac{\partial}{\partial x} \rho(\mathbf x,t) = \frac{\partial}{\partial x}\psi^*(\mathbf x,-t) = \big(\partial_1\psi^*)(\mathbf x,-t)$$ while for the time variable, $$\big(\partial_4\rho\big)(\mathbf x,t) = \frac{\partial}{\partial t}\rho(\mathbf x,t) = \frac{\partial}{\partial t}\psi^*(\mathbf x,-t) = \color{red}{-}\big(\partial_4\psi^*\big)(\mathbf x,-t)$$ where we've used the chain rule, c.f. the first example in $(1)$. As a result, plugging $\rho$ into $(2)$ yields

$$\left[i\partial_4 + \frac{1}{2m}(\partial_1^2+\partial_2^2+\partial_3^2)-V\right]\rho(\mathbf x,t) $$ $$= \left[-i\partial_4 + \frac{1}{2m}(\partial_1^2+\partial_2^2+\partial_3^2)-V\right]\psi^*(\mathbf x,-t) = 0$$

where we've used the fact that $(3)$ holds for the function $\psi^*$, regardless of the dummy variables I subsequently choose to plug in.

Answer 3

There is also the following answer:

whenever we switch from $\psi(x,t)$ to $\psi^{*}(x,t)$ we have to formally change the operators to act on the correct space. In other words, if we have a PDE defined on the ket space and if we want to replace the ket by a bra, we also need to replace the operators for the ket space to its Hermitian adjoints.

Think of it this way: if you have a matrix operator acting on the column vector and if you switch the column vector to a row vector you will need to complex conjugate the matrix, otherwise, the inner product will not be conserved in general and since inner products correspond to probabilities in quantum mechanics it follows that you will no longer describe the given system.

So, how do we find the Hermitian adjoint of the operator $i\partial_t$, we do the following calculation

since $\langle\psi(x,t)|\psi(x,t)\rangle = 1$ because all states must be normalized to unity assuming $\psi(x,t)$ satisfies the TD Shrodinger equation. Now, if we take the time derivative we get $$ \partial_t \langle\psi(x,t)|\psi(x,t)\rangle =0 $$ but this is equivalent to $$ (\langle\psi(x,t)|\partial_t)\psi(x,t)\rangle + \langle\psi(x,t)|(\partial_t\psi(x,t)\rangle)=0 $$ where the parenthesis indicates whether the differential operator is acting on the ket or bra. The above is also equivalent to $$ (\langle\psi(x,t)|\partial_t)\psi(x,t)\rangle = - \langle\psi(x,t)|(\partial_t\psi(x,t)\rangle) $$ Now we utilize the bilinear property of the inner product and move the negative sign in the inner product and therefore we get $$ (\langle\psi(x,t)|\partial_t)\psi(x,t)\rangle = \langle\psi(x,t)|(-\partial_t\psi(x,t)\rangle) $$ What the above tells us is that the Hermitian adjoint of $\partial_t$ is the negative of itself i.e. $(\partial_t)^{\dagger} = -\partial_t $ which then means that if we replace the $\psi(x,t)$ by $\psi^*(x,t)$ then we should replace $i\partial_t$ by $-i\partial_t$. Also, since the time is replace by $-t$, the LHS is equivalent to $$-\partial_t \psi^*(x,-t)=-(\partial_t\psi^*(x,-t))\frac{d(-t)}{dt}$$ using the chain rule. At this point, the minus signs cancel and you are left with the correct form of the time derivative. But so far we have been talking about the LHS of TDSE and the RHS remains intact under the change of bra to ket because the Hamiltonian operator is Hermitian or self-adjoint (as mathematicians like to call it).