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我试图理解为什么我在使用梯度磁带获取函数的导数时会出错。尝试取幂关于 T 的导数，定义为：

    import tensorflow as tf
    import numpy as np
    from scipy.fft import fft, fftfreq, fftn
    import tensorflow.python.ops.numpy_ops.np_config as np_config
    np_config.enable_numpy_behavior()

    #####Initialize Values######

    s1 = np.array([[0,1,0],
                   [1,0,1],
                   [0,1,0]])

    s2 = np.array([[0,-1j,0],
                   [1j,0,-1j],
                   [0,1j,0]])

    s3 = np.array([[1,0,0],
                  [0,0,0],
                  [0,0,-1]])

    spin1 = (1/np.sqrt(2))*s1
    spin2 = (1/np.sqrt(2))*s2
    spin3 = (1/np.sqrt(2))*s3

    spin1 = tf.constant(spin1)
    spin2 = tf.constant(spin2)
    spin3 = tf.constant(spin3)

    a = tf.constant(1.0)
    b = tf.constant(1.0)
    c = tf.constant(1.0)
    d = tf.constant(1.0)

    v = tf.constant(1.0)     # ~N(0,sigma_v)
    w = tf.constant(1.0)     # ~N(0,sigma_w)

    c0_0 = tf.complex(tf.constant(1.0), tf.constant(0.0))
    c1_0 = tf.complex(tf.constant(1.0), tf.constant(0.0))

    ###### Define Functions########

    def getDE(T):
        D = a*T+b+v
        E = c*T+d+w
        return D,E

    def H(D,E):
        return D*(spin3**2 - 2/3) + E*(spin1**2-spin2**2)

    def psi(t,eigenvalues,eigenvec1, eigenvec2):
        c_0 = np.array(np.exp(-1j*(eigenvalues[0])*t)*c0_0)
        c_0.shape = (N,1)
        c_1 = np.array(np.exp(-1j*(eigenvalues[1])*t)*c1_0)
        c_1.shape = (N,1)
        return c_0*(eigenvec1.T)+c_1*(eigenvec2.T)

    def forward(T):
        T = tf.Variable(T)
        with tf.GradientTape() as tape:
            D,E = getDE(T)
            H_tf = H(D,E)
            eigenvalues, eigenstates = tf.linalg.eig(H_tf)
            eigenvec1 = eigenstates[:,0]
            eigenvec2 = eigenstates[:,1]
            wave = psi(t,eigenvalues,eigenvec1, eigenvec2)
            a = np.abs(tf.signal.fft2d(wave))**2
            Power = np.full([100,1], None)
            for i in range(N):
                Power[i,:] = a[i,:].conj().T@a[i,:]
        
        return tape.gradient(Power,T)

如果有人可以告诉我我是否正确执行此操作，或者是否有更好的方法可以执行此操作，因为我对 python 中的自动微分不是很熟悉。

在前向函数中，波对 T 的导数似乎有效，但是一旦我执行 fft，我就会收到以下错误：

WARNING:tensorflow:调用 GradientTape.gradient 时，目标张量的 dtype 必须是浮动的（例如 tf.float32），得到了 dtype('O')

    AttributeError                            Traceback (most recent call last)
    ~\AppData\Local\Temp/ipykernel_352/3452884380.py in <module>
    ----> 1 T_hat = forward(17.0)
          2 print(T_hat)

    ~\AppData\Local\Temp/ipykernel_352/2053063608.py in forward(T)
         13             Power[i,:] = a[i,:].conj().T@a[i,:]
         14 
    ---> 15     return tape.gradient(Power,T)

    ~\anaconda3\envs\tensorflow-gpu\lib\site-packages\tensorflow\python\eager\backprop.py in 
    gradient(self, target, sources, output_gradients, unconnected_gradients)
       1072                           for x in nest.flatten(output_gradients)]
       1073 
    -> 1074     flat_grad = imperative_grad.imperative_grad(
       1075         self._tape,
       1076         flat_targets,

    ~\anaconda3\envs\tensorflow-gpu\lib\site- 
   packages\tensorflow\python\eager\imperative_grad.py in imperative_grad(tape, target, 
    sources, output_gradients, sources_raw, unconnected_gradients)
         69         "Unknown value for unconnected_gradients: %r" % unconnected_gradients)
         70 
    ---> 71   return pywrap_tfe.TFE_Py_TapeGradient(
         72       tape._tape,  # pylint: disable=protected-access
         73       target,

    AttributeError: 'numpy.ndarray' object has no attribute '_id'
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